Final Examination in Linear Algebra: 18.06

Home , Block matrix

May 18, 1998 9:00{12:00 Professor Strang

Grading 1

Your name is:

Please circle your recitation:

1 M2 2-132 M. Nevins 2-588 3-4110 monica@math

2 M3 2-131 A. Voronov 2-246 3-3299 voronov@math

3 T10 2-132 A. Edelman 2-380 3-7770 edelman@math

4 T12 2-132 A. Edelman 2-380 3-7770 edelman@math

5 T12 2-131 Z. Spaso jevic 2-101 3-4470 zoran@math

6 T1 2-131 Z. Spaso jevic 2-101 3-4770 zoran@math

7 T2 2-132 Y. Ma 2-333 3-7826 yanyuan@math

Answer all 8 questions on these pages. This is a closed book exam. Calculators are not

needed in any way and therefore not allowed to be fair to all. Grades are known only to

your recitation instructor. Best wishes for the summer and thank you for taking 18.06. GS

1 If A isa5by 4 matrix with linearly indep endent columns, nd each of these explicitly :

a 3 p oints The nullspace of A.

b 3 p oints The dimension of the left nullspace N A .

c 3 p oints One particular solution x to Ax = column 2 of A.

p p

d 3 p oints The general complete solution to Ax = column 2 of A.

e 3 p oints The reduced row echelon form R of A. 2

2 a 5 p oints Find the general complete solution to this equation Ax = b:

2 3 3 2 2 3

1 1 2 x 2

6 7 7 6 6 7

= :

6 7 7 6 6 7

1 1 2 x 2

4 5 5 4 4 5

x 2 2 2 4

b 3 p oints Find a basis for the column space of the 3 by 9 blo ck matrix [A 2AA ]. 3

3 a 5 p oints The command N = null A pro duces a matrix whose columns are a

basis for the nullspace of A. What matrix describ e its prop erties is then pro duced

by B = null N ?

b 3 p oints What are the shap es how many rows and columns of those matrices

N and B , if A is m by n of rank r ? 4

4 Find the determinants of these three matrices:

a 2 p oints

2 3

0 0 0 1

6 7

0 0 2 0

6 7

A =

6 7

0 3 0 0

4 5

1 2 3 4

b 2 p oints

2 3

0 A

4 5

B = 8 by 8, same A

I I

c 2 p oints

2 3

A A

4 5

C = 8 by 8, same A

I I 5

5 If p ossible construct 3by 3 matrices A, B , C , D with these prop erties:

a 3 p oints A is a symmetric matrix. Its row space is spanned by the vector

1; 1; 2 and its column space is spanned by the vector 2; 2; 4.

b 3 p oints All three of these equations have no solution but B 6=0:

2 3 2 3 2 3

1 0 0

6 7 6 7 6 7

Bx = Bx = Bx = :

6 7 6 7 6 7

0 1 0

4 5 4 5 4 5

0 0 1

c 3 p oints C is a real square matrix but its eigenvalues are not all real and not all

pure imaginary.

d 3 p oints The vector 1; 1; 1 is in the row space of D but the vector 1; 1; 0 is

not in the nullspace. 6

6 Supp ose u ;u ;u is an orthonormal basis for R and v ;v is an orthonormal basis

1 2 3 1 2

for R .

a 5 p oints What is the rank, what are all vectors in the column space, and

what is a basis for the nullspace for the matrix B = u v + v ?

1 1 2

T T T

b 5 p oints Supp ose A = u v + u v . Multiply AA and simplify. Show that this

1 2

is a pro jection matrix by checking the required prop erties.

c 4 p oints Multiply A A and simplify. This is the identity matrix! Prove this for

example compute A Av and then nish the reasoning.

1 7

7 a 4 p oints If these three p oints happ en to lie on a line y = C + Dt, what system

Ax = b of three equations in two unknowns would b e solvable?

y =0 at t =1; y =1 at t =0; y = B at t =1:

Which value of B puts the vector b =0;1;Binto the column space of A?

b 4 p oints For every B nd the numb ers C and D that give the b est straight line

y = C + Dt closest to the three p oints in the least squares sense.

c 4 p oints Find the pro jection of b =1;0;0 onto the column space of A.

d 2 p oints If you apply the Gram-Schmidt pro cedure to this matrix A, what is

the resulting matrix Q that has orthonormal columns? 8

8 a 5 p oints Find a complete set of eigenvalues and eigenvectors for the matrix

2 3

2 1 1

6 7

A = :

6 7

1 2 1

4 5

1 1 2

b 6 p oints, 1 each Circle all the prop erties of this matrix A:

A is a pro jection matrix

A is a p ositive de nite matrix

A is a Markov matrix

A has determinant larger than trace

A has three orthonormal eigenvectors

A can be factored into A = LU

2 3

6 7

c 4 p oints Write the vector u = as a combination of eigenvectors of A, and

4 5

100

compute the vector u = A u .

100 0 9

Final Examination in Linear Algebra: 18.06

May 18, 1998 Solutions Professor Strang

1. a zero vector f0g

b 5 4= 1

2 3

6 7

c x =

4 5

d x = x b ecause N A=f0g.

3 2

1 0 0 0

7 6

0 1 0 0

7 6

0 0 1 0

= e R =

7 6

5 4

0 0 0 1

0 0 0 0

2 3 2 3 2 3

1 1 2 1 1 2 1 1 0

4 5 4 5 4 5

1 1 2 0 0 2 0 0 1

2. a A = ! U = ! R = :

2 2 2 0 0 0 0 0 0

The free variable is x . The complete solution is

2 3 2 3

2 1

4 5 4 5

0 1

x = x + x = + x :

p n 2

0 0

2 3 2 3

1 2

4 5 4 5

b A basis is 1 and 2

2 2

3. a The columns of B are a basis for the row space of A b ecause the row space is

the orthogonal complementofthe nullspace.

b N is n by n r ; B is n by r .

4. a det A =6

b det B =6

c det C =0

2 3

1 1 2

4 5

5. a A = 1 1 2

2 2 4

2 3

1 1 1

4 5

b B = 1 1 1

1 1 1

2 3

1 1 0

4 5

c C = 1 1 0

0 0 0

2 3

1 1 1

4 5

d D = 1 0 0

1 0 0

All four matrices are only examples many other correct answers exist.

6. a rank B = 1; all multiples of u are in the column space; the vectors v v and

1 1 2

v are a basis for the nullspace.

T T T T T T T T T

b AA = u v + u v v u + v u = u u + u u since v v = 0. AA is

1 2 1 2 1 2 2

1 2 1 2 1 2 1

T 2

symmetric and it equals AA :

T T T T T T

u u + u u u u + u u =u u +u u

1 2 1 2 1 2

The eigenvalues of AA are 1; 1; 0

T T T T

c A A = v v + v v since u u =0.

1 2 2

1 2 1

T T T

A Av = v v + v v v = v

1 1 2 1 1

1 2

T T T

A Av = v v + v v v = v :

2 1 2 2 2

1 2

2 T

Since v ;v are a basis for R , A Av = v for all v .

1 2 2

2 3 2 3

1 1 0

4 5 4 5

7. a Ax = b is 1 0 = 1 . This is solvable if B =2.

1 1 B

T T

b A Ax = A b is

3 0 C 1+B

= :

0 2 B

1+B B

Then C = and D =

3 2

T T

c A Ax = A b is

2 3 2 3

1 1 1=6

3 0 x 1 1=3

4 5 4 5

A = = p = Ax = 1 0 = 1=3 :

0 2 x 1 1=2

1 1 5=6

p p

2 3

1= 3 1= 2

4 5

d Q = columns were already orthogonal, now orthonormal

1= 3 0

p p

1= 3 1= 2

8. a =1;1;4. Eigenvectors can be

3 2 3 2 3 2

1 1 1

5 4 5 4 5 4

0 1 1

1 1 0

could also b e chosen orthonormal b ecause A = A

b Circle all the prop erties of this matrix A:

A is a pro jection matrix

A is a p ositive de nite matrix

A is a Markov matrix

A has determinant larger than trace

A has three orthonormal eigenvectors

A can b e factored into A = LU

2 3 2 3 2 3

2 1 1

4 5 4 5 4 5

0 1 1

= +

1 0 1

2 3 2 3 2 3 2 3

100

+1 2 1 1 4

100 100 100

100

4 5 4 5 4 5 4 5

0 1 1 4 1

= 1 +4 = : A

100

1 0 1 4 3

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Zzx] [ `{h t¯[ n`±¾£É ¾-wdn hvhv`bad¥ `{a nvt h¤lij eg[ihv`bh¤xEl^adhv`bh `{ad¥ lij

o D DporB D B B Ü DpB DpodB D

`b¥ a¦ª x l^nvh©lj Zm¼ hp[it ed[ihv`bh¤`{hqdh c1jmlin `{adwdq [iadcÀliq wdq yé]

F©Ñ

B B B D odB B B DporB D D D

Zz¼ n [in lwl^hhv`{eru x6l^nn x [iadh nvhE#c w arcd`{ad¥l^a [ l^nvc n°il^q

orB B BD ¢ B B D ¢êB B B ¢ko D B

wd`bxÎ `{¥ aª x l^nh6yé]

DpodBB B B D

¢z ¨

~ ~

Z\[^] Zv¿ë Âm] Z¿ Â¦] Zv¿ Â¦] y

G GÙì>í8î

Ü ~ Ü ~ Ü

ïð

Zzeg] y

FñGòìíóôí8î

ïð

Zzx] ¼ `b¥ a¦ª[iubq hklij [in [iadc ghvl¯É y

F G

N SP

odBB B B B ~

ë í

I T

J U

e hvqdedhwg[ix lijÝ x6l^arhv`{h `bad¥lijª x l^nh hv[ `bhjß°8`{ar¥

õ 5648ö:9<;>=@? ACBED¤÷ B©DvodB B D B D D

ø ù

K W

N Sûú M

Z\[^] _`badc[ t¯[ n`±¾ l^h x6liu{qdt afhvwg[x `{h y

QüÀN

D ¢ko B B ÷

Zzeg] _`badc£[ial^n l^adl^nt[u¢eg[ih`{h2jml^n y

Dvo ÷

Zzx] _`badc wdnvl@ý x `{l^a»t¯[ n`±¾(þÙwdnl@ý x `bad¥Àl^a l u j a8qru{u{hwg[ix Zzadl

DporB B D D B D D DporB B D B D

xEl^u{qrt¯afhwg[ix ]lij y

DporB Bÿ

Zzcg] _`badc u [ih h qg[in hhl^u{q `{lia l

DpodB B D Ð B D D

IJ TVU

J U

K W

[n h Z\[^]

¢z ¨

;> D o B Ë DvB B B D ¢ B

y Z\[^] ¼ x6lûbqdt adhlij h l^qdu{cÈe lYu{`ba [nvub°}`{arc w adc a hlû{q `{lârh6 l Ä

G F

M N Sú

odB o BD ¢ B B B B D D D

_gl^n ¾d[t¯wdu y

FñG¢¡

B B

~¤£

~ ~

[adc y Zzeg] ¬¤wrwdub°8`{ar¥ np[it¨§ x t¯`bc wdnl-x hh l x6l^ubqdt adh2lij °8` u{crh

Ò F

¡ ¡

DpodB¦¥ o D B D DvodB B

£ £

Zzx] e t¯[ n`±¾ l^h x6l^ubqdt adh[in ª x l^nvh¨`{a:wg[in Zme ]y¼ wrnvl@ý x `bl^a:t[ §

ACBED¦ BDporB D ¢ko B BDporB B D D odB B D

n`±¾´lâ l x6lûbqdt aûhwg[ix lij `{h y `{adx u j a8qru{u{hwg[ix `bh l^n l^¥ilâg[iu

F ¸ Ò

D D DpodB B BfDpodB B D B DpodB Dpo

y x6lit¯wdu t a lij xEl^u{qrt¯ahvwd[ix ` h2wdnl@ý x `bl^at[ nv` ¾`bh

G ¡

B B D DporB B D B D D

ó¨í

~ £

ï ï

í í

Zzcg] ¼ hvliu{q `{l^a l `{h y

F¸FG F¸ £G

¡ ¡

odB D D

£ £

qdwrwl^h

564:9<;>=@? B

IJ TVU

J U

[ihkadl hvl^ubq `bl^a

o D

K W

edq

I T

J U

[h`ba ad` ub°¯t[ia¦°fhl^u{q `{liadh

o Ë DpB D

K W

Z\[^] _`badc([u{u,wl^hvh`{edu `barjml^nt[ `{l^a1[el^q À[iadcyÈZz¼ np[iadÎ1[adc

B D D odB DporB

h [iw lij yé]

o B

Zzeg] _`badcÀ[ia ¾d[t¯wdu lij¨hqdx [t[ nv` ¾ ` ÀC[iadc ´[iu{u[ih¤ht[iubu[h

B B o D ¢ Dvo

wl^hvh`{edu y

IJ TVU

J U

Zzx] Íl cdl°^l^q£Îadl [ `bh2adl `{a aqdubu{hwg[ix lj

F Ñ

¢ ¢"Dpo D D DporB B

K W

¢z ¨

Z\[^] [iadc y

Q M¨! !ÆQ !

y Zzeg] ` # [iadc" µ

G G G FHG

P Q N

Dvo"

Zzx] ¼ x6liu{qdt ahvwg[ix [iadc u j aqdu{ubhvwg[ix [in [iu [°8hkl^n l^¥^liag[iuÃy

odB B DpodB B D B B ¢ Dvo

a [ix x6[ih ¥^`±ª [iu{u `{arjmlinvt¯[ `bl^aÆ°^liq x[a [iel^q `{¥ a¦ªi[u{q h}[adc

ö 5648Û:9<;>=@? B o B B DpodB D DDvodBÈB B B

`b¥ a¦ª x l^nvhE a t¯[ n`±¾ [ih jml^ubu{l `{ad¥ wdnvl^w n °µ

B B B D ¢kodB DpodB D o DpodB ¢ B D

Z\[^] ¼ wl nh [iwdwdnl¦[ix ¹ nvl t¯[ n`±¾y

F%$

odB ¢ B oDpodB B D

Zzeg] ¼ t[ nv` ¾`bh2h°8t t nv`bx¤wl^hv` `bª c ad` y

odB D BD D B BEË DvB

Zzx] ¼ t¯[ n`±¾`{h2adl cr`¦[i¥^liag[iu{`b¹[ieru y

odB D D B

Zzcg] ¼ t¯[ n`±¾ [ih jml^nvt n [iadc [in ª x linvh`{aÀÝ y

F§G'&)( & (

odB D o DporB ¢korB B B B D

Z+*ôliq£t `{¥ [ia l n °[ia ¾r[it wdu yé]

oDê¢ DD D B B

Z ] `{h2h`{t `{u{[in l [ cd`{[i¥^l^ad[iu¢t¯[ n`±¾ ` cr`¦[i¥^liag[iu a nv` h d[iadc y

P P N

B D D ¢ Dvo B D B

¢z ¨

Z\[^] ¼ `b¥ a¦ª[iubq h[in e a [adc y

SP P

odBB B B B BD ¢êBEB

Zzeg] ¼ `{¥ a¦ª[iu{q hº[in wl^h` `±ª y¼ n [n u{`ba [nvub°¤`{adc w adc a ^l^n l^¥il^ag[iu `b¥ a¦ª x l^nvhEy

odB B B B B D B odB B B B B B B D Dpo B B B D

Zzx] ¼ n `bhº[¤n w [ c `b¥ a¦ª[iubq [iarc [ `b¥ a¦ª[iubq [hj nºub`{a [invu±°©`{arc w adc a `b¥ a,§

odB B B B DvB B B B Dpo DºB B Bêo BE¢ B B B B B D¨B B

ª x l^nvh [a£` h2t%qru `bwdu{`bx6` ° [ih[nl-l lij x [inv[ix nv`bh `{xwliub°8adl^t `¦[iu\y

B D Dpo D D D D DvodB o DvB D

Zzcg] ¼ `{¥ aª[iubq h¤[in Z `{¥ a¦ª x linvh[in [iu{u<ª x linvhl^n l^¥^liag[iu l ][iarc Z

( & (

odBB B B B DpodB B B B D B B D Dvo D DporB

`b¥ a¦ª x l^nvh[in t%qdu `bwdu hslij ]y

B B B D B¤DpodB D B

Z ] ¼ `{¥ a¦ªi[u{q h2[in [iadc [iarc n [in n ub`{a [invu±°`{adc w adc a `{¥ a¦ª x linvhZ l

P N

B odB¤B B B B DpodB B BDpo B6B B B B B DkB B B D D ¢

jml^n.- [adc}lia jml^n.- ]y

G G

P N B

a [h q arx lij,aqdt%e nh©`{a jmliu{u{l `bad¥ [°µ21 31~ ¢[adc

G G

M P>¶iN /

56487:9<;>=@? 0¤BEË B B6Ð B B B DpodB ¢ ¢

1 Z1 ~,Ä61 ] yZ\Ì¨[ix aqdt%e n`bh [ª nv[i¥ lij lÀwdn ª8`{liqdh

¶^N

o B DpodB B B DporBD ¢ B

$54 $54 $

aqdt%e nh6yé]

I T I T

J U J U

1 ~ 1

$54 $54

Z\[^] qrw[ t[ nv` ¾ l¯¥ jmnvl^t l y

Ò F

NüÀN

BD D D BD D

K W K W

1 1 ~

$ $54

Zzeg] _`badc£[ia ¾-wdu{`bx6` jml^nvt qdu¦[ jml^n71 y

B D

Þ :

Zzx] [ `bh ub`{t ` lij81 [ih%9

o D DpodB D

¢z ¨

~ô~

Z\[^] y

FñGòì

Zzeg] [ih `{¥ a¦ªi[u{q h [adc ` x6linvn hvwl^adcd`bad¥ `b¥ a¦ª x l^nvh¤Z ][iarcZ ]y n`

