Matrix Calculations: Determinants and Basis Transformation

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Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen

A. Kissinger

Institute for Computing and Information Sciences Radboud University Nijmegen

Version: autumn 2017

A. Kissinger Version: autumn 2017 Matrix Calculations 1 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Outline

Determinants

Change of basis

Matrices and basis transformations

A. Kissinger Version: autumn 2017 Matrix Calculations 2 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Last time

• Any linear map can be represented as a matrix:

f (v) = A · v g(v) = B · v

• Last time, we saw that composing linear maps could be done by multiplying their matrices:

f (g(v)) = A · B · v

• Matrix multiplication is pretty easy: 1 2 1 −1 1 · 1 + 2 · 0 1 · (−1) + 2 · 4 1 7 · = = 3 4 0 4 3 · 1 + 4 · 0 3 · (−1) + 4 · 4 3 13 ...so if we can solve other stuﬀ by matrix multiplication, we are pretty happy.

A. Kissinger Version: autumn 2017 Matrix Calculations 3 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Last time

• For example, we can solve systems of linear equations: A · x = b ...by ﬁnding the inverse of a matrix: x = A−1 · b

• There is an easy shortcut formula for 2 × 2 matrices: a b 1 d −b A = =⇒ A−1 = c d ad − bc −c a ...as long as ad − bc 6= 0. • We’ll see today that “ad − bc” is an example of a special number we can compute for any square matrix (not just 2 × 2) called the determinant.

A. Kissinger Version: autumn 2017 Matrix Calculations 4 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Determinants

What a determinant does For an n × n matrix A, the determinant det(A) is a number (in R) It satisﬁes: det(A) = 0 ⇐⇒ A is not invertible ⇐⇒ A−1 does not exist ⇐⇒ A has < n pivots in its echolon form

Determinants have useful properties, but calculating determinants involves some work.

A. Kissinger Version: autumn 2017 Matrix Calculations 6 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Determinant of a 2 × 2 matrix

a b • Assume A = c d • Recall that the inverse A−1 exists if and only if ad − bc 6= 0, and in that case is: d −b A−1 = 1 ad−bc −c a

• In this 2 × 2-case we deﬁne: a b ab det = = ad − bc c d cd

• Thus, indeed: det(A) = 0 ⇐⇒ A−1 does not exist.

A. Kissinger Version: autumn 2017 Matrix Calculations 7 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Determinant of a 2 × 2 matrix: example

• Example: 0.8 0.1 8 1 P = = 1 0.2 0.9 10 2 9 • Then: 8 9 1 2 det(P) = 10 · 10 − 10 · 10 72 2 = 100 − 100 70 7 = 100 = 10 • We have already seen that P−1 exists, so the determinant must be non-zero.

A. Kissinger Version: autumn 2017 Matrix Calculations 8 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Determinant of a 3 × 3 matrix

  a11 a12 a13 • Assume A = a21 a22 a23 a31 a32 a33 • Then one deﬁnes:

a11 a12 a13

det A = a21 a22 a23

a31 a32 a33

a22 a23 a12 a13 a12 a13 =+ a11 · − a21 · + a31 · a32 a33 a32 a33 a22 a23

• Methodology:

• take entries ai1 from ﬁrst column, with alternating signs (+,-) • take determinant from square submatrix obtained by deleting the ﬁrst column and the i-th row

A. Kissinger Version: autumn 2017 Matrix Calculations 9 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Determinant of a 3 × 3 matrix, example

1 2 −1 3 4 2 −1 2 −1 5 3 4 = 1 − 5 + −2 0 1 0 1 3 4 −2 0 1 = 3 − 0 − 5 2 − 0 − 2 8 + 3

= 3 − 10 − 22

= −29

A. Kissinger Version: autumn 2017 Matrix Calculations 10 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen The general, n × n case

a12 ··· a1n a11 ··· a1n a22 ··· a2n a32 ··· a3n . . =+ a · . . − a · . . 11 . . 21 . . . . an1 ... ann an2 ... ann an2 ... ann

a12 ··· a1n ··· . . + a31 ··· ···± an1 . . ··· a(n−1)2 ... a(n−1)n

(where the last sign ± is+ if n is odd and- if n is even)

Then, each of the smaller determinants is computed recursively.