P S¨P>¶iN Pi²P P^²S¤N

o B B B ¢ Dpo B B B B D DpB

~ ~

`{ad` `{[iu,h [ [ih%[Yu{`{a [inx6l^t ed`{ag[ `{lia(lij `b¥ a¦ª x l^nvhEµ ¿ ÂÄ y

ì ì

DporB D D DpB B D DporBB B B D ~

ð ð

~ ~ ~ ~ ~ ~

¬¤wrwdub°8`{ar¥ ¥ Z ] ¿ Â-Ä Z ] ghlB1 Z Z ] ]y

F ¡<;#=?>A@ ì G G

$ $ $ $

P S P2S S

¢ B BED

ð $

;

Þ : Þ

Zzx] ¬¤h%9 C1 y

Z\[^] qdwrwl^h `bhk[ t¯[ n`±¾lijnp[iadÎ [iadcu

Ò F

O¯üYO Q

D 56487:9<;>=@? B D BD

IJ TVU

®~ ~

J U

®~z

J U

®~z

K W

®~

e x6lijz[x linvhêlj<` h nvh nvl y¨Ì¾-wdu¦[`{a ° yZ l xElijz[ix l^nh

FG Ò

BDporB D D Ë D ¢ ¢ko DpodB D

¥^`±ª [%jml^nvt qdu¦[ jml^nk[aqdubu{hvwd[ix ª x l^n ]

B B B D ÿ

Í`ba µê¼ nh x6l^t wlia a lij [adc h xEl^adcÀx6lit¯wl^a a lij [in

F F

D orB¨Ë D B D DvodB B B D B

c nt¯`bag[ia h2ljsZmcd`FE n a ]st¯[ n`{x h6y [ [in h t[ nv`bx hs[iadc °

BDpB D B B D D B o D BDvodB B D B ¢ko

cdl ° [ª ¹ nvlfc nvt `{ag[a h Zm¼ nc1[iadc x6l^t wlia a h¤lij

Ñ F

Q O

DvodB o B B BEDpB D odB Dvo B D

jml^ubu{l h`{t `{u{[invu±°^hvl°^l^q x[iaýqrh [adh nÖjml^n h [iadc arc x6lit¯wl^a a h6yé]

P N

¢ D ¢ B DvodB D B D

Zzeg] Gêl^t wdq c nt¯`bag[ia lij

DvBDvodB BEDvB D

TVU IJ

U J

P P P P

U J

P SP P SP

U J

P P SP SP

W K

P SP SP P

Í`ba µH*ôl^qt¯`b¥ arcf` x6l^a¦ª ad` a lqdh jz[ix [ x6l^ubqdt¯arhs[in

D o¦D Ë D B B D2D B¤DpodB D Dpo D DpodB B

l^n l^¥il^ag[iu\y

Dpo

¢z ¨

Z\[^] ¼ nh x6lit¯wl^a a lj `bh¤c `{x `{h¤¹ nl}e x6[iqdh [ih©np[iarÎ y%¼ h xEl^adc

F F F

odB%Ë D B D BED ¢ko o B B B o orB B

x6lit¯wl^a a lij `{h c nt¯`bag[ia lij t¯[ n`±¾l^e [i`ba c}e°£n wdu{[ix6`bad¥ nvh nvl

B D DpodB BDpB D DporB D D B B DvodB¨Ë D ¢

lij ` h xEl^adc£nvl y¨¼ `{hkt¯[ n`±¾ [ihkn w [ c}nl h6 hvl ` h2c nvt `{ad[ia `{h2¹ nvldy

¢ Dvo£DvodB B ¢ o D o B B DpB ¢ D BEDvB D B

Zzeg] c É yI*ôliqx[iaxEl^t¯wrq `{he° ednvq jml^nx y¬¤adl n [°`bh l©arl [ É

G G

PR O

BD DpBsDvo DpB B DvodB ¢ D DpBsDvo D

hvl `b¥ a¦ª[iu{q h2lijÉ§[in y `{arx np[x lijÉÕ`{h2¹ nvld `b¥ a¦ª[iubq hk[in -

Ò G

N N

DvodB©B B B B J B¤DporB¤D B B DpodBB B B B

[iarc dhl¯c É y

N S¤N S¤N PR BD

FiÒaÐ EÜaÑiÒaØiÓÒ iÒ ÄiÒeaÖ AÐgebÖa: ½8º¼6

ÅaÝ ½8¸ ½999 ½:¿¼ ß 4:¿¼ ÈÖÓfe××ÓÖ ËØÖaÒg

YÓÙÖ ÒaÑe i×:

GÖadiÒg ½

ËecÖeØ CÓ de ´ÓÔØiÓÒaÐµ:

ÈÐea×e ciÖcÐe ÝÓÙÖ ÖeciØaØiÓÒ:

½µ ÅÓÒ ¾ß¿ ¾¹½¿½ Ëº ÃÐeiÑaÒ 5µ ÌÙe× ½¾ß½ ¾¹½¿½ Ëº ÃÐeiÑaÒ

¾µ ÅÓÒ ¿ß4 ¾¹½¿½ Ëº ÀÓÐÐaÒdeÖ 6µ ÌÙe× ½ß¾ ¾¹½¿½ Ëº ÃÐeiÑaÒ

¿µ ÌÙe× ½½ß½¾ ¾¹½¿¾ Ëº ÀÓÛ×ÓÒ 7µ ÌÙe× ¾ß¿ ¾¹½¿¾ Ëº ÀÓÛ×ÓÒ

4µ ÌÙe× ½¾ß½ ¾¹½¿¾ Ëº ÀÓÛ×ÓÒ

AÒ×ÛeÖ aÐÐ 9 ÕÙe×ØiÓÒ× ÓÒ Øhe×e Ôage×º Ìhi× i× a cÐÓ×ed bÓÓk eÜaÑº CaÐcÙÐaØÓÖ× aÖe ÒÓØ

Òeeded iÒ aÒÝ ÛaÝ aÒd ØheÖefÓÖe ÒÓØ aÐÐÓÛed ´ØÓ be faiÖ ØÓ aÐÐµº GÖade× aÖe kÒÓÛÒ ÓÒÐÝ

ØÓ ÝÓÙÖ ÖeciØaØiÓÒ iÒ×ØÖÙcØÓÖ ´ÛhÓ i× fÖee ØÓ Ô Ó×Ø ÛiØh ×ecÖeØ cÓ de× aÒd ØÓ ÖeØÙÖÒ eÜaÑ× iÒ

Ô eÖ×ÓÒµº ËÓÐÙØiÓÒ× ÛiÐÐ be Ô Ó×Øed ÓÒ Øhe Ûeb iÒ a feÛ daÝ×º Be×Ø Ûi×he× fÓÖ Øhe ×ÙÑÑeÖ aÒd

ØhaÒk ÝÓÙ fÓÖ ØakiÒg ½8º¼6º GË

½ ËÙÔÔ Ó×e Øhe ÑaØÖiÜ A i× Øhe ÔÖÓ dÙcØ

¿ ¿ ¾ ¾

½ ¼ ¿ 4 ½ ¼ ¼

7 7 6 6

: A =

7 7 6 6

¼ ½ ½ ¼ ¾ ½ ¼

5 5 4 4

¼ ¼ ¼ ¼ 5 4 ¿

´aµ ´¿ Ô ÓiÒØ×µ GiÚe a ba×i× fÓÖ Øhe ÒÙÐÐ×Ôace Óf Aº

´bµ ´¿ Ô ÓiÒØ×µ GiÚe a ba×i× fÓÖ Øhe ÖÓÛ ×Ôace Óf Aº

´cµ ´¾ Ô ÓiÒØ×µ EÜÔÖe×× ÖÓÛ ¿ Óf A a× a cÓÑbiÒaØiÓÒ Óf ÝÓÙÖ ba×i× ÚecØÓÖ×

iÒ ÝÓÙÖ aÒ×ÛeÖ ØÓ ´bµº

´dµ ´¿ Ô ÓiÒØ×µ ÏhaØ i× Øhe diÑeÒ×iÓÒ Óf Øhe ÒÙÐÐ×Ôace Óf A ?

¾ ËÙÔÔ Ó×e A i× a 5 bÝ 7 ÑaØÖiÜ¸ aÒd AÜ = b ha× a ×ÓÐÙØiÓÒ fÓÖ eÚeÖÝ ÖighØ ×ide bº

´aµ ´¿ Ô ÓiÒØ×µ ÏhaØ dÓ Ûe kÒÓÛ ab ÓÙØ Øhe cÓÐÙÑÒ ×Ôace Óf A?

´bµ ´¿ Ô ÓiÒØ×µ ÏhaØ dÓ Ûe kÒÓÛ ab ÓÙØ Øhe ÖÓÛ× Óf A?

´cµ ´¿ Ô ÓiÒØ×µ ÏhaØ dÓ Ûe kÒÓÛ ab ÓÙØ Øhe ÒÙÐÐ×Ôace Óf A?

´dµ ´¿ Ô ÓiÒØ×µ ÌÖÙe ÓÖ faÐ×e ´ÛiØh Öea×ÓÒµ:

Ìhe cÓÐÙÑÒ× Óf A aÖe a ba×i× fÓÖ Øhe cÓÐÙÑÒ ×Ôace Óf Aº

YÓÙÖ aÒ×ÛeÖ× cÓÙÐd ÖefeÖ ØÓ diÑeÒ×iÓÒ»ba×i×»ÐiÒeaÖ iÒdeÔ eÒdeÒce»×ÔaÒÒiÒg a ×Ôaceº

¿ A××ÙÑe ØhaØ A i× iÒÚeÖØibÐe aÒd Ô eÖÑÙØaØiÓÒ× aÖe ÒÓØ Òeeded iÒ eÐiÑiÒaØiÓÒ if Ô Ó××ibÐeº

? ´aµ ´¾ Ô ÓiÒØ×µ AÖe Øhe ÔiÚÓØ× Óf A eÕÙaÐ ØÓ

ÔiÚÓØ× Óf A

Áf Ýe×¸ giÚe a Öea×ÓÒ; if ÒÓ¸ giÚe aÒ eÜaÑÔÐeº

´bµ ´¿ Ô ÓiÒØ×µ Á× Øhe ÔÖÓ dÙcØ Óf ÔiÚÓØ× Óf A eÕÙaÐ ØÓ ?

ÔÖÓ dÙcØ Óf ÔiÚÓØ× Óf A

Áf Ýe×¸ giÚe a Öea×ÓÒ; if ÒÓ¸ giÚe aÒ eÜaÑÔÐeº

´cµ ´¿ Ô ÓiÒØ×µ AÔÔÐÝ bÐÓ ck eÐiÑiÒaØiÓÒ ØÓ Øhe ¾Ò bÝ ¾Ò ÑaØÖiÜ

¿ ¾

A Á

5 4

: Å =

Á ¼

ÅÙÐØiÔÐÝ bÐÓ ck ÖÓÛ ½ bÝ a ×ÙiØabÐe ÑaØÖiÜ aÒd add ØÓ bÐÓ ck ÖÓÛ ¾º

ÏhaØ ÑaØÖiÜ aÔÔ eaÖ× iÒ Øhe ´¾; ¾µ bÐÓ ck?

´dµ ´¿ Ô ÓiÒØ×µ ÏhaØ i× Øhe deØeÖÑiÒaÒØ Óf Å ? EÜÔÐaiÒ Øhe ¬ÒaÐ ÔÐÙ× ÓÖ ÑiÒÙ× ×igÒº

4 ËÙÔÔ Ó×e A ha× eigeÒÚaÐÙe× = ¿¸ = ½¸ = ¼ ÛiØh cÓÖÖe×Ô ÓÒdiÒg eigeÒÚecØÓÖ×

½ ¾ ¿

¾ ¿ ¾ ¿ ¾ ¿

½ ¼ ¼

6 7 6 7 6 7

Ü = ; Ü = ; Ü = :

6 7 6 7 6 7

½ ½ ¼

½ ¾ ¿

4 5 4 5 4 5

½ ½ ½

´aµ ´¿ Ô ÓiÒØ×µ ÀÓÛ dÓ ÝÓÙ kÒÓÛ ØhaØ Øhe ØhiÖd cÓÐÙÑÒ Óf A cÓÒØaiÒ× aÐÐ ÞeÖÓ×?

´bµ ´¿ Ô ÓiÒØ×µ FiÒd Øhe ÑaØÖiÜ Aº

½ Ì

´cµ ´¿ Ô ÓiÒØ×µ BÝ ØÖaÒ×Ô Ó×iÒg Ë AË = ^¸ ¬Òd Øhe eigeÒÚecØÓÖ× Ý ; Ý ; Ý Óf A º

½ ¾ ¿

´Á aÑ ÐÓ ÓkiÒg fÓÖ ×Ô eci¬c ÚecØÓÖ× Ðike Ü ; Ü ; Ü ab ÓÚeºµ

½ ¾ ¿

5 Ìhi× ÔÖÓbÐeÑ cÓÑÔÙØe× Øhe ÔÐaÒe Þ = CÜ · DÝ · E ØhaØ i× cÐÓ×e×Ø ´iÒ Øhe Ðea×Ø ×ÕÙaÖe×

×eÒ×eµ ØÓ Øhe×e fÓÙÖ Ñea×ÙÖeÑeÒØ×:

AØ Ü = ½; Ý = ¼ Ñea×ÙÖeÑeÒØ giÚe× Þ = ½

AØ Ü = ½; Ý = ¾ Ñea×ÙÖeÑeÒØ giÚe× Þ = ¿

AØ Ü = ¼; Ý = ½ Ñea×ÙÖeÑeÒØ giÚe× Þ = 5

AØ Ü = ¼; Ý = ¾ Ñea×ÙÖeÑeÒØ giÚe× Þ = ¼

´aµ ´¿ Ô ÓiÒØ×µ ÏÖiØe dÓÛÒ Øhe ÐiÒeaÖ ×Ý×ØeÑ AÜ = b ÛiØh ÙÒkÒÓÛÒ ÚecØÓÖ Ü = ´C; D; E µ

ØhaØ ÛÓÙÐd giÚe a ÔÐaÒe gÓiÒg eÜacØÐÝ ØhÖÓÙgh Øhe fÓÙÖ giÚeÒ Ô ÓiÒØ× | eÜceÔØ ØhaØ

Øhi× ÔaÖØicÙÐaÖ ×Ý×ØeÑ ha× ÒÓ ×ÓÐÙØiÓÒº

´bµ ´¿ Ô ÓiÒØ×µ ËhÓÛ ØhaØ Øhi× ×Ý×ØeÑ AÜ = b ha× ÒÓ ×ÓÐÙØiÓÒ!

b b b

´cµ ´¿ Ô ÓiÒØ×µ FiÒd Øhe b e×Ø Ðea×Ø ×ÕÙaÖe× ×ÓÐÙØiÓÒ Üb = ´C; D; E µº

´dµ ´¿ Ô ÓiÒØ×µ Ìhe eÖÖÓÖ ÚecØÓÖ e = ´e ; e ; e ; e µ iÒ Øhe ÙÒdeÖÐÝiÒg ÔÖÓ jecØiÓÒ ÔÖÓbÐeÑ

½ ¾ ¿ 4

i× Ô eÖÔ eÒdicÙÐaÖ ØÓ Ûhich ÚecØÓÖ×? YÓÙ dÓÒ³Ø haÚe ØÓ cÓÑÔÙØe e bÙØ ÝÓÙ dÓ haÚe ØÓ

×aÝ Ûhich ×Ô eci¬c ÒÙÑeÖicaÐ ÚecØÓÖ× iØ i× Ô eÖÔ eÒdicÙÐaÖ ØÓº

¿ ¾ ¾ ¿

4 ¿

7 6 6 7

½ ½

7 6 6 7

aÖe ÓÖØhÓÒÓÖÑaÐº 6 ´aµ ´¿ Ô ÓiÒØ×µ Ìhe ÚecØÓÖ× Õ = aÒd Õ =

7 6 6 7

¿ 4

½ ¾

5 4 4 5

5¼

¼ 5

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Your name is:

Please circle your recitation:

1) M2 2-131 P.-O. Persson 2-088 2-1194 persson 2) M2 2-132 I. Pavlovsky 2-487 3-4083 igorvp 3) M3 2-131 I. Pavlovsky 2-487 3-4083 igorvp 4) T10 2-132 W. Luo 2-492 3-4093 luowei 5) T10 2-131 C. Boulet 2-333 3-7826 cilanne 6) T11 2-131 C. Boulet 2-333 3-7826 cilanne 7) T11 2-132 X. Wang 2-244 8-8164 xwang 8) T12 2-132 P. Cliﬀord 2-489 3-4086 peter 9) T1 2-132 X. Wang 2-244 8-8164 xwang 10) T1 2-131 P. Cliﬀord 2-489 3-4086 peter 11) T2 2-132 X. Wang 2-244 8-8164 xwang

The ten questions are worth 10 points each. Thank you for taking 18.06! 1 The 4 by 6 matrix A has all 2’s below the diagonal and elsewhere all 1’s:   1 1 1 1 1 1      2 1 1 1 1 1  A =      2 2 1 1 1 1    2 2 2 1 1 1

(a) By elimination factor A into L (4 by 4) times U (4 by 6).

(b) Find the rank of A and a basis for its nullspace (the special solutions would be good).

2 2 Suppose you know that the 3 by 4 matrix A has the vector s = (2, 3, 1, 0) as a basis for its nullspace.

(a) What is the rank of A and the complete solution to Ax = 0?

(b) What is the exact row reduced echelon form R of A?

3 3 The following matrix is a projection matrix:   1 2 −4 1     P =  2 4 −8  . 21   −4 −8 16

(a) What subspace does P project onto?

(b) What is the distance from that subspace to b = (1, 1, 1)?

4 4 (a) Suppose the product of A and B is the zero matrix: AB = 0. Then the (1) space of A contains the (2) space of B. Also the (3) space of B contains the (4) space of A. Those blank words are

(1) (2) (3) (4)

(b) Suppose that matrix A is 5 by 7 with rank r, and B is 7 by 9 of rank s. What are the dimensions of spaces (1) and (2) ? From the fact that space (1) contains space (2) , what do you learn about r + s?

5 5 Suppose the 4 by 2 matrix Q has orthonormal columns.

(a) Find the least squares solution xb to Qx = b. (b) Explain why QQT is not positive deﬁnite.

6     1 5     3     6 Let S be the subspace of R spanned by  2  and  4 .     2 −2

(a) Find an orthonormal basis q 1, q 2 for S by Gram-Schmidt.

(b) Write down the 3 by 3 matrix P which projects vectors perpendicularly onto S.

7 3 7 (a) If v 1, v 2, v 3 form a basis for R then the matrix with those three columns is .

3 (b) If v 1, v 2, v 3, v 4 span R , give all possible ranks for the matrix with those four columns. .

3 (c) If q 1, q 2, q 3 form an orthonormal basis for R , and T is the transformation that

projects every vector v onto the plane of q 1 and q 2, what is the matrix for T in this basis? Explain.

8 8 Suppose the n by n matrix An has 3’s along its main diagonal and 2’s along the diagonal below and the (1, n) position:   3 0 0 2      2 3 0 0  A4 =   .    0 2 3 0    0 0 2 3

Find by cofactors of row 1 or otherwise the determinant of A4 and then the determi-

nant of An for n > 4.

9 9 There are six 3 by 3 permutation matrices P .

(a) What numbers can be the determinant of P ? What numbers can be pivots?

(b) What numbers can be the trace of P ? What four numbers can be eigenvalues of P ?

10 10 Suppose A is a 4 by 4 upper triangular matrix with 1, 2, 3, 4 on its main diagonal. (You could put all 1’s above the diagonal.)

(a) For A − 3I, which columns have pivots? Which components of the eigenvector

x 3 (the special solution in the nullspace) are deﬁnitely zero?

(b) Using part (a), show that the eigenvector matrix S is also upper triangular.

11 Course 18.06, Fall 2002: Final Exam, Solutions

1 (a) 1 0 0 0 1 1 1 1 1 1  2 1 0 0 0 −1 −1 −1 −1 −1 A = LU =     . 2 0 1 0 0 0 −1 −1 −1 −1 2 0 0 1 0 0 0 −1 −1 −1 (b) Four pivots ⇒ rank of A = 4. The row reduced echolon form is

1 0 0 0 0 0 0 1 0 0 0 0 R =   . 0 0 1 0 0 0 0 0 0 1 1 1

The two special solutions (0, 0, 0, −1, 1, 0), (0, 0, 0, −1, 0, 1) form a basis for the nullspace.

2 (a) The dimension of the nullspace is 1, so the rank of A is 4 − 1 = 3. The complete solution to Ax = 0 is x = c · (2, 3, 1, 0) for any constant c. (b) The row reduced echelon form has 3 pivots and the special solution x = (2, 3, 1, 0):

1 0 −2 0 R = 0 1 −3 0 . 0 0 0 1

3 (a) The projection matrix P projects onto the column space of P which is the line c·(1, 2, −4). (b) The vector from b to the subspace is

1 −1 22 1 1 e = b − P b = 1 − −2 = 23   21   21   1 4 17

and the distance is √ 1 p 1302 kek = 222 + 232 + 172 = . 21 21 (c) Since P projects onto a line, its three eigenvalues are 0, 0, 1. Since P is symmetric, it has a full set of (orthogonal) eigenvectors, and is then diagonalizable.

4 (a) When AB = 0, every column of B is in the nullspace of A. So the null space of A contains the column space of B. Also the left null space of B contains the row space of A. (b) The dimension of the nullspace of A is n−r = 7−r. The dimension of the column space of B is s. Since the ﬁrst contains the second, 7 − r ≥ s, or r + s ≤ 7.

5 (a) The least squares solution xˆ to Qx = b is

xˆ = (QTQ)−1QTb = (I)−1QTb = QTb. (b) One approach: QT is 2 by 4 so there are free variables and many solutions to QTx = 0. Then QQTx = 0 and QQT is singular; not positive deﬁnite. Second approach: The rank of Q is 2, so the rank of QQT must be ≤ 2. But QQT is a 4 by 4 matrix, so the dimension of its nullspace is 2. This means that it has 2 zero eigenvalues, and QQT is not positive deﬁnite. (c) The singular values of Q are the square roots of the eigenvalues of QTQ = I, that is, all 1. 6 (a) Let a = (1, 2, 2) and b = (5, 4, −2). The orthonormal vectors are 1 a 1 q = = 2 , 1 kak 3   2 T q 1 b  4   2  b − T q 1 q 1 q 1 1 q 2 = =  2  =  1  k · k p 2 2 2 3 4 + 2 + (−4) −4 −2 where k · k means the norm of the numerator. (b) Let Q = q 1 q 2 . The projection P is then  5 4 −2 1 P = QQT = 4 5 2 9   −2 2 8

(c) The properties of a projection matrix are P 2 = P and P T = P . This gives (P b)T(b − P b) = bTP T(b − P b) = bT(P b − P 2b) = 0.

3 7 (a) If v 1, v 2, v 3 is a basis for R then the matrix with those three columns is invertible (non-singular, full rank). 3 3 (b) If v 1, v 2, v 3, v 4 span R then the column space is R . The only possible rank for the matrix with those four columns is 3. (c) The transformations of the three basis vectors are

T (q 1) = q 1

T (q 2) = q 2

T (q 3) = 0

so the transformation matrix T in the basis q 1, q 2, q 3 is 1 0 0 T = 0 1 0 . 0 0 0

3 0 0 2 3 0 0 2 3 0 2 3 0 0 = 3 2 3 0 − 2 0 2 3 = 3 · 27 − 2 · 8 = 65. 0 2 3 0 0 2 3 0 0 2 0 0 2 3 n n−1 n For general n > 4, the determinant is |An| = 3 + (−1) 2 . 9 (a) The determinant of a permutation matrix P is 1 or −1. The only possible pivot is 1. (b) For 3 by 3 permutations, the trace of P is 0, 1, or 3. The eigenvalues are 1 and −1 when two rows are exchanged. Otherwise

−λ 0 1 √ 3 1 3 2πi/3 4πi/3 1 −λ 0 = −λ + 1 = 0 ⇒ λ = 1, − ± i (or e , e ), 2 2 0 1 −λ √ √ 1 3 1 3 so the four possible eigenvalues are 1, −1, − 2 + i 2 , and − 2 − i 2 . 10 (a) 1 1 1 1 −2 1 1 1 0 2 1 1  0 −1 1 1 A =   ,A − 3I =   0 0 3 1  0 0 0 1 0 0 0 4 0 0 0 1 Columns 1, 2, and 4 have pivots. Because of the nonzero bottom-right element in A−3I, the fourth component of x 3 is deﬁnitely zero. (b) In the same way as above, the special solutions for the matrices A − 1I, A − 2I, A − 3I, and A − 4I must have 3, 2, 1, and 0 zeros as the last components. The eigenvector matrix S is then upper triangular. Questions from 18.06 Final, Fall 2003

1. Suppose A = LU where 1 0 0 5 0 5 1 L =  2 1 0 , U = 0 3 3 0 . −2 3 1 0 0 0 0     (a) What are the dimensions of the 4 fundamental subspaces associated with A?

(b) Give a basis for each of the 4 fundamental subspaces.

1 N(A)

R(A)

C(A)

N(AT )

2 2. Let F be the subspace of R4 given by

F = {(x, y, z, w) : x − y + 2z + 3w = 0}.

Let P be the projection matrix for projecting onto F . (Many of the subquestions can be answered independently of the others.)

(a) Give an orthonormal basis {v1, · · · , vk} for the orthogonal complement to F .

(b) Find an orthonormal basis {w1, · · · , wl} for F . Explain how you proceed.

3 The following questions refer to the projection matrix P for projecting onto F . (c) What are the eigenvalues of P ? Give them with their multiplicities.

1 1 (d) What is the projection of   onto F ? 1   1  

4 3. (a) Write down the 2×2 rotation matrix, R(θ), that rotates R2 in the counterclockwise direction by an angle θ (this matrix is a function of θ).

(b) Compute the eigenvalues of R(θ). For which value(s) of θ are the eigenvalues real?

5 (d) Write down the singular value decomposition of R(θ).

6 4. (a) Give two 3 × 3 matrices A and B such that AB is not equal to BA.

(b) Suppose A and B are n × n matrices with the same set of linearly independent eigenvectors v1, v2, · · · , vn. However, the eigenvalues might be diﬀerent: vi is the eigenvector for the eigenvalue λi of A and the eigenvector for the eigenvalue µi of B. Show that AB = BA.

7 du dt 0 3 u 5. Consider the diﬀerential equation dv = . dt 2 −1 v

(a) Solve the diﬀerential equation and express u(t), v(t) as functions of u(0) and v(0).

8 p u (b) Find a linear transformation = T such that the differential equation simplifies q v dp into two independent differential equations in p and in q (one relating dt and p, the other dq relating dt and q)

(d) Are there initial conditions u(0), v(0) that would make u(t) go to 0? If yes, give one such value for u(0) and u0(0).

9 6. True of False. Circle the appropriate answer. “True” means “always true”, and “false” means “sometimes false”. Justify each answer brieﬂy.

(a) The product of the pivots when performing Gauss-Jordan is equal to the determinant if we do not have to permute rows. True or False.

(b) Let A be an m × n matrix whose columns are independent. Then AAT is positive deﬁnite. True or False.

(c) If A is a symmetric matrix then the singular values are the absolute values of the nonzero eigenvalues. True or False.

(d) There exists a 5 × 5 unitary matrix with eigenvalues 1, 1 + i, 1 − i, i and −i. True or False.

(e) Suppose V and W are two vector spaces of dimension n. If T is a linear transformation from V to W with only the 0 vector in the kernel, then for any basis of V , there exists an orthonormal basis of W such that the resulting matrix representing T is upper triangular. True or False.

10 7. Consider the following matrix A:

0.5 0.4 0.2 A = 0.4 0.5 0.2 . 0.1 0.1 0.6   (c) Can you immediately tell one of the eigenvalues of A (without computing them)? Explain.

(d) Compute the determinant of A.

(e) Find the eigenvalues of A and the corresponding eigenvectors. (Check your answer.)

11 1 (f) Two out of the 3 eigenvectors of A should be orthogonal to 1. How could you have 1 explained this before computing the eigenvectors?  

(g) Write an exact expression for A100.

12 8. Let 1 0 0 A = 0 3 0 . 0 0 5   For each of the following matrices, either complete it (ﬁnd values for the non-diagonal elements) so that it becomes similar to A, or explain why it is impossible to complete it to a matrix similar to A. Circle whether you are able to complete it or not to a matrix similar to A.

(a) 2 . . B = . 2 . . . 4   Able to complete it to similar?: Yes No If yes, give a completion. If not, why not?

13 (b) 3 . . C = . 3 . . . . 3   Able to complete it to similar?: Yes No If yes, give a completion. If not, why not?

14 15 18.06 Professor Strang Final Exam December 19, 2005

Grading 1 Your PRINTED name is: 2 3 4 5 Please circle your recitation: 6 7 1) M 2 2-131 P. Lee 2-087 2-1193 lee 8 2) M 2 2-132 T. Lawson 4-182 8-6895 tlawson 4) T 10 2-132 P-O. Persson 2-363A 3-4989 persson 5) T 11 2-131 P-O. Persson 2-363A 3-4989 persson 6) T 11 2-132 P. Pylyavskyy 2-333 3-7826 pasha 7) T 12 2-132 T. Lawson 4-182 8-6895 tlawson 8) T 12 2-131 P. Pylyavskyy 2-333 3-7826 pasha 9) T 1 2-132 A. Chan 2-588 3-4110 alicec 10) T 1 2-131 D. Chebikin 2-333 3-7826 chebikin 11) T 2 2-132 A. Chan 2-588 3-4110 alicec 12) T 3 2-132 T. Lawson 4-182 8-6895 tlawson 1 (12 pts.) This question is about the matrix A = I + E where E is the all-ones matrix ones(4, 4): 2 1 1 1 1 1 1 1     1 2 1 1 1 1 1 1 A =   = I +   .      1 1 2 1   1 1 1 1           1 1 1 2   1 1 1 1      (a) By elimination ﬁnd the pivots of A.

(b) Factor A into LDLT (if that is possible).

2 2 (12 pts.) Keep the same matrix A as in Problem 1.

(a) Find the matrix P that projects any vector in R4 onto the subspace spanned by the ﬁrst column of A.

(b) Describe the nullspace of I − P and the nullspace of P A.

3 3 (12 pts.) Now suppose A = I + bE, with the same E = ones(4, 4).

(a) What are the eigenvalues of E ?

(b) If b = 2, what is the determinant of A ?

4 4 (16 pts.) Suppose A is an 8 by 8 invertible matrix. Throw away any 3 columns of A to get an 8 by 5 matrix B.

(a) You will correctly think that B has rank 5. Give a mathematical reason why this is true.

(b) Tell all you know about the nullspace of BT and the reduced row echelon form rref(B).

(c) Give as much information as possible about the eigenvalues and eigenvectors of BTB and BBT (those are separate questions).

5 5 (12 pts.) Suppose Q is an m by n matrix with QTQ = I. Write down the most important facts about

(a) The columns of Q

(b) m and n and the rank of Q

6 0 0 0 1   1 0 0 0 6 (12 pts.) (a) The eigenvalues of A =   are .    0 1 0 0       0 0 1 0    (b) An orthogonal set of 4 eigenvectors is .

diagonalizable permutation nonsingular Jordan matrix orthogonal projection skew-symmetric

7 7 (12 pts.) Suppose A is 2 by 3 with this Singular Value Decomposition UΣV T. U and V are orthogonal matrices:

T v1 4 0 0  T  A = u1 u2   v2 . h i 0 0 0    T     v3    (a) Find a basis for the nullspace of A.

(b) Find all solutions to the equation Ax = u1.

8 8 (12 pts.) Suppose A (3 by 3) has eigenvalues λ1, λ2, λ3 and independent eigenvectors

x1, x2, x3.

du (a) What is the general form of the solutions to u +1 = Au and = Au ? k k dt (Two questions)

(b) Suppose every solution to uk+1 = Auk approaches a multiple c x1 as

k → ∞ (c depends on u0). What does this tell you about λ1, λ2, λ3 ? du (c) For some 3 by 3 matrices, the complete solution to = Au does not dt have the form you gave in part (a). What can go wrong ? Give an example of such a matrix A.

9 1.(a) 3 4 5 2, 2 , 3 , 4

T  1   2   1  1 3 1  2 1   2   2 1  (b)A =  1 1   4   1 1   2 3 1   3   2 3 1  1 1 1 5 1 1 1 2 3 4 1 4 2 3 4 1

 4 2 2 2  2.(a) aaT 1  2 1 1 1  Trick: computation easiest using P = T =   , a a 7  2 1 1 1  2 1 1 1 T a = (2, 1, 1, 1) .

(b) The nullspace of I − P consists of all multiples of a. (One view is that x = P x. Another view is that it is the orthogonal complement of the nullspace of P which are all vectors orthogonal to a.)

3.(a) E is rank one symmetric with trace 4, so the eigenvalues are 0, 0, 0, 4.

(b) The eigenvalues of A are 1, 1, 1, and 1 + 2 · 4 = 9 so the determinant is 9.

4.(a) The columns of B are independent since the columns of A are. There- fore the span of B is ve dimensional. (B has no free columns.)

(b) The nullspace of BT is a three dimensional subspace of R8.It is the orthogonal complement of the column space of B in R8. The rref of B looks like the rst ve columns of I8.

(c) BT B is a 5 × 5 matrix with positive eigenvalues that are the squares of the singular values of B.BBT has the same ve positive eigenvalues and three more 0 eigenvalues as well.

5.(a) The columns of Q are n orthonormal vectors in Rm.

1 (b) n ≤ m and the rank of Q is n.

6.(a) 1, i, −1, −i

 1 1 1 1   1 −i −1 i  (b) The four columns of the DFT matrix F4 =    1 −1 1 −1  1 i 1 −i

7.(a) The nullspace of A has basis v2 and v3.

(b) x = v1/4 + c1v2 + c2v3

8.(a) k k k and λ1t λ2t λ3t uk = c1λ1 x1 +c2λ2 x2 +c3λ3 x3 u(t) = c1e x1 +c2e x2 +c3e x3

(b)λ1 = 1 and |λi| < 1, for i = 2, 3.

(c) The matrix may not have a complete set of independent eigenveectors.  0 1 0  An example is a three by three Jordan block:  0 0 1  . 0 0 0

2 18.06 Professor Strang Final Exam May 16, 2005

Grading 1 Your PRINTED name is: 2 Closed book / 10 wonderful problems. 3 Thank you for taking this course ! 4 5 Please circle your recitation: 6 7 1) M 2 2-131 A. Chan 2-588 3-4110 alicec 8 2) M 3 2-131 A. Chan 2-588 3-4110 alicec 9 3) M 3 2-132 D. Testa 2-586 3-4102 damiano 10 4) T 10 2-132 C.I. Kim 2-273 3-4380 ikim 5) T 11 2-132 C.I. Kim 2-273 3-4380 ikim 6) T 12 2-132 W.L. Gan 2-101 3-3299 wlgan 7) T 1 2-131 C.I. Kim 2-273 3-4380 ikim 8) T 1 2-132 W.L. Gan 2-101 3-3299 wlgan 9) T 2 2-132 W.L. Gan 2-101 3-3299 wlgan

There is an extra blank page at the end. n 1(10pts.) Suppose P1,...,Pn are points in R . The coordinates of Pi are (ai1,ai2,...,ain).

We want to ﬁnd a hyperplane c1x1 +···+cnxn = 1 that contains all n points Pi.

(a) What system of equations would you solve to ﬁnd the c’s for that hyperplane ?

(b) Give an example in R3 where no such hyperplane exists (of this form), and an example which allows more than one hyperplane of this form.

(c) Under what conditions on the points or their coordinates is there not a unique interpolating hyperplane with this equation ?