A. Kissinger Version: autumn 2017 Matrix Calculations 11 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Applications

• Determinants detect when a matrix is invertible • Though we showed an ineﬃcient way to compute determinants, there is an eﬃcient algorithm using, you guessed it...Gaussian elimination! • Solutions to non-homogeneous systems can be expressed directly in terms of determinants using Cramer’s rule (wiki it!) • Most importantly: determinants will be used to calculate eigenvalues in the next lecture

A. Kissinger Version: autumn 2017 Matrix Calculations 12 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Vectors in a basis

Recall: a basis for a vector space V is a set of vectors B = {v 1,..., v n} in V such that: 1 They uniquely span V , i.e. for all v ∈ V , there exist unique ai such that: v = a1v 1 + ... + anv n

Because of this, we use a special notation for this linear combination:   a1  .   .  := a1v 1 + ... + anv n a n B

A. Kissinger Version: autumn 2017 Matrix Calculations 14 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Same vector, different outfits The same vector can look different, depending on the choice of basis. Consider the standard basis: S = {(1, 0), (0, 1)} vs. another basis: 100 100 B = , 0 1

Is this a basis? Yes... 100 100 • It’s independent because: has 2 pivots. 0 1

• It’s spanning because... we can make every vector in S using linear combinations of vectors in B: 1 1 100 0 100 100 = = − 0 100 0 1 1 0

2 ...so we can also make any vector in R . A. Kissinger Version: autumn 2017 Matrix Calculations 15 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Same vector, diﬀerent outﬁts

1 0 100 100 S = , B = , 0 1 0 1

Examples:

100 1 300 2 = = 0 0 1 1 S B S B

1 1 0 −1 = 100 = 0 0 1 1 S B S B

A. Kissinger Version: autumn 2017 Matrix Calculations 16 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Why???

• Many find the idea of multiple bases confusing the first time around. 2 • S = {(1, 0), (0, 1)} is a perfectly good basis for R . Why bother with others? 1 Some vector spaces don’t have one “obvious” choice of basis. Example: subspaces S ⊆ Rn. 2 Sometimes it is way more efficient to write a vector with respect to a different basis, e.g.:  93718234  1  −438203  1      110224  0   =   −5423204980 0  .  . . . S B 3 The choice of basis for vectors affects how we write matrices as well. Often this can be done cleverly. Example: JPEGs, MP3s, search engine rankings, ... A. Kissinger Version: autumn 2017 Matrix Calculations 17 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Transforming bases, part I

• Problem: given a vector written in B = {(100, 0), (100, 1)}, how can we write it in the standard basis? Just use the deﬁnition: x 100 100 100x + 100y = x · + y · = y 0 1 y B S

• Or, as matrix multiplication: 100 100 x 100x + 100y · = 0 1 y y | {z } |{z} | {z } T B⇒S in basis B in basis S

• Let T B⇒S be the matrix whose columns are the basis vectors B. Then T B⇒S transforms a vector written in B into a vector written in S.

A. Kissinger Version: autumn 2017 Matrix Calculations 19 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Transforming bases, part II

• How do we transform back? Need T S⇒B which undoes the matrix T B⇒S . −1 • Solution: use the inverse! T S⇒B := (T B⇒S ) • Example:

100 100−1 1 −1 (T )−1 = = 100 B⇒S 0 1 0 1

• ...which indeed gives:

1 −1 a a−100b 100 · = 100 0 1 b b

A. Kissinger Version: autumn 2017 Matrix Calculations 20 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Transforming bases, part IV

• How about two non-standard bases? 100 100 −1 1 B = { , }C = { , } 0 1 2 2

a a0 • Problem