2 2(10pts.) (a) Find a complete set of “special solutions” to Ax = 0 by noticing the pivot variables and free variables (those have values 1 or 0). ⎡ ⎤ ⎢ 12345⎥ ⎢ ⎥ A ⎢ ⎥ . = ⎣ 12346⎦ 00000

(b) and (c) Prove that those special solutions are a basis for the nullspace N(A). What two facts do you have to prove ?? Those are parts (b) and (c) of this problem.

3 3(10pts.) (a) I was looking for an m by n matrix A and vectors b, c such that Ax = b has no solution and ATy = c has exactly one solution. Why can I not ﬁnd A, b, c ?

(b) In Rm, suppose I gave you a vector b and a vector p and n linearly

independent vectors a1,a2,...,an.IfIclaimthatp is the projection of b onto the subspace spanned by the a’s, what tests would you make to see if this is true ?

4 4(10pts.) (a) Find the determinant of ⎡ ⎤ ⎢ 1211⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1111⎥ B = ⎢ ⎥ . ⎢ ⎥ ⎣ 1121⎦ 1112

(b) Let A be the 5 by 5 matrix ⎡ ⎤ ⎢ 21111⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 12111⎥ ⎢ ⎥ A ⎢ ⎥ . = ⎢ 11211⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 11121⎦ 11112

Find all ﬁve eigenvalues of A by noticing that A − I hasrank1and the trace of A is .

5 5(10pts.) (a) Complete the matrix A (ﬁll in the two blank entries) so that A has

eigenvectors x1 =(3, 1) and x2 =(2, 1): ⎡ ⎤ 26 A = ⎣ ⎦

(b) Find a diﬀerent matrix B with those same eigenvectors x1 and x2,and 10 with eigenvalues λ1 =1andλ2 =0. What is B ?

6 2 3 6(10pts.) We can ﬁnd the four coeﬃcients of a polynomial P (z)=c0 +c1z+c2z +c3z 2 3 if we know the values y1,y2,y3,y4 of P (z) at the four points z =1,i,i ,i .

(a) What equations would you solve to ﬁnd c0,c1,c2,c3 ?

(b) Write down one special property of the coeﬃcient matrix.

7 7(10pts.) Suppose S is a 4-dimensional subspace of R7,andP is the projection matrix onto S.

(a) What are the seven eigenvalues of P ?

(b) What are all the eigenvectors of P ? du (c) If you solve dt = −Pu (notice minus sign) starting from u(0), the solution u(t) approaches a steady state as t →∞. Can you describe that limit vector u(∞)?

8 8(10pts.) Suppose my favorite −1, 2, −1 matrix swallowed extra zeros to become ⎡ ⎤ − ⎢ 2010⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ 0201 ⎥ A = ⎢ ⎥ . ⎢ − ⎥ ⎣ 1020⎦ 0 −102

(a) Find a permutation matrix P so that ⎡ ⎤ − ⎢ 2 100⎥ ⎢ ⎥ ⎢ − ⎥ T ⎢ 1200⎥ B = PAP = ⎢ ⎥ . ⎢ − ⎥ ⎣ 0021 ⎦ 00−12

(b) What are the 4 eigenvalues of B ? Is this matrix diagonalizable or not ?

(c) How do you know that A has the same eigenvalues as B ?ThenA is positive deﬁnite—what function of u, v, w, z is therefore positive except when u = v = w = z =0?

9 9(10pts.) (a) Describe all vectors that are orthogonal to the nullspace of this singular matrix A. You can do this without computing the nullspace. ⎡ ⎤ ⎢ 137⎥ ⎢ ⎥ A ⎢ ⎥ . = ⎣ 226⎦ 214

(b) If you apply Gram-Schmidt to the columns of this A, what orthonormal vectors do you get ?

(c) Find a “reduced” LU factorization of A,withonly2 columns in L and 2 rows in U. Can you write A as the sum of two rank 1 matrices ?

10 10 (10 pts.) Suppose the singular value decomposition A = UΣV T has ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ 1111⎥ −122 1000 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − − ⎥ 1 ⎢ ⎥ ⎢ ⎥ 1 ⎢ 1 111 ⎥ U = ⎢ 2 −12⎥ Σ=⎢ 0400⎥ V = ⎢ ⎥ . 3 ⎣ ⎦ ⎣ ⎦ 2 ⎢ − − ⎥ ⎣ 111 1 ⎦ 22−1 0000 1 −1 −11

(a) Find the eigenvalues of ATA.

(b) Find a basis for the nullspace of A.

(d) Find a singular value decomposition of −AT.

11 12 18.06 - Final Exam, Monday May 16th, 2005 solutions

1. (a) We want the coordinates (ai1,...,ain)ofPi to satisfy the equation c1x1 + ...+ cnxn =1. Thus the system of equations is Ac = ones:

c1a11 + c2a12 + ...+ cna1n =1

c1a21 + c2a22 + ...+ cna2n =1 ...

c1an1 + c2an2 + ...+ cnann =1

(b) There is no plane of the given form, if one of the points Pi is the origin. More than one plane contains the Pi’s if the three points are on a line not through the origin. (c) There is not a unique solution precisely when det A = 0. This means geometrically that n the points Pi lie in an (n − 1)−dimensional subspace of R . 2. (a) Subtracting the ﬁrst row from the second, we ﬁnd the matrix ⎡ ⎤ 12345 U = ⎣ 00001⎦ . 00000

(In the row reduced echelon form R, the 5 changes to 0.) The pivot variables are the ﬁrst and the last, while x2,x3,x4 are the free variables. Thus the “special solutions” to Ax =0 are ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −2 −3 −4 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ , ⎢ 1 ⎥ , ⎢ 0 ⎥ . ⎣ 0 ⎦ ⎣ 0 ⎦ ⎣ 1 ⎦ 0 0 0

(b) and (c) We need to prove that these three vectors are linearly independent and they span the nullspace. By considering the second, third and fourth coordinates, a combination of the vectors adding to zero must have zero coeﬃcients. The vectors span the nullspace, since the dimension of the nullspace is three (note that the rank of the matrix A is 2). 3. (a) If Ax = b has no solution, the column space of A must have dimension less than m.The rank is r

1 the scalar products a1 · (b − p),...,an · (b − p) all equal zero. The second condition we check by considering the (n +1)× m matrix whose ﬁrst n rows are the coordinates of the ai’s and whose last row consists of the coordinates of p. The vector p is in the span of the ai’s if and only if the last row becomes zero in elimination. 4. (a) To compute the determinant, subtract the second row from all the other rows: ⎡ ⎤ 0100 ⎡ ⎤ 111 ⎢ 1111⎥ det B =det⎢ ⎥ = − det ⎣ 010⎦ = −1 . ⎣ 0010⎦ 001 0001

(b) λ =1, 1, 1, 1, 6. Since A − I has all equal rows, it has rank one. It follows that it has four zero eigenvalues. The eigenvalues of A are the eigenvalues of A − I increased by one, so A has the eigenvalue 1 with multiplicity four. The trace of A equals 10 so 10 − 4=6is the other eigenvalue. (c) A is symmetric, and thus so is A−1. The cofactor formula gives:

−1 1+3 det B (A )13 =(−1) , det A and det A = 6 since it equals the product of the eigenvalues of A. We conclude that the (1, 3) and the (3, 1) entries of A−1 are both equal to −1/6.

5. (a) The answer is 26 A = . −17 Reason: 26 3 12 3 = = λ1 . ab 1 3a + b 1

We deduce that λ1 =4and3a + b = 4. Similarly, since x2 is an eigenvector we have 26 2 10 2 = = λ2 . ab 1 2a + b 1

We deduce that λ2 = 5 and therefore that 2a+b = 5. We conclude that a = −1andb =7. −1 (b) B = SΛS , where the columns of S are the vectors x1 and x2, and Λ is the diagonal matrix with entries 1 and 0: 32 10 1 −2 3 −6 B = = . 11 00 −13 1 −2 Then Λ10 = Λ and therefore B10 = SΛ10S−1 = SΛS−1 = B. 6. (a) We would solve the equations

c0 + c1 + c2 + c3 = y1 c0 + ic1 − c2 − ic3 = y2 c0 − c1 + c2 − c3 = y3 c0 − ic1 − c2 + ic3 = y4 .

2 and the matrix of coeﬃcients is ⎡ ⎤ ⎡ ⎤ 1111 1111 ⎢ 1 i −1 −i ⎥ ⎢ 1 ii2 i3 ⎥ F = ⎢ ⎥ = ⎢ ⎥ . ⎣ 1 −11−1 ⎦ ⎣ 12 i2 i4 i6 ⎦ 1 −i −1 i 13 i3 i6 i9

(b) F has orthogonal columns and it is symmetric. It is also a Vandermonde matrix: each column consists of the ﬁrst four powers of a number (starting from the zero-th power). (c) Since the columns of F are orthogonal and non-zero, the matrix is invertible. Its inverse is F/4. The determinant of this Vandermonde matrix is equal to the product of the diﬀerences of 1,i,i2,i3: det F =(i − 1)(−1 − 1)(−1 − i)(−i − 1)(−i − i)(−i +1)=−16i.

7. (a) The seven eigenvalues of P are 1, 1, 1, 1, 0, 0, 0. (b) The eigenvectors with eigenvalue 1 are the non-zero vectors in S. The eigenvectors with eigenvalue 0 are the non-zero vectors in the orthogonal complement of S. (c) The solution u(t) to the diﬀerential equation has the form

−t u(t)=v1e + v2 ,

where v1 is in S and v2 is in the orthogonal complement of S.Thenu(∞)=v2,whichisthe projection of u(0) onto the orthogonal complement of S.

8. (a) The required matrix is ⎡ ⎤ 1000 ⎢ 0010⎥ P = ⎢ ⎥ . ⎣ 0100⎦ 0001

(b) Since B is block diagonal, its eigenvalues are the eigenvalues of the diagonal blocks. In our case, the two blocks are the same and the eigenvalues of each block are 3 and 1. Thus the eigenvalues of B are 3, 3, 1, 1. (c) Since P is a permutation matrix, it is orthogonal and therefore P T = P −1. The matrix B is thus similar to the matrix A and we conclude that A and B havethesameeigenvalues. The function of u, v, w, z which is positive except if u = v = w = z =0isthus ⎡ ⎤ u ⎢ v ⎥ uvwzA ⎢ ⎥ u2 v2 w2 z2 − uw − vz . ⎣ w ⎦ =2 + + + z

9. (a) The vectors orthogonal to the nullspace of A are the rows of A. Since we know that the matrix A is singular and it is clearly not rank one, it follows that the rank of A is two. The

3 ﬁrst two rows are independent and therefore the orthogonal complement of the nullspace of A is spanned by the two vectors ⎡ ⎤ ⎡ ⎤ 1 2 ⎣ 3 ⎦ and ⎣ 2 ⎦ . 7 6

(b) We get the vectors ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 2 0 1 1 ⎣ 2 ⎦ , √ ⎣ 0 ⎦ and ⎣ 0 ⎦ . 3 2 5 −1 0

(c) The “reduced” LU decomposition, from ignoring the zero row in U,is ⎡ ⎤ 10 137 Answer A = ⎣ 21⎦ . 5 0 −4 −8 2 4 Here are the details: Starting elimination we ﬁnd ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 100 137 13 7 ⎣ −210⎦ ⎣ 226⎦ = ⎣ 0 −4 −8 ⎦ , (1) −201 214 0 −5 −10 and proceeding further we ﬁnd ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 100 13 7 137 ⎣ 010⎦ ⎣ 0 −4 −8 ⎦ = ⎣ 0 −4 −8 ⎦ . − 5 − − 0 4 1 0 5 10 000

Collecting all the information together we obtain ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 137 100 100 137 ⎣ 226⎦ = ⎣ 210⎦ ⎣ 010⎦ ⎣ 0 −4 −8 ⎦ , 5 214 201 0 4 1 000 and multiplying the ﬁrst two matrices on the right-hand side we deduce that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 137 100 137 ⎣ 226⎦ = ⎣ 210⎦ ⎣ 0 −4 −8 ⎦ . 5 214 2 4 1 000

Since the last row of the last matrix is all zero, we conclude that ⎡ ⎤ ⎡ ⎤ 137 10 137 ⎣ 226⎦ = ⎣ 21⎦ . 5 0 −4 −8 214 2 4

4 This is the “reduced” LU factorization of A. Multiplying columns of L by rows of U,thisis ⎡ ⎤ ⎡ ⎤ 1 0 A = ⎣ 0 ⎦ 137+ ⎣ 1 ⎦ 0 −4 −8 . 5 0 4

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Your PRINTED name is: SOLUTIONS

Grading 1 Please circle your recitation: 2 3 4 1) T10 2-131 K.Meszaros 2-333 3-7826 karola 5 2) T10 2-132 A.Barakat 2-172 3-4470 barakat 6 3) T11 2-132 A.Barakat 2-172 3-4470 barakat 7 4) T11 2-131 A.Osorno 2-229 3-1589 aosorno 8 5) T12 2-132 A.Edelman 2-343 3-7770 edelman 9 6) T12 2-131 K.Meszaros 2-333 3-7826 karola 7) T1 2-132 A.Edelman 2-343 3-7770 edelman 8) T2 2-132 J.Burns 2-333 3-7826 burns 9) T3 2-132 A.Osorno 2-229 3-1589 aosorno 1 (4+7=11 pts.) Suppose A is 3 by 4, and Ax =0 has exactly 2 special solutions:      1   −2           1   −1  x1 =   and x2 =        1   0  0 1

(a) Remembering that A is 3 by 4, ﬁnd its row reduced echelon form R.

(b) Find the dimensions of all four fundamental subspaces C(A), N(A), C(AT), N(AT).

You have enough information to ﬁnd bases for one or more of these subspaces—ﬁnd those bases.

2 Solution.

(a) Each special solution tells us the solution to Rx =0 when we set one free variable =1 and the others =0. Here, the third and fourth variables must be the two free variables,  1 0 ∗ ∗      and the other two are the pivots: R =  0 1 ∗ ∗  0000    1 0 −1 2      Now multiply out Rx1 =0 and Rx2 =0 to ﬁnd the ∗’s: R =  0 1 −1 1  00 00 (The ∗’s are just the negatives of the special solutions’ pivot entries.)

(b) We know the nullspace N(A) has n − r =4 − 2=2 dimensions: the special solutions

x1, x2 form a basis.

The row space C(AT) has r = 2 dimensions. It’s orthogonal to N(A), so just pick

two linearly-independent    vectors orthogonal to x1 and x2 to form a basis: for example, 1 0          0   1      x3 =   , x4 =  .  1   1   −   −  2 1 (Or: C(AT) = C(RT) is just the row space of R, so the ﬁrst two rows are a basis. Same thing!)

The column space C(A) has r =2 dimensions (same as C(AT)). We can’t write down a basis because we don’t know what A is, but we can say that the ﬁrst two columns of A are a basis.

The left nullspace N(AT) has m − r = 1 dimension; it’s orthogonal to C(A), so any vector orthogonal to the ﬁrst two columns of A (whatever they are) will be a basis.

3 2 (6+3+2=11 pts.) (a) Find the inverse of a 3 by 3 upper triangular matrix U, with nonzero entries a, b, c, d, e, f. You could use cofactors and the formula

for the inverse. Or possibly Gauss-Jordan elimination.  a b c      Find the inverse of U =  0 d e . 0 0 f (b) Suppose the columns of U are eigenvectors of a matrix A. Show that A is also upper triangular.

4 Solution.

(a) By elimination: (We keep track of the elimination matrix E on one side, and the product EU on the other. When EU = I, then E = U −1.)      a b c 1 0 0   1 b/a c/a 1/a 0 0           0 d e 0 1 0   0 1 e/d 0 1/d 0  0 0 f 0 0 1 001 001/f    100 1/a −b/ad (be − cd)/adf  h i     −1  010 0 1/d −e/df  = I U 001 0 0 1/f

By cofactors: (Take the minor, then “checkerboard” the signs to get the cofactor matrix, then transpose and divide by det(U)= adf.)          a b c   df 0 0   df 0 0   df −bf be − cd                   0 d e   bf af 0   −bf af 0   0 af −ae   0 0 f be − cd ae ad be − cd −ae ad 0 0 ad  1/a −b/ad (be − cd)/adf      −1  0 1/d −e/df  = U 00 1/f

(b) We have a complete set of eigenvectors for A, so we can diagonalize: A = UΛU −1. We know U is upper-triangular, and so is the diagonal matrix Λ, and we’ve just shown that U −1 is upper-triangular too. So their product A is also upper-triangular.

T (c) The columns aren’t orthogonal! (For example, the product u1 u2 of the ﬁrst two columns is ab + 0d + 0 · 0 = ab, which is nonzero because we’re assuming all the entries are nonzero.)

5 3 (3+3+5=11 pts.) (a) A and B are any matrices with the same number of rows. What can you say (and explain why it is true) about the comparison of h i rank of A rank of the block matrix A B

(b) Suppose B = A2. How do those ranks compare ? Explain your reasoning.

Solution.

(a) All you can say is that rank A ≤ rank [A B]. (A can have any number r of pivot columns, and these will all be pivot columns for [A B]; but there could be more pivot columns among the columns of B.)

(b) Now rank A = rank [A A2]. (Every column of A2 is a linear combination of columns 2 of A. For instance, if we call A’s ﬁrst column a1, then Aa1 is the ﬁrst column of A . So there are no new pivot columns in the A2 part of [A A2].)

(c) The nullspace N(A) has dimension n − r, as always. Since [A A] only has r pivot columns — the n columns we added are all duplicates — [A A] is an m-by-2n matrix of rank r, and its nullspace N ([A A]) has dimension 2n − r.

6 4 (3+4+5=12 pts.) Suppose A is a 5 by 3 matrix and Ax is never zero (except when x is the zero vector).

(a) What can you say about the columns of A ?

(b) Show that ATAx is also never zero (except when x =0) by explaining this key step:

If ATAx =0 then obviously xTATAx =0 and then (WHY ?) Ax =0.

(c) We now know that ATA is invertible. Explain why B = (ATA)−1AT is a one-sided inverse of A (which side of A ?). B is NOT a 2-sided inverse of A (explain why not).

Solution.

(a) N(A)=0 so A has full column rank r = n =3: the columns are linearly independent.

(b) xTATAx = (Ax)T Ax is the squared length of Ax. The only way it can be zero is if Ax has zero length (meaning Ax =0).

(c) B is a left inverse of A, since BA =(ATA)−1ATA = I is the (3-by-3) identity matrix. B is not a right inverse of A, because AB is a 5-by-5 matrix but can only have rank 3. (In fact, BA = A(ATA)−1AT is the projection onto the (3-dimensional) column space of A.)

7 5 (5+5=10 pts.) If A is 3 by 3 symmetric positive deﬁnite, then Aqi = λiqi with

positive eigenvalues and orthonormal eigenvectors qi.

Suppose x = c1q1 + c2q2 + c3q3.

(a) Compute xTx and also xTAx in terms of the c’s and λ’s.

(b) Looking at the ratio of xTAx in part (a) divided by xTx in part (a), what c’s would make that ratio as large as possible ? You can assume T T λ1 < λ2 <...<λn. Conclusion: the ratio x Ax/x x is a maximum when x is .

Solution.

(a)

T T T T x x = (c1q1 + c2q2 + c3q3 )(c1q1 + c2q2 + c3q3)

2 T T T 2 T = c1q1 q1 + c1c2q1 q2 + ··· + c3c2q3 q2 + c3q3 q3

2 2 2 = c1 + c2 + c3.

T T T T x Ax = (c1q1 + c2q2 + c3q3 )(c1Aq1 + c2Aq2 + c3Aq3)

T T T = (c1q1 + c2q2 + c3q3 )(c1λ1q1 + c2λ2q2 + c3λ3q3)

2 T T T 2 T = c1λ1q1 q1 + c1c2λ2q1 q2 + ··· + c3c2λ2q3 q2 + c3λ3q3 q3

2 2 2 = c1λ1 + c2λ2 + c3λ3.

2 2 2 2 2 2 (b) We maximize (c1λ1 + c2λ2 + c3λ3)/(c1 + c2 + c3) when c1 = c2 = 0, so x = c3q3 is a

multiple of the eigenvector q3 with the largest eigenvalue λ3.

(Also notice that the maximum value of this “Rayleigh quotient” xTAx/xTx is the largest eigenvalue itself. This is another way of ﬁnding eigenvectors: maximize xTAx/xTx numerically.)

8 6 (4+4+4=12 pts.) (a) Find a linear combination w of the linearly independent vectors v and u that is perpendicular to u. h i (b) For the 2-column matrix A = u v , ﬁnd Q (orthonormal columns) and R (2 by 2 upper triangular) so that A = QR.

Solution.

(a) You could just write down w =0u +0v =0 — that’s perpendicular to everything! But a more useful choice is to subtract oﬀ just enough u so that w = v −cu is perpendicular to u. That means 0= wTu = vTu − cuTu, so c =(vTu)/(uTu) and

vTu w = v − ( )u. uTu

(b) We already know u and w are orthogonal; just normalize them! Take q1 = u/|u| and  |u|  q2 = w/|w|. Then solve for the columns r1, r2 of R: Qr1 = u so r1 = , and   0

 c|u|  T T Qr2 = v so r2 = . (Where c =(v u)/(u u) as before.) |w|

Then Q = [q1 q2] and R = [r1 r2].

9 7 (4+3+4=11 pts.) (a) Find the eigenvalues of      0001   0010           1000  2  0001  C =   and C =   .      0100   1000  0010 0100

(b) Those are both permutation matrices. What are their inverses C−1 and (C2)−1 ?

10 Solution.

(a) Take the determinant of C − λI (I expanded by cofactors): λ4 − 1=0. The roots of this “characteristic equation” are the eigenvalues: +1, −1, i, −i.

The eigenvalues of C2 are just λ2 = ±1 (two of each).

(Here’s a “guessing” approach. Since C4 = I, all the eigenvalues λ4 of C4 are 1: so λ =1, −1, i, −i are the only possibilities. Just check to see which ones work. Then the eigenvalues of C2 must be ±1.)

(b) For any permutation matrix, C−1 = CT: so    0100      −1  0010  C =      0001  1000

and (C2)−1 = C2 is itself.

Add 1 to every eigenvalue to get the eigenvalues of C + I (if C = SΛS−1, then C + I = S(Λ + I)S−1): 2(0)(1 + i)(1 − i)=0.

(Or let λ = −1 in the characteristic equation det(C − λI).)

Add 2 to get the eigenvalues of C +2I (or let λ = −2): 3(1)(2 + i)(2 − i)=15.

11 8 (4+3+4=11 pts.) Suppose a rectangular matrix A has independent columns.

(a) How do you ﬁnd the best least squares solution xb to Ax = b ? By taking those steps, give me a formula (letters not numbers) for xb and also for p = Axb.

(b) The projection p is in which fundamental subspace associated with A ? The error vector e = b − p is in which fundamental subspace ?

(c) Find by any method the projection matrix P onto the column space of A:    1 0       3 0  A =   .    0 −1  0 −3

Solution.

(a)

Ax = b

Least-squares “solution”: ATAxˆ = ATb

ATA is invertible: xˆ = (ATA)−1ATb

and p = Axˆ is: Axˆ = A(ATA)−1ATb

(b) p = Axˆ is a linear combination of columns of A, so it’s in the column space C(A). The error e = b − p is orthogonal to this space, so it’s in the left nullspace N(AT).    

T −1 T T  10 0   1/10 0  1 (c) I used P = A(A A) A . Since A A = , its inverse is = 10 I, 0 10 0 1/10 and    1300      1  3900  P =   10    0013  0039

12 9 (3+4+4=11 pts.) This question is about the matrices with 3’s on the main diagonal, 2’s on the diagonal above, 1’s on the diagonal below.        3200  3 2 0   h i      3 2     1320  A1 = 3 A2 = A3 =  1 3 2  An =      ·  1 3  0 1 3  0 1 3 0 0 · ·

(a) What are the determinants of A2 and A3 ?

(b) The determinant of An is Dn. Use cofactors of row 1 and column 1 to

ﬁnd the numbers a and b in the recursive formula for Dn:

(∗) Dn = a Dn−1 + b Dn−2 .

Dn a b Dn−1   =     . Dn−1 1 0 Dn−2

>From the eigenvalues of that matrix, how fast do the determinants

Dn grow ? (If you didn’t ﬁnd a and b, say how you would answer part

13 Solution.

(a) det(A2)=3 · 3 − 1 · 2=7 and det(A3) = 3det(A2) − 2 · 1 · 3=15.

(b) Dn =3Dn−1 +(−2)Dn−2. (Show your work.)

(c) The trace of that matrix A is a =3, and the determinant is −b =2. So the characteristic equation of A is λ2 − aλ − b =0, which has roots (the eigenvalues of A) p a ± a2 − 4(−b) 3 ± 1 λ± = = =1 or 2. 2 2

n n n Dn grows at the same rate as the largest eigenvalue of A , λ+ =2 .

The ﬁnal point: D5 =3D4 +2D3 = 3(3D3 +2D2)+2D3 = 11D3 +6D2 = 207.

14 18.06 FINAL EXAM May 24, 2007

Your PRINTED name is: SOLUTIONS

Please circle your recitation: Grading

(1) M 2 2-131 A. Osorno 1 (2) M 3 2-131 A. Osorno (3) M 3 2-132 A. Pissarra Pires 2 (4) T 11 2-132 K. Meszaros (5) T 12 2-132 K. Meszaros 3 (6) T 1 2-132 Jerin Gu (7) T 2 2-132 Jerin Gu 4

Total: Problem 1 (10 points)

  3 2 1 1   Let A = 6 6 3 3.   3 4 2 2

(a) Calculate the dimensions of the 4 fundamental subspaces associated with A.

(b) Give a basis for each of the 4 fundamental subspaces.   1   (c) Find the complete solution of the system A x = 3.   2

Solution 1

(a) The rank is 2, so the dimensions are: C(A) Ã r = 2 C(AT ) Ã r = 2 N(A) Ã n − r = 4 − 2 = 2 N(AT ) Ã m − r = 3 − 2 = 1.

(b) We can get these by elimination or by inspection: ³ ´T ³ ´T C(A) Ã 3 6 3 , 2 6 4 ³ ´T ³ ´T C(AT ) Ã 3 2 1 1 , 6 3 3 1 ³ ´T ³ ´T N(A) Ã 0 −1/2 1 0 , 0 −1/2 0 1 ³ ´T N(AT ) Ã 1 −1 1

      0 0 0             1/2 −1/2 −1/2 (c) x = x + x =   + c   + c  . particular nullspace   1   2    0   1   0  0 0 1 Problem 2 (10 points)

Consider the system of linear equations:   x + y + z = 1  2x + z = 2    −x + y + az = b In parts (a)(c) below circle correct answers. Explain your answers. (a) For a = 1, b = −1, the system has: (1) exactly one solution (2) innitely many solutions (3) no solutions

(b) For a = 0, b = 1, the system has: (1) exactly one solution (2) innitely many solutions (3) no solutions

(d) Solve the system for a = b = 1.

Solution 2   1 1 1 1     If we eliminate the augmented matrix we get 0 −2 −1 0 .   0 0 a b + 1 (a) Exactly one solution: The matrix is invertible. (b) No solutions: Get a row of zeroes in the matrix with no zero in the augmented column. (c) Innitely many solutions: Get a row of zeroes with a zero in the augmented column. ³ ´T (d) Using back substitution we get x = 0 −1 2 . Problem 3 (10 points)

Let L be the line in R3 spanned by the vector (1, 1, 1)T . Let P be the projection matrix for the projection onto the line L. (a) What are the eigenvalues of the matrix P ? (Indicate their multiplicities.)

(b) Find an orthonormal basis of the orthogonal complement L⊥ to the line L.

(d) Calculate the projection of the vector (1, 2, 3)T onto the orthogonal complement L⊥.

Solution 3

(a) P is a projection matrix onto a subspace of dimension 1, so the eigenvalues are 1, 0, 0.  √   √   1/ 2   1/ 6     √  (b)  0  , −2/ 6.  √   √  −1 2 1/ 6   2     (c) p = 2.   2   −1   (d) The projection onto L⊥ is b − p =  0 .   1 Problem 4 (10 points)   1 2 3     Let A = 2 2 2.   3 2 1 In parts (a)(c) below circle correct answers. Explain your answers. (a) The matrix A is singular: True False (b) The matrix A + 2I is singular: True False (c) The matrix A is positive denite: True False (d) Find all eigenvalues of A and the corresponding eigenvectors. (e) Find an orthogonal matrix Q and a diagonal matrix Λ such that A = QΛQT . du(t) (f) Solve the system of dierential equations = A u(t), u(0) = (1, 0, 0)T . dt

Solution 4

³ ´T (a) True ( 1 −2 1 is in the nullspace). ³ ´T (b) True ( 1 0 −1 is in the nullspace). (c) False The matrix is singular so has 0 as eigenvalue. ³ ´T (d) A is singular, so 0 is an eigenvalue with eigenvector 1 −2 1 . ³ ´T A + 2I is singular, so -2 is an eigenvalue with eigenvector 1 0 −1 . ³ ´T We can get the last eigenvalue by looking a the trace: 6. The eigenvector is 1 1 1 . (e)  √ √ √     √ √ √  1/ 6 1/ 62 1/ 3 0 0 0 1/ 6 −2/ 6 1/ 6        √ √     √ √  A = −2/ 6 0 1/ 3 0 −2 0 1/ 2 0 −1/ 2 .  √ √ √     √ √ √  1/ 6 −1/ 2 1/ 3 0 0 6 1/ 3 1/ 3 1/ 3 | {z } | {z } | {z } Q Λ QT        1   1  1       (f) u(t) = eAtu(0) = QeΛtQT u(0) = 1 −2 + 1 e−2t  0  + 1 e6t 1. 6   2   3   1 −1 1 Problem 5 (10 points)   1 1 1 1     Let A = 2 2 2 2.   3 3 3 3 (a) What is the rank of A? (b) Calculate the matrix AT A. Find all its eigenvalues (with multiplicities). (c) Calculate the matrix AAT . Find all its eigenvalues (with multiplicities). (d) Find the matrix Σ in the singular value decomposition A = UΣV T .

Solution 5

(a) 1   14 14 14 14     14 14 14 14 (b) AT A =  . It was rank 1 so the eigenvalues are 56, 0, 0, 0.   14 14 14 14 14 14 14 14    4 8 12   (c) AAT =  8 16 24. It was rank 1 so the eigenvalues are 56, 0, 0.   12 24 36 √  56 0 0 0     (d) Σ =  0 0 0 0.   0 0 0 0 Problem 6 (10 points)

Let An be the tridiagonal n×n-matrix with 2's on the main diagonal, 1's immediately above the main diagonal, 3's immediately below the main diagonal, and 0's everywhere else:   2 1 0 0 ··· 0    ..  3 2 1 0 . 0    ..  0 3 2 1 . 0   An =  ..  , 0 0 3 2 . 0   ......  ......    0 0 0 0 ··· 2

(a) Express the determinant det(An) in terms of det(An−1) and det(An−2).

(b) Explicitly calculate det(An), for n = 1,..., 6.

Solution 6

(a) Using cofactors twice we get det(An) = 2 det(An−1) − 3 det(An−2).

(b) . det(A1) = det[2] = 2 2 1 det(A2) = det   = 1. 3 2 det(A3) = 2 · 1 − 3 · 2 = −4. det(A4) = 2 · (−4) − 3 · 1 = −11. det(A5) = 2 · (−11) − 3 · (−4) = −10. det(A6) = 2 · (−10) − 3 · (−11) = 13. Problem 7 (10 points)

Calculate the determinant of the following 6 × 6-matrix:   1 1 1 0 1 1     1 1 1 1 0 1     1 1 1 1 1 0 A =     0 1 1 1 1 1     1 0 1 1 1 1 1 1 0 1 1 1

Solution 7

This determinant could be computed using cofactors or doing row operations to simplify and then cofactors. it could also be computed as follows.     0 0 0 1 0 0 0 1 1 1 1 1         0 0 0 0 1 0 1 0 1 1 1 1         0 0 0 0 0 1 1 1 0 1 1 1 A =    .     1 0 0 0 0 0 1 1 1 0 1 1         0 1 0 0 0 0 1 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 1 0 | {z } | {z } P B is a permutation matrix with determinant -1. P   1 1 1 1 1 1     1 1 1 1 1 1     1 1 1 1 1 1 B =   − I.   1 1 1 1 1 1     1 1 1 1 1 1 1 1 1 1 1 1 The eigenvalues of the matrix with all 1's are 6, 0, 0, 0, 0, 0, 0 so the eigenvalues of B are 5, -1, -1, -1, -1, -1, so the determinant of B is -5. Thus the determinant of A is 5. Problem 8 (10 points)   3 4 (a) Calculate A100 for A =  . 4 −3   1 −1 (b) Calculate B100 for B =  . 1 1 (c) What will happen with the house (shown below) when we apply the linear transformation   1 −1 T (v) = Bv for B =  ? Will it be dilated? Draw the picture of the transformed 1 1 house.

Solution 8     25 0 5100 0 (a) A2 =   so A100 =  . 0 25 0 5100       −4 0 (−4)25 0 −425 0 (b) B4 =   so B100 =   =  . 0 −4 0 (−4)25 0 −425

(c) From part (b) we see that B4 is stretching by 4 and rotating by π. Thus B is rotating √ by π/4 and stretching by 2. Practice 18.06 Final Questions with Solutions

17th December 2007

Notes on the practice questions

The final exam will be on Thursday, Dec. 20, from 9am to 12noon at the Johnson Track, and will most likely consist of 8–12 questions. The practice problems below mostly concentrate on the material from exams 1–2, since you already have practice problems for exam 3. The real final will have plenty of eigenproblems! These questions are intended to give you a flavor of how I want you to be able to think about the material, and the flavor of possible questions I might ask. Obviously, these questions do not exhaust all the material that we covered this term, so you should of course still study your lecture notes and previous exams, and review your homework. Solutions for these practice problems should be posted on the 18.06 web site by 12/15.

List of potential topics:

Material from exams 1, 2, and 3, and the problem sets (and lectures) up to that point. Definitely not on final: finite-difference approximations, sparse matrices and iterative methods, non-diagonalizable matrices, generalized eigenvectors, principal components analysis, choosing a basis to convert a linear operator into a matrix, numerical linear algebra and error analysis. Key ideas:

• The four subspaces of a matrix A and their relationships to one another and the solutions of Ax = b. • Gaussian elimination A → U → R and backsubstitution. Elimination = invertible row operations = multiplying A on the left by an invertible matrix. Multiplying on left by an invertible matrix preserves N(A), and hence we can use elimination to ﬁnd the nullspace. Also, it thus preserves C(AH ) = N(A)⊥. Also, it thus preserves the linear independence of the columns and hence the pivot columns in A are a basis for C(A). Conversely, invertible column operations = multiplying A on the right by an invertible matrix, hence preserving N(AH ) and C(A). • Solution of Ax = b when A is not invertible: exactly solvable if b ∈ C(A), particular solutions, not unique if N(A) 6= {0}. Least-squares solution AH Axˆ = AH b if A full column 2 rank, equivalence to minimizing kAx − bk , relationship Axˆ = PAb to projection matrix H −1 H H H PA = A(A A) A onto C(A). The fact that rank(A A) = rank A = rank A , hence AH Axˆ = AH b is always solvable for any A, and AH A is always invertible (and positive- deﬁnite) if A has full column rank. • Vector spaces and subspaces. Dot products, transposes/adjoints, orthogonal complements. Linear independence, bases, and orthonormal bases. Gram-Schmidt and QR factorization.

1 Unitary matrices QH = Q−1, which preserve dot products x · y = (Qx) · (Qy) and lengths kxk = kQxk. • Determinants of square matrices: properties, invariance under elimination (row swaps ﬂip sign), = product of eigenvalues. The trace of a matrix = sum of eigenvalues. det AB = (det A)(det B), trace AB = trace BA, trace(A + B) = trace A + trace B. • For an eigenvector, any complicated matrix or operator acts just like a number λ, and we can do anything we want (inversion, powers, exponentials...) using that number. To act on an arbitrary vector, we expand that vector in the eigenvectors (in the usual case where the eigenvectors form a basis), and then treat each eigenvector individually. Finding eigenvectors and eigenvalues is complicated, though, so we try to infer as much as we can about their properties from the structure of the matrix/operator (Hermitian, Markov, etcetera). SVD as generalization of eigenvectors/eigenvalues.

Problem 1

Suppose that A is some real matrix with full column rank, and we do√ Gram-Schmidt on it to√ get T T an the following orthonormal√ basis for C(A): q1 = (1, 0, 1, 0, −1) / 3, q2 = (1, 2, −1, 0, 0) / 6, q = (−2, 1, 0, 0, 2)T / 9. 3 √ √ √ (a) Suppose we form the matrix B whose columns are 3q1, 6q2, and 9q3. Which of the four subspaces, if any, are guaranteed to be the same for A and B? (b) Find a basis for the left nullspace N(AT ). (c) Find a basis for the row space C(AT ). Solution: (a) All four subspaces for A and B are the same. Obviously B has the same column space as A, i.e. C(A) = C(B). It follows that they have the same left nullspace, N(AT ) = N(BT ), which are just the orthogonal complement of column space. Since both A and B has full column rank, N(A) = N(B) = {0}. It follows that the row spaces C(AT ) = C(BT ).  1 1 −2  0 2 1    (b) As we have observed above, N(AT ) = N(BT ), where B =  1 −1 0 . To ﬁnd a    0 0 0  −1 0 2 basis for BT , we do Gauss elimination:

 1 0 1 0 −1 1 0 1 0 −1 1 0 1 0 −1   1 2 −1 0 0  0 2 −2 0 1  0 2 −2 0 1  −2 1 0 0 2 0 1 2 0 0 0 0 3 0 −1/2 1 0 0 0 −5/6 0 2 0 0 2/3  , 0 0 3 0 −1/2 So we may take two vectors (0, 0, 0, 1, 0)T and (5, −2, 1, 0, 6)T as a basis of N(AT ). (c) A is a 5 × 3 matrix with full column rank, i.e. rank(A) = 3. So the row space have dimension dim(C(AT )) = 3. However, C(AT ) lies in R3. So C(AT ) = R3. We may take the vectors (1, 0, 0)T , (0, 1, 0)T , (0, 0, 1)T as a basis.

2 Problem 2

Perverse physicist Pat proposes a permutation: Pat permutes the columns of some matrix A by some random sequence of column swaps, resulting in a new matrix B. (a) If you were given B (only), which of the four subspaces of A (if any) could you find? (i.e. which subspaces are preserved by column swaps?)  1 −1 2  (b) Suppose B =  2 1 3  and b = (0, 1, 2)T . Check whether Ax = b is solvable, −1 5 0 and if so whether the solution x is unique. (c) Suppose you were given B and it had full column rank. You are also given b. Give an explicit formula (in terms of B and b only) for the minimum possible value of kAx − bk2. Solution: (a) Only column swaps are performed, so the column space is not changed. As a consequence, the left nullspace is also not changed. So given B, we can find C(A) and N(AT ). (b) Since C(A) = C(B), the solvability of Ax = b is equivalent to the solvability of Bx = b. We do Gauss elimination for the second equation:  1 −1 2 0 1 −1 2 0 1 −1 2 0   2 1 3 1 0 3 −1 1 0 3 −1 1  . −1 5 0 2 0 4 2 2 0 0 10/3 2/3 So the equation Bx = b, and thus the equation Ax = b, is solvable. Moreover, the solution x is unique. (c) All the vectors Ax are exactly the vectors in the column space C(A) = C(B). Thus the minimum possible value of kAx − bk2 is exactly the minimum possible value of kBx − bk2. To find the latter one, we need the vector Bx to be the vector in C(B) which is closest to the vector b. In other words, Bx should be the projection of b on C(B). Since B is of full column rank, the projection is Bx = B(BT B)−1BT b. So the minimum value we want is kB(BT B)−1BT b − bk2

Problem 3

Which of the following sets are vector spaces (under ordinary addition and multiplication by real numbers)? (a) Given a vector x ∈ Rn, the set of all vectors y ∈ Rn with x · y = 3. R ∞ (b) The set of all functions f(x) whose integral −∞ f(x)dx is zero. (c) Given a subspace V ⊆ Rn and an m × n matrix A, the set of all vectors Ax for all x ∈ V . (d) Given a line L ⊆ Rn and an m × n matrix A, the set of all vectors Ax for all x ∈ L. (e) The set of n × n Markov matrices. 1 (f) The set of eigenvectors with |λ| < 2 of an n × n Markov matrix A. Solution: (a) This is not a vector space, since it doesn’t contain the zero vector. (b) This is a vector space. It is a subset of the vector space consisting all functions. If f and g both has whole integral zero, af + bg also has whole integral zero. So it is a vector subspace. (c) This is a vector space. It is just the column space of A, which is a vector subspace of Rm.

3 (d) This is not a vector space. A line in Rn doesn’t have to pass the origin, so the set of all vectors Ax need not contain the zero vector. (e) This is not a vector space. The zero matrix is not a Markov matrix. (f) This is not a vector space in general. If the Markov matrix A has more than one eigenvalue whose absolute value are less than 1/2, then the corresponding set of eigenvectors is not a vector space, since the summation of two eigenvectors corresponding to diﬀerent eigenvalues is not a eigenvector.

Problem 4

The rows of an m × n matrix A are linearly independent. (a) Is Ax = b necessarily solvable? (b) If Ax = b is solvable, is the solution necessarily unique? (c) What are N(AH ) and C(A)? Solution: (a) Yes. Since the rows of A are linearly independent, rank(A) = m. So the column space of A is an m-dimensional subspace of Rm, i.e., is Rm itself. It follows that for any b, the equation Ax = b is always solvable. (b) No, the solution maybe not unique. Since rank(A) = m, the nullspace is n−m dimensional. Thus the solution is not unique if n > m, and is unique if n = m. (It will never have that n < m, otherwise the rows are not linearly independent.) (c) We have seen in part (a) that C(A) = Rm. So N(AT ) = {0}. Since N(AH ) consists those points whose conjugate lies in N(AT ), we see N(AH ) = {0}.

Problem 5

Make up your own problem: give an example of a matrix A and a vector b such that the solutions of Ax = b form a line in R3, b 6= 0, and all the entries of the matrix A are nonzero. Find all solutions x. Solution: Such a matrix must be an m × 3 matrix whose nullspace is one dimensional. In other words, the rank is 3 − 1 = 2. We may take A to be an 2 × 3 matrix whose rows are linearly independent. As an example, we take 1 1 1 1 A = , b = . 1 1 2 1

To ﬁnd all solutions, we do elimination

1 1 1 1 1 1 1 1 1 1 0 1 , 1 1 2 1 0 0 1 0 0 0 1 0 so the solutions are given by x = (1 − t, t, 0)T .

4 Problem 6

Clever Chris the chemist concocts a conundrum: in ordinary least-squares, you minimize kAx−bk2 by solving AH Axˆ = AH b, but suppose instead that we wanted to minimize (Ax − b)H C(Ax − b) for some Hermitian positive-definite matrix C? (a) Suppose C = BH B for some matrix B. Which of the following (if any) must be properties of B: (i) Hermitian, (ii) Markov, (iii) unitary, (iv) full column rank, (v) full row rank? (b) In terms of B, A, and b, write down an explicit formula for the x that minimizes (Ax − b)H C(Ax − b). (c) Rewrite your answer from (b) purely in terms of C, A, and b. (d) Suppose that C was only positive semi-definite. Is there still a minimum value of (Ax − b)H C(Ax − b)? Still a unique solution x? Solution: (a) For C = BH B a Hermitian positive-definite matrix, B must be (iv) full column rank. To show this, we only need to notice that x · Cx = x · BH Bx = kBxk2. Since C is positive definite, B has no nonzero nullspace. It follows that B is of full column rank. B don’t have to be Hermitian or Markov or unitary or full row rank. For example, we may 1 0 1 0 take B = 0 2 . Then C = BH B = .   0 4 0 0 (b) We have

(Ax − b)H C(Ax − b) = (Ax − b)H BH B(Ax − b) = (BAx − Bb)H (BAx − Bb) = kBAx − Bbk2.

To minimize this, we need to solve the equation (BA)H (BA)ˆx = (BA)H Bb. Since B and A are of full column rank, BA is of full column rank. So (BA)H (BA) is invertible. The solution is given byx ˆ = (AH BH BA)−1AH BH V b. (c) Since BH B = C, we havex ˆ = (AH CA)−1AH Cb. (d) If C is only positive semi-deﬁnite, we still have C = BH B for some B, but B need not to be of full column rank. As above, we have

(Ax − b)H C(Ax − b) = kBAx − Bbk2.

So the minimum value of (Ax − b)H C(Ax − b) still exists. However, since B may be not of full column rank, the matrix need not be invertible. The solution x need not be unique. For example, if we take C to be the zero matrix, any x will minimize (Ax − b)H C(Ax − b).

Problem 7

True or false (explain why if true, give a counter-example if false). (a) For n×n real-symmetric matrices A and B, AB and BA always have the same eigenvalues. [Hint: what is (AB)T ?] (b) For n × n matrices A and B with B invertible, AB and BA always have the same eigenvalues. [Hint: you can write det(AB − λI) = det((A − λB−1)B). Alternative hint: think about similar matrices.] (c) Two diagonalizable matrices A and B with the same eigenvalues and eigenvectors must be the same matrix.

5 (d) Two diagonalizable matrices A and B with the same eigenvalues must be the same matrix. (e) For n × n matrices A and B with B invertible, AB and BA always have the same eigenvectors. Solution: (a) True. Since A and B are real-symmetric matrices, (AB)T = BT AT = BA. But (AB)T has the same eigenvalue as AB. So AB and BA have the same eigenvalues. (b) True. Since B is invertible, we have AB = B−1(BA)B. So BA is similar to AB, and they must have the same eigenvalues. (c) True. Let Λ be the diagonal matrix consists of eigenvalues, and S be the matrix of eigenvectors (ordered as the eigenvalue matrix). Then we have A = S−1ΛS and also B = S−1ΛS. So A = B. (d) False. 1 0 1 2 For example, and are both diagonalizable, with the same eigenvalues, but 0 2 0 2 they are diﬀerent. (e) False. 0 1 1 1 1 0 0 1 For example, let A = and B = . Then AB = and BA = . 0 0 1 0 0 0 0 1 The eigenvectors of AB are (0, 1)T and (1, 0)T , while the eigenvectors or BA are (0, 1)T and (1, 1)T .

Problem 8

You are given the matrix  0 −1 0 1   0 1 0 1  A =   .  1 0 1 1  1 0 0 −1

(a) What is the sum of the eigenvalues of A? (b) What is the product of the eigenvalues of A? (c) What can you say, without computing them, about the eigenvalues of AAT ? Solution: (a) The sum of the eigenvalues of A equals trace of A. So the sum of the eigenvalues of A is 0 + 1 + 1 − 1 = 1. (b) The product of the eigenvalues of A equals the determinant of A. We do row transforms

0 −1 0 1  2 0 1 2  2 0 1 2  2 0 1 2  0 1 0 1  0 1 0 1  0 1 0 1  0 1 0 1          , 1 0 1 1  1 0 1 1  0 0 1/2 0  0 0 1/2 0  1 0 0 −1 1 0 0 −1 0 0 −1/2 −2 0 0 0 −2

where in the ﬁrst step we add the other rows to the ﬁrst row. It follows that the product of eigenvalues equals det(A) = −2. (c) AAT is a real symmetric matrix, so its eigenvalues are all real and nonnegative. Since A is nonsingular, all eigenvalues of AAT are positive. The product of all eigenvalues of AAT are

6 det(AAT ) = det(A) det(AT ) = det(A)2 = 4. The sum of all eigenvalues of AAT is trace(AAT ) = P 2 i,j aij = 9.

Problem 9

You are given the quadratic polynomial f(x, y, z):

f(x, y, z) = 2x2 − 2xy − 4xz + y2 + 2yz + 3z2 − 2x + 2z.

(a) Write f(x, y, z) in the form f(x, y, z) = xT Ax − bT x where x = (x, y, z)T , A is a real- symmetric matrix, and b is some constant vector. (b) Find the point (x, y, z) where f(x, y, z) is at an extremum. (c) Is this point a minimum, maximum, or a saddle point of some kind? Solution: (a) We have

 2 −1 −2  2  A = −1 1 1  , b =  0  . −2 1 3 −2

(b) We compute the partial derivatives to ﬁnd the extremum point:

∂f = 4x − 2y − 4z − 2 = 0 ∂x ∂f = −2x + 2y + 2z = 0 ∂y ∂f = −4x + 2y + 6z + 2 = 0 ∂x The equation is just 2Ax = b. The solution to the the equations above is x = 1, y = 1, z = 0. So the extreme point is (1, 1, 0). (c) We look for the pivots of A:

 2 −1 −2 2 −1 −2 −1 1 1  0 1/2 0  . −2 1 3 0 0 1

Since A is positive deﬁnite, the extreme point is a minimum.

Problem 10

2 Suppose A is some diagonalizable matrix. Consider the vector y(t) = eA tx for some vector x.

(a) If A is 3 × 3 with eigenvalues λ1, λ2, and λ3 with eigenvectors x1, x2, and x3, and x = x1 + 3x2 + 4x3, what is y(t)?

(b) If limt→∞ y(t) = 0 for every vector x, what does that tell you about the eigenvalues of A 2 and of eA ? (c) In terms of A, give a system of diﬀerential equations that y(t) satisﬁes, and the initial condition.

7 (d) In terms of A, give a linear recurrence relation that y(t) satisﬁes for t = k∆t for integers k and some ﬁxed ∆t. Solution: 2 2 2 2 A2t (a) A has eigenvalues λ1, λ2 and λ3 with eigenvectors x1, x2 and x3, so e has eigenvalues λ2t λ2t λ2t e 1 , e 2 and e 3 with the same eigenvectors. Thus

A2t A2t λ2t λ2t λ2t y(t) = e x = e (x1 + 3x2 + 4x3) = e 1 x1 + 3e 2 x2 + 4e 3 x3.

2 2 2 (b) If limt→∞ y(t) = 0 for every vector x, we must have Re λ1 < 0, Re λ2 < 0 and Re λ3 < 0. λ2 λ2 λ2 λ2 The eigenvalues e 1 , e 2 and e 3 all sastisﬁes |e | < 1. (c) y(t) satisﬁes the system y0 = A2y. The initial condition is y(0) = x. A2t (d) Denote by yk = y(k∆t). Then equation y(t) = e x gives the recurrence relation A2∆t yk+1 = e yk with initial condition y0 = x.

Problem 11

Suppose A is an m × n matrix with full row rank. Which of the following equations always have a solution (possibly non-unique) for any b? (a) Ax = b (b) AH x = b (c) AH Ax = b (d) AAH x = b (e) AH Ax = AH b (f) AAH x = Ab Solution: (a) The equation Ax = b has a solution for any b, as explained in problem 4. (b) The equation AH x = b may has no solution, since the column space of AH is an m- dimensional subspace in the whole space Rn. (c) The equation AH Ax = b may has no solution, since the C(AH A) = C(AH ) is an m- dimensional subspace in Rn. (That C(AH A) = C(AH ) comes from the fact N(AH A) = N(A) which we proved in class.) (d) The equation AAH x = b will always has a unique solution, since AAH is an m×m matrix whose rank is m, i.e. AAH is invertible. (e) The equation AH Ax = AH b will always has a solution. In fact, any solution to Ax = b (from part (a)) is always a solution to AH Ax = AH b. (f) The equation AAH x = Ab will always has a unique solution, since AAH is invertible, as explained in part (d).

Additional Practice Problems

Be sure to look at: (i) Exams 1, 2, and 3. (ii) The practice problems for exam 3. (The above problems mostly cover non-eigenvalue stuﬀ.) (iii) Ideally, also review your homework problems.

8 18.06 Professor Edelman Final Exam December 18, 2008

Grading 1 Your PRINTED name is: 2 3 4 Please circle your recitation: 5 6 7 1) T 10 2-131 J.Yu 2-348 4-2597 jyu 8 2) T 10 2-132 J. Aristo 2-492 3-4093 jeffa 9 3) T 10 2-255 Su Ho Oh 2-333 3-7826 suho 4) T 11 2-131 J. Yu 2-348 4-2597 jyu 5) T 11 2-132 J. Pascale 2-492 3-4093 jpascale 6) T 12 2-132 J. Pascale 2-492 3-4093 jpascale 7) T 12 2-131 K. Jung 2-331 3-5029 kmjung 8) T 1 2-131 K. Jung 2-331 3-5029 kmjung 9) T 1 2-136 V. Sohinger 2-310 4-1231 vedran 10) T 1 2-147 M Frankland 2-090 3-6293 franklan 11) T 2 2-131 J. French 2-489 3-4086 jfrench 12) T 2 2-147 M. Frankland 2-090 3-6293 franklan 13) T 2 4-159 C. Dodd 2-492 3-4093 cdodd 14) T 3 2-131 J. French 2-489 3-4086 jfrench 15) T 3 4-159 C. Dodd 2-492 3-4093 cdodd 1 (9 pts.) Given real numbers a, b and c, nd x, y and z such that the matrix B below is guaranteed to be singular with real eigenvalues and orthogonal eigenvectors.

   a b a + b      B =  b c b + c  . x y z

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3 2 (12 pts.) The matrix    0 1 0      A =  0 0 1  . 0 0 p

(a) What are the eigenvalues of A (possibly in terms of p)?

(b) If p is not 0, nd an eigenvector that is not in the nullspace.

(d) Find a nonzero solution u(t) to du/dt = ( A + 2009 I)u. Check that your answer is correct. (Note that A + 2009 I is the matrix above with the upcoming new year added to the diagonal elements. )

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5 3 (8 pts.) A 4x4 square matrix A has singular values 3, 2, 1, and 0. Find an eigenvalue of A. Briey explain your answer.

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7 4 (9 pts.) The square matrix A has QR decomposition A = QR where Q is orthogonal and R is upper triangular with diagonal elements all equal to 1.

(a) What is the determinant of ATA?

(b) What are all the pivots of ATA?

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9 5 (5 pts.) All matrices in this question are n×n. We have that C = A−1BX . Propose an X which guarantees that B and C have the same eigenvalues.

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11 6 (12 pts.) All you are told about a 3 × 3 matrix A is that ve of the nine entries are 1, and the other four are 0. For the ranks below, exhibit a matrix A with this property, or else briey (but convincingly) argue that it is impossible.

(a) A has rank 0.

(b) A has rank 2.

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13 7 (10 pts.) (Do two of the three problems below. Please avoid any confusion to the graders as to which two you chose.)

(a) All you are told about a 100 by 100 matrix is that all of its entries are even integers. Must the determinant be odd? Must the determinant be even? Argue your answer convincingly.

(b) Give an example, if possible, of a 100 by 100 matrix with odd integer entries but an even determinant.

(c) All you are told about a 100 by 100 matrix is that all of its entries are odd integers. Must the determinant be odd? Must the determinant be even? Argue your answer convincingly.

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15 8 (15 pts.) The functions of the form

x 2x f(x) = c1 + c2e + c3e ,

form a three dimensional vector space V .

(a) The transformation d/dx can be written as a 2x3 matrix when the domain is specied to have basis {1, e x, e 2x}, and the range has basis {ex, e 2x}. Write down this 2x3 matrix.

(b) On the above three dimensional vector space V , is the evaluation of f at x = 7 a linear transformation from that space to R?

(c) On the above three dimensional vector space V , is the transformation R that takes to x a linear transformation from that three f(x) 0 f(t)dt dimensional space to itself (from V to V ) ?

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17 9 (20 pts.) Suppose an n by n matrix has the property that its nullspace is equal to its column space.

(a) Can the matrix be the zero matrix?

(b) Possibly in terms of n, what is the rank of the matrix? (c) What are the eigenvalues of this matrix? (Briey explain your answer.

Hint: It might be useful to consider applying A more than once in some way.)

(d) Give an example of a 2 by 2 such matrix.

(e) Perhaps using the previous case twice somehow, give an example of a 4 x 4 such matrix.

18 Have a great holiday vacation! Thank you for taking linear algebra.

19 18.06 FINAL SOLUTIONS

a b a + b Problem 1. (10 points) B = b c b + c . We know that symmetric matrices have x y z real eigenvalues and orthogonal eigenvectors. So we set x = a + b and y = b + c. This leaves only the singularity of B. For this, we note that setting z = x +y = a+2b+c makes the third column a sum of the first two, thus ensuring singularity. 0 1 0 Problem 2. (4 points each). A = 0 0 1. 0 0 p −λ 1 0  a) To find the eigenvalues of A, we compute det(A−λI) = det  0 −λ 1  = 0 0 p − λ (λ 2)(p − λ) (because A is upper triangular). So the eigenvalues are 0 and p. a a pa a  b  b) If p 6= 0, the we wish to find b so that Ab = pb. But Ab =  c , c c pc c pc a  1  so we need b = pa and c = pb; and so b =  p  works. c p2 c) The singular vaues of A are found by first computing the eigenvalues of AT A = 0 0 00 1 0 0 0 0  1 0 10 0 1 = 0 1 p . As this is upper triangular, the eigenval- 0 1 p 0 0 p 0 0 1 + p2 p ues are 0, 1 and 1 + p2. So the singular values are 0, 1 and 1 + p2. 2009 1 0  d) In general, du/dt = Bu is solved by eBt u(0). For us, B =  0 2009 1  = 0 0 2009 + p 2009 0 0  0 1 0 A+2009I. To compute eBt , we note that eBt = exp(t  0 2009 0 +t 0 0 1) = 0 0 2009 + p 0 0 0 e2009 0 0  0 1 0 0 1 0  0 e2009 0 exp(t 0 0 1). To compute exp(t 0 0 1), we note 0 0 epe2009 0 0 0 0 0 0 0 1 02 0 0 1 0 1 03 0 1 0 that 0 0 1 = 0 0 0 and 0 0 1 = 0. So exp(t 0 0 1) = I + 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 t t2/2 e2009 te2009 e2009t2/2 t 0 0 1+(t2/2)0 0 0 = 0 1 t , and finally eBt =  0 e2009 te2009 . 0 0 0 0 0 0 0 0 1 0 0 epe2009 e2009 te2009 e2009t2/21 e2009 So our answer is  0 e2009 te2009 0 =  0 . 0 0 epe2009 0 0 1 18.06 FINAL SOLUTIONS 2

Problem 3. (8 points) A is 4 × 4 and has singular values {3,2,1,0}. As the product of the singular values is (up to sign) the determinant, we get that det(A) = 0, so A has nontrivial nullspace, and so 0 is an eigenvalue. Problem 4. (3 points each). A = QR where Q is orthogonal and R is upper triangular with 1’s on the diagonal. a) det(AT A) = det(RT QT QR) = det(RT R) = det(R)2 = 1 since det(R) = 1 by the as- sumption on R. b) The equation AT A = RT R tells us that (R−1)T (AT A) = R; and since (R−1)T is lower triangular, this is exactly the elimination of AT A. So the pivots are all equal to 1. c) Yes, since Q−1(QR)Q = RQ. Problem 5. (10 points) C = A−1BX . We know that similar matrices have the same eigenvalues, so putting X = A forces C and B to have the same eigenvalues. Problem 6. (4 points each) A is 3 × 3 and has four 0’s and ﬁve 1’s. a) A has rank 0 is impossible- it isn’t the zero matrix. 1 1 0 b) A has rank 2: A = 1 1 0. 1 0 0 1 0 0 c) A has rank 3: A = 1 1 0. 1 0 1 Problem 7. (10 points each) A is 100 × 100. a) A has all even integers as entries. Therefore each column of A has the form 2c where c is a vector of integers. So we set C = (1/2)A. Then det(A) = det(2C) = 2100det(C); so det(A) is an even integer (note that det(C) really is an integer because all the entries of C are integers and the det can be computed by the big formula). b) This time, we use the big formula to compute det(C) = ∑sign(σ)a1σ(1)a2σ(2) ··· anσ(n). This is a sum containing 100! terms. Now, 100! is an even number, and each term in the sum is odd (as a product of odd integers). Since the sum of two odd numbers is even, the sum of an even number of odd numbers is even; so this sum is an even integer. Problem 8. (5 points each) We consider the vector space V of functions of the form x 2x x 2x c1 + c2e + c3e , with basis {1,e ,e }. x 2x d d x x a) d/dx takes V to the space W spanned by {e ,e }. We have that dx (1) = 0, dx e = e 0 1 0 , and d e2x = 2e2x. So the linear transformation of d/dx in the given bases is . dx 0 0 2 b) We conisder the transformation φ from V to R deﬁned by f → f (7). This is linear: φ( f + g) = ( f + g)(7) = f (7) + g(7) = φ( f ) + φ(g), and φ(c f ) → c f (7) = cφ( f ). x c) No. 0 1 = x is a function not in V. ´ 18.06 Professor Strang Final Exam May 20, 2008

Grading 1 Your PRINTED name is: 2 3 4 Please circle your recitation: 5 6 7 1) M 2 2-131 A. Ritter 2-085 2-1192 afr 8 2) M 2 4-149 A. Tievsky 2-492 3-4093 tievsky 9 3) M 3 2-131 A. Ritter 2-085 2-1192 afr 10 4) M 3 2-132 A. Tievsky 2-492 3-4093 tievsky 5) T 11 2-132 J. Yin 2-333 3-7826 jbyin 6) T 11 8-205 A. Pires 2-251 3-7566 arita 7) T 12 2-132 J. Yin 2-333 3-7826 jbyin 8) T 12 8-205 A. Pires 2-251 3-7566 arita 9) T 12 26-142 P. Buchak 2-093 3-1198 pmb 10) T 1 2-132 B. Lehmann 2-089 3-1195 lehmann 11) T 1 26-142 P. Buchak 2-093 3-1198 pmb 12) T 1 26-168 P. McNamara 2-314 4-1459 petermc 13) T 2 2-132 B. Lehmann 2-089 2-1195 lehmann 14) T 2 26-168 P. McNamara 2-314 4-1459 petermc

Thank you for taking 18.06. If you liked it, you might enjoy 18.085 this fall. Have a great summer. GS 1 (10 pts.) The matrix A and the vector b are     1 1 0 2 3         A =  0 0 1 4  b =  1      0 0 0 0 0 (a) The complete solution to Ax = b is x = .

T (b) A y = c can be solved for which column vectors c = (c1, c2, c3, c4) ? (Asking for conditions on the c’s, not just c in C(AT).)

Solution (10 points) a) The complete solution is a particular solution xp plus any vector in the nullspace xn. Since the matrix A is already reduced, we can just read the special solutions oﬀ: [−1, 1, 0, 0]T and [−2, 0, −4, 1]T . To ﬁnd a particular solution to Ax = b, we put any numbers (we may as well choose 0) in for the free variables. This yields the two equations x1 = 3 and x3 = 1, so T xp = [3, 0, 1, 0] . In the end we get       3 −1 −2              0   1   0  xcomp =   + c1   + c2   (1)        1   0   −4        0 0 1

T b) You can do this computation by hand by augmenting A with the column (c1, c2, c3, c4) and row reducing. The solution is given by the equations that correspond to 0 rows in the reduced matrix. A quicker way is to note that AT y = c has a solution whenever c is in the column space C(AT ), i.e. the row space of A. This is perpendicular to the nullspace. Thus, we can find the equations by taking a basis for the nullspace and using the components as coefficients in our equations. We find equations −c1 + c2 = 0 and −2c1 − 4c3 + c4 = 0. c) Because these c are in the row space, they are perpendicular to vectors in the nullspace of A, and in particular are perpendicular to the special solutions.

2 2 (8 pts.) (a) Suppose q1 = (1, 1, 1, 1)/2 is the ﬁrst column of Q. How could you

ﬁnd three more columns q2, q3, q4 of Q to make an orthonormal basis ? (Not necessary to compute them.)

(b) Suppose that column vector q1 is an eigenvector of A: Aq1 = 3q1. (The other columns of Q might not be eigenvectors of A.) Define T = Q−1AQ so that AQ = QT . Compare the first columns of AQ and QT to discover what numbers are in the first column of T ?

Solution (8 points)

4 a) First, we ﬁnd additional vectors v2, v3 and v4 that (along with q1) make up a basis of R .

Then we run Gram-Schmidt on q1, v2, v3, v4. b) Using the column picture of multiplication, we see that the ﬁrst column of AQ will be

Aq1 = 3q1. Similarly, if we denote the ﬁrst column of T by (t1, t2, t3, t4), then the ﬁrst column

of QT will be t1q1 +t2q2 +t3q3 +t4q4. Since these two are equal, we get an equality of vectors

3q1 = t1q1 + t2q2 + t3q3 + t4q4 (2)

Since the qi are linearly independent, we must have t1 = 3 and the other ti = 0, showing that the ﬁrst column of T is (3, 0, 0, 0).

T We can also note that the ﬁrst column of T is equal to 3Q q1, which yields the same answer.

3 3 (12 pts.) Two eigenvalues of this matrix A are λ1 = 1 and λ2 = 2. The ﬁrst two

pivots are d1 = d2 = 1.   1 0 1     A =  0 1 1  .   1 1 0

(a) Find the other eigenvalue λ3 and the other pivot d3.

(b) What is the smallest entry a33 in the southeast corner that would make A positive semideﬁnite ? What is the smallest c so that A + cI is positive semideﬁnite ?

(c) Starting with one of these vectors u0 = (3, 0, 0) or (0, 3, 0) or (0, 0, 3), 1 and solving uk+1 = 2 Auk, describe the limit behavior of uk as k → ∞ (with numbers).

Solution (12 points)

a) The sum of the eigenvalues is the trace, so 1 + 2 + λ3 = 2. Thus λ3 = −1. The product of

the pivots is the determinant, which is the product of the eigenvalues as well. So d3 = −2. Note that this means that A is not positive-deﬁnite. b) We can test positive-deﬁniteness using the determinant method. The two top-left determinants of A are both positive, so we just need to check the third one. We obtain the relation: 1(c − 1) + 1(−1) ≥ 0 (3) so the smallest value of c is 2. For the second part, we test whether the eigenvalues are non-negative. The eigenvalues of A + cI are just the eigenvalues of A plus c. So when c = 1 all the eigenvalues will be non-negative.

4 1 c) The matrix 2 A is a Markov matrix. Because it has some 0 entries, we don’t automatically 1 know that it has a unique steady state vector. However, since the eigenvalues of 2 A are 1/2, −1/2 and 1, it does have a unique steady state vector (only one eigenvalue has absolute value 1). To ﬁnd it, we calculate the eigenvector of A with eigenvalue 2 by taking the nullspace of A − 2I:   −1 0 1     A − 2I =  0 −1 1  (4)   1 1 −2   −1 0 1      0 −1 1  (5)   0 0 0

The nullspace is generated by the special solution (1, 1, 1). So, a vector u will have limit

∞ 1 1 1 A u equal to c( 3 , 3 , 3 ), where c is the sum of the components of u. In particular, the vectors (3, 0, 0), etc., all go to (1, 1, 1).

5 4 (10 pts.) Suppose Ax = b has a solution (maybe many solutions). I want to prove two facts:

T A. There is a solution xrow in the row space C(A ). B. There is only one solution in the row space.

(a) Suppose Ax = b. I can split that x into xrow + xnull with xnull in the nullspace. How do I know that Axrow = b ? (Easy question) ∗ ∗ (b) Suppose xrow is in the row space and Axrow = b. I want to prove that ∗ ∗ xrow is the same as xrow. Their diﬀerence d = xrow −xrow is in which subspaces ? How to prove d = 0 ?

(c) Compute the solution xrow in the row space of this matrix A, by solving for c and d:         1 1 1 2 3 14           xrow =   with xrow = c  2 + d  1  . 1 1 −1 9     3 −1

Solution (10 points)

a) We have A(xrow + xnull) = A(xrow) + A(xnull) = A(xrow) + 0, so A(xrow) = b.

∗ ∗ b) Suppose both A(xrow) = b and A(xrow) = b. Then xrow − xrow is in the row space (since it is a linear combination of vectors in the row space) and is in the nullspace (since multiplying by A will give us 0). Because the row space and nullspace are orthogonal complements, the only vector that is in both is the 0 vector: any vector in both will have |x|2 = x · x = 0.

6 c) Substituting in the given expressions for Axrow = b we ﬁnd     1 1     1 2 3   c 14      2 1    =   (6) 1 1 −1   d 9 3 −1 or       14 0 c 14     =   (7) 0 3 d 9

We ﬁnd (c, d) = (1, 3), so xrow = (4, 5, 0). Remark: essentially what we are doing here is projecting onto the row space.

7 5 (10 pts.) The numbers Dn satisfy Dn+1 = 2Dn − 2Dn−1. This produces a ﬁrst-order

system for un = (Dn+1,Dn) with this 2 by 2 matrix A:       Dn+1 2 −2 Dn   =     or un = Aun−1 . Dn 1 0 Dn−1

(a) Find the eigenvalues λ1, λ2 of A. Find the eigenvectors x1, x2 with

second entry equal to 1 so that x1 = (z1, 1) and x2 = (z2, 1).

(b) What is the inner product of those eigenvectors ? (2 points)

ﬁnd c1 and c2 and a formula for Dn.

Solution (10 points) a) The eigenvalues of   2 −2 A =   (8) 1 0

2 satisfy the equation λ − 2λ + 2 = 0, so λ1 = 1 + i and λ2 = 1 − i. We ﬁnd the eigenvectors by taking the appropriate nullspaces:   1 − i −2 A − λ1I =   (9) 1 −1 − i has nullspace generated by x1 = (1 + i, 1), and   1 + i −2 A − λ2I =   (10) 1 −1 + i has nullspace generated by x2 = (1 − i, 1). If you pick a diﬀerent vector in the nullspace, you just rescale so that the bottom entry is 1.

H 2 H b) The inner product is x1 x2 = (1−i) +1 = 1−2i, or its conjugate expression x2 x1 = 1+2i.

8 n n c) If u0 = c1x1 + c2x2, then un = c1λ1 x1 + c2λ2 x2. A matrix always acts on its eigenvectors

in a diagonal way. In particular, (2, 2) = x1 + x2. So we ﬁnd     n 1 + i n 1 − i un = (1 + i)   + (1 − i)   (11) 1 1

n n with second entry Dn = (1 + i) + (1 − i) .

9 6 (12 pts.) (a) Suppose q1, q2, a3 are linearly independent, and q1 and q2 are already

orthonormal. Give a formula for a third orthonormal vector q3 as a

linear combination of q1, q2, a3.

(b) Find the vector q3 in part (a) when       1 1 1             1  1  1  −1   2  q1 =   q2 =   a3 =   2   2      1   1   3        1 −1 4

two vectors q1 and q2. You can give a formula for P using q1 and q2 or give a numerical answer.

Solution (12 points) a) This is the Gram-Schmidt process. We deﬁne

w3 = a3 − (q1 · a3)q1 − (q2 · a3)q2 (12)

and then set q3 = w3/kw3k. Note that we do not need denominators in the expression for

w3 because the qi are already unit vectors.

b) Substituting in, we ﬁnd

w3 = a3 − 5q1 − (−1)q2 = (−1, −1, 1, 1) (13)

1 Renormalizing we get q3 = 2 (−1, −1, 1, 1). c) The projection matrix P is exactly the expression we used for Gram-Schmidt: P =

T T q1q1 + q2q2 . There are other more complicated expressions which are also correct. We can start at the most general and simplify to get this one; if A has columns q1 and q2 then T −1 T T T T P = A(A A) A = A(I)A = q1q1 + q2q2 where we used the column-row picture of multiplication for the last step.

10 7 (12 pts.) (a) Find the determinant of this N matrix.   1 0 0 4      2 1 0 3  N =      3 0 1 2    4 0 0 1

(b) Using the cofactor formula for N −1, tell me one entry that is zero or tell me that all entries of N −1 are nonzero.

Solution (12 points) a) There are many ways to do this. Perhaps the easiest is cofactors along the top row:   2 1 0     det(N) = 1(1) − 4 det  3 0 1  = 1 − 4(4) = −15 (14)   4 0 0

Here I found the determinant of the 3 by 3 by swapping the columns to get an upper triangular matrix with diagonal entries 1, 1, 4. b) The cofactor formula is A−1 = CT / det(A) (we know that A is invertible from part a). To check for 0 entries we can ignore the det(A) part. We just need to ﬁnd some cofactors that are 0, and we can arrange this by crossing out rows and columns that will give us a smaller matrix with a column of 0s. Some choices are C21, C23, C24, C31, C32, and C34. The corresponding 0 entries of the inverse are the transposes, so we get the entries (1, 2), (3, 2), (4, 2), (1, 3), (2, 3), (4, 3).

11 c) The matrix N − I has two columns that are all 0s, and the other two columns are clearly independent, so it has rank 2. So N − I has eigenvalue 0 with multiplicity 2. This tells us that N has eigenvalue 1 with multiplicity 2. Calling the other eigenvalues λ1 and λ2, we can ﬁnd them solving the trace and determinant equations:

1 + 1 + λ1 + λ2 = 4 (15)

(1)(1)λ1λ2 = −15 (16)

Thus λ1 = 5 and λ2 = −3.

12 8 (8 pts.) Every invertible matrix A is the product A = QH of an orthogonal matrix Q and a symmetric positive deﬁnite matrix H. I will start the proof:

A has a singular value decomposition A = UΣV T. Then A = (UV T)(V ΣV T).

(a) Show that UV T is an orthogonal matrix Q (what is the test for an orthogonal matrix ?).

(b) Show that V ΣV T is a symmetric positive deﬁnite matrix. What are its eigenvalues and eigenvectors ? Why did I need to assume that A is invertible ?

Solution (8 points) a) To test that Q = UV T is orthogonal, we must show that QT Q = I. But QT Q = (UV T )T UV T = VU T UV T = V (I)V T = I. We used the fact that U and V are orthogonal matrices. b) The matrix H = V ΣV T is deﬁnitely symmetric, as HT = V ΣT V T = V ΣV T because Σ is diagonal. Furthermore, note that the expression H = V ΣV T is a diagonalization of H. This means that H has eigenvalues given by the entries of Σ and eigenvectors equal to the columns of V . To show that H is positive-deﬁnite, we just need to show that the diagonal entries of Σ are all positive. Now, we know that they are all non-negative, because the SVD always gives us non-negative singular values. We must also show that none of the singular values are zero. Remember that the singular values are equal to the square roots of the eigenvalues of AT A. However, because A is invertible, the matrix AT A is also invertible, and so can’t have any eigenvalues equal to 0. So no singular value is 0 either.

13 9 (7 pts.) (a) Find the inverse L−1 of this real triangular matrix L:   1 0 0     L =  a 1 0    0 a 1

You can use formulas or Gauss-Jordan elimination or any other method.

(b) Suppose D is the real diagonal matrix D = diag(d, d2, d3). What are the conditions on a and d so that the matrix A = LDLT is (three separate questions, one point each)

(i) invertible ? (ii) symmetric ? (iii) positive deﬁnite ?

Solution (7 points) a) I’ll do Gauss-Jordan elimination.     1 0 0 1 0 0 1 0 0 1 0 0          a 1 0 0 1 0   0 1 0 −a 1 0  (17)     0 a 1 0 0 1 0 a 1 0 0 1   1 0 0 1 0 0      0 1 0 −a 1 0  (18)   0 0 1 a2 −a 1 b) Note that L is invertible no matter what a is, and D is invertible so long as d 6= 0. So A = LDLT will be invertible whenever d 6= 0. If d = 0, then of course A can’t be invertible. The matrix A is always symmetric, since AT = (LDLT )T = LDT LT = LDLT . Because A is always symmetric, to check positive-deﬁniteness we just need to check that the pivots are all positive. But A = LDLT is the “pivot” decomposition for A. So the pivots of A are d, d2, d3, and we need d > 0.

14 10 (11 pts.) This problem uses least squares to ﬁnd the plane C + Dx + Ey = b that best ﬁts these 4 points:

x = 0 y = 0 b = 2 x = 1 y = 1 b = 1 x = 1 y = −1 b = 0 x = −2 y = 0 b = 1

(a) Write down 4 equations Ax = b with unknown x = (C,D,E) that would hold if the plane went through the 4 points. Then write down the

equations to solve for the best (least squares) solution xb = (C,b D,b Eb).

3 4 (b) Find the best xb and the error vector e (is the vector e in R or R ?).

the best plane to ﬁt these four new points (p1, p2, p3, p4) ? What will be the new error vector ?

Solution (11 points) a) The equations are of the form C + 0D + 0E = 2, etc., or in matrix form     1 0 0   2   C        1 1 1     1     D  =   (19)        1 1 −1     0    E   1 −2 0 1

T T Of course this system is not solvable. The best solution is given by A Axb = A b.

15 b) We have   4 0 0   T   A A =  0 6 0  (20)   0 0 2 and   4   T   A b =  −1  (21)   1 It is a diagonal system, so we immediately ﬁnd (C,D,E) = (1, −1/6, 1/2). The error vector is the diﬀerence of the real b and the approximate values we get for our plane: e = b − Axb. T Since Axb = [1, 4/3, 1/3, 4/3] , we get e = (1, −1/3, −1/3, −1/3).

c) We know p = Axb is the projection of b onto the column space of A. So the system Ax = p is solvable exactly; we don’t need any approximations. The best fit plane will be the same plane as in part b: 1 − x/6 + y/2 = b (we changed the b-coordinates of the points so that they lie on this plane, so of course it is the best fit). The error vector will become 0 because it is an exact fit.

16 18.06 Final Solution Hold on Tuesday, 19 May 2009 at 9am in Walker Gym. Total: 100 points.

Problem 1: A sequence of numbers f0, f1, f2,... is deﬁned by the recurrence

fk+2 = 3fk+1 − fk,

with starting values f0 = 1, f1 = 1. (Thus, the ﬁrst few terms in the sequence are 1, 1, 2, 5, 13, 34, 89,....) fk+1 (a) Deﬁning uk = , re-express the above recurrence as uk+1 = Auk, and fk give the matrix A.

(b) Find the eigenvalues of A, and use these to predict what the ratio fk+1/fk of successive terms in the sequence will approach for large k.

(c) The sequence above starts with f0 = f1 = 1, and |fk| grows rapidly with k. Keep f0 = 1, but give a diﬀerent value of f1 that will make the sequence (with the same recurrence fk+2 = 3fk+1 − fk) approach zero (fk → 0) for large k.

Solution (18 points = 6+6+6) (a) We have

f 3 −1 f 3 −1 k+2 = k+1 ⇒ A = . fk+1 1 0 fk 1 0

2 (b) Eigenvalues√ of A are roots√ of det(A − λI) = λ − 3λ + 1 = 0. They are 3 + 5 3 − 5 λ1 = and λ2 = . Note that λ1 > λ2, so the ratio fk+1/fk will 2 √ 2 3 + 5 approach λ = for large k. 1 2

(c) Let v1, v2 be the eigenvectors with eigenvalues λ1 and λ2, respectively. So, k k we can write u0 = c1v1 + c2v2 and then uk = c1λ1v1 + c2λ2v2. If we need fk → 0,

1 we have to make c1 = 0. In other words, u0 must be proportional to the eigenvector v2. √ 3 + 5   √  −1 3 − 5 A − λ I =  2 √  ⇒ v = . 2  3 − 5 2  2  1 − 1 2 √ 3 − 5 Hence, we need to take f = so that f will approach zero for large k. 1 2 k

2 1 0 −1 Problem 2: For the matrix A = 1 1 1  with rank 2, consider the system of 2 1 0 equations Ax = b.

(i) Ax = b has a solution whenever b is orthogonal to some nonzero vector c. Explicitly compute such a vector c. Your answer can be multiplied by any overall constant, because c is any basis for the space of A. 9 (ii) Find the orthogonal projection p of the vector b = 9 onto C(A). (Note: 9 The matrix ATA is singular, so you cannot use your formula P = A(ATA)−1AT to obtain the projection matrix P onto the column space of A. But I have repeatedly discouraged you from computing P explicitly, so you don’t need to be reminded anyway, right?)

(iii) If p is your answer from (ii), then a solution y of Ay = p minimizes what? [You need not answer (ii) or compute y for this part.]

Solution (18 points = 7+7+4) (i) The system of equations Ax = b has a solution if and only if b lies in the column space of A, which is orthogonal to the left nullspace of A. We solve for a (nonzero) vector c in the left nullspace using Gaussian elimination, as follows.

 1 1 2 1 1 2 1 0 1 −1 T A =  0 1 1 ; 0 1 1 ; 0 1 1 ⇒ c = −1 . −1 1 0 0 2 2 0 0 0 1

The answer can by any nonzero multiple of c, which will be a basis for the left nullspace of A.

(ii) Method 1: Since c is a basis of the orthogonal complement of the column space C(A), the projection of b onto C(A) can be computed as

9 −1  6  cT bc −9 p = b − = 9 − −1 = 6 . kck2   3     9 1 12

3 Method 2: (not recommended) We know that p is the best linear approximation of b. So we solve     y1 9 T T A A y2 = A 9 , y3 9       6 3 0 y1 36 3 2 1 y2 = 18 0 1 2 y3 0

We can get a particular solution y = (6, 0, 0)T. (There are other solutions too.) Hence,         y1 1 0 −1 6 6 p = A y2 = 1 1 1  0 =  6  . y3 2 1 0 0 12

(iii) Since p is the orthogonal projection of b onto C(A), A solution y of Ay = p minimizes the distance kAy − bk.

4 Problem 3: True or false. Give a counter-example if false. (You need not provide a reason if true.)

(a) If Q is an orthogonal matrix, then det Q = 1.

(b) If A is a Markov matrix, then du/dt = Au approaches some ﬁnite constant vector (a “steady state”) for any initial condition u(0).

(d) If S and T are subspaces of R2, then their union (points in either S or T ) is also a subspace.

(e) The rank of AB is less than or equal to the ranks of A and B for any A and B.

(f) The rank of A + B is less than or equal to the ranks of A and B for any A and B.

Solution (12 points = 2+2+2+2+2+2) (a) False. For example, Q = (−1) is an orthogonal matrix: QTQ = (−1)(−1) = (1). REMARK: In general, for a real orthogonal matrix Q, det Q = ±1. This is because det(QTQ) = det(I) = 1 ⇒ det(Q)2 = det(QT) det(Q) = 1.

(b) False. Be careful here that we are discussing differential equations but not the powers of A. For example, A = (1), the differential equation has solution u = cet for some constant c, which does not approach to any finite constant vector. REMARK: It is true that for the Markov process uk+1 = Auk, uk approaches some finite constant vector (a “steady state”) for any initial condition u0.

(d) False. For example, S and T are the x- and y-axes. Then (1, 1) = (1, 0) + (0, 1) is a linear combination of points in the union of S and T , but does not lie in the union itself. So the union of S and T is not a subspace.

(e) Ture. One may see this by arguing as follows. Since the column space of AB is a subspace of the column space of A, the rank of AB is less than or equal to

5 the rank of A. Similarly, since the row space of AB is a subspace of the row space of B, the rank of AB is less than or equal to the rank of B.

0 0 1 0 1 0 (f) False. A = ,B = both have rank 1. But A + B = 0 1 0 0 0 1 has rank 2. REMARK: It is true that rank(A + B) ≤ rankA + rankB.

6 Problem 4: Consider the matrix 1 1 1  1 −1 −1 A =   1 0 −3 1 0 −1

(a) Find an orthonormal basis for C(A) using Gram-Schmidt, forming the columns of a matrix Q.

(b) Write each step of your Gram-Schmidt process from (a) as a multiplication of A on the (left or right) by some invertible matrix. Explain how the product of these invertible matrices relates to the matrix R from the QR factorization A = QR of A.

(c) Gram-Schmidt on another matrix B (of the same size as A) gives the same orthonormal basis (the same Q) as in part (a). Which of the four subspaces, if any, must be the same for the matrices AAT and BBT?[You can do this part without doing (a) or (b).]

Solution (18 points = 6+6+6) T 1 T (a) From u1 = (1, 1, 1, 1) , we get q1 = u1/ku1k = 2 (1, 1, 1, 1) .

T v2 = (1, −1, 0, 0) , T T u2 = v2 − q1 v2q1 = v2 = (1, −1, 0, 0) ,

1 T q2 = v2/kv2k = √ (1, −1, 0, 0) ; 2 T v3 = (1, −1, −3, −1) , T T T u3 = v3 − q1 v3q1 − q1 v3q1 = v3 + u1 − u2 = (1, 1, −2, 0) ,

1 T q3 = v3/kv3k = √ (1, 1, −2, 0) . 6 Hence, we have  √ √  1/2 1/ √2 1/√6 1/2 −1/ 2 1/ 6  Q =  √  . 1/2 0 −2/ 6 1/2 0 0

7 (b) Each step of the Gram Schmidt process from (a) is a multiplication of A on the right as follows.       1/2 0 0 1/2 0 0 1 0√ 0 A ; A  0 1 0 ; A  0 1 0 0 1/ 2 0 0 0 1 0 0 1 0 0 1       1/2 0 0 1 0√ 0 1 0 1 ; A  0 1 0 0 1/ 2 0 0 1 −1 0 0 1 0 0 1 0 0 1         1/2 0 0 1 0√ 0 1 0 1 1 0 0 ; A  0 1 0 0 1/ 2 0 0 1 −1 0 1 0√  = Q. 0 0 1 0 0 1 0 0 1 0 0 1/ 6

The product of these invertible 3 × 3 matrices is exactly R−1.

(c) Since the Gram-Schmidt of A and B gives the same outcome, the column space of A and B are the same. We know that A and AAT have the same column space, and B and BBT have the same column space. Hence AAT and BBT have the same column space. Moreover, since left nullspace is always orthogonal to the column space, AAT and BBT have the same left nullspace too. Also, notice that AAT and BBT are symmetric matrices, their row spaces are the same as the column spaces, and their nullspaces are the same as the left nullspaces. Therefore, all four subspaces of AAT are the same as BBT.

8 Problem 5: The complete solution to Ax = b is

 1  1 −2 x =  0  + c 1 + d  0  −1 0 1 for any arbitrary constants c and d.

(i) If A is an m × n matrix with rank r, give as much true information as possible about the integers m, n, and r.

(ii) Construct an explicit example of a possible matrix A and a possible right-hand side b with the solution x above. (There are many acceptable answers; you only have to provide one.)

Solution (16 points = 8+8) (i) Since we can multiply A with x, n = 3. Also, since the nullspace of A is 2-dimensional, r = n − 2 = 1. There is no restriction on m except that m ≥ r = 1.

(ii) We construct a minimal one, namely, A = (a1 a2 a2) is 1 × 3. For this, we 1 −2 need A 1 = 0 and A  0  = 0. That is 0 1

a  1 1 0 1 0 a = . −2 0 1  2 0 a3

 1  A special solution is A = (1 −1 2). In this case, b = Ax = (1 −1 2)  0  = (−1). −1 So, an example is 1 −1 2 x = (−1).

9 Problem 6: Consider the matrix  1 −1 −1 A = −1 1 −1 −1 −1 1

(i) A has one eigenvalue λ = −1, and the other eigenvalue is a double root of det(A − λI). What is the other eigenvalue? (Very little calculation required.)

(ii) Is A defective? Why or why not?

(iii) Using the above A, suppose we want to solve the equation

du = Au + cu dt where c is some real number, for some initial condition u(0).

(a) For what values of c will the solutions u(t) always to go zero as t → ∞? (b) For what values of c will the solutions u(t) typically diverge (ku(t)k → ∞) as t → ∞? (c) For what values of c will the solutions u(t) approach a constant vector (possibly zero) as t → ∞?

Solution (18 points = 6+6+6 (2+2+2)) (i) Let λ1 = −1 and let λ2 = λ3 denote the double roots. Then from the trace of A, we have λ1 + 2λ2 = trace(A) = 3. Hence, λ2 = 2.

(ii) A is not defective. There are two ways to see it. For one way, since A is symmetric, it is always non-defective; for another way, we compute A − λ2I = −1 −1 −1 −1 −1 −1, which has rank 1 and hence its nullspace is 2-dimentional. −1 −1 −1

(iii) The key point here is that A + cI would have eigenvalues λ1 + c and λ2 + c (with multiplicity 2). An alternative point of view is as follows. If we write the initial condition u(t) = c1(t)v1 + c2(t)v2 + c3(t)v3, then the diﬀerential equation becomes dc (t) dc (t) dc (t) 1 v + 2 v + 3 v = c λ v + c λ v + c λ v + cc v + cc v + cc v . dt 1 dt 2 dt 3 1 1 1 2 2 2 3 3 3 1 1 2 2 3 3

10 We have  dc1(t)  = c λ + cc , ⇒ c = e(λ1+c)t;  dt 1 1 1 1  dc (t) 2 (λ2+c)t = c2λ2 + cc2, ⇒ c2 = e ;  dt  dc3(t)  = c λ + cc , ⇒ c = e(λ3+c)t; dt 3 3 3 3

(a) If we require u(t) always go zero as t → ∞, λ1 + c < 0, λ2 + c = λ3 + c < 0. Hence, we require c < −2.

(b) If the solution u(t) typically diverge, we need either λ1 + c > 0 or λ2 + c = λ3 + c > 0. Hence, we require c > −2.

(c) If we allow the solution to approach to some constant vector, we allow to have the extreme case of (a), that is to say c ≤ −2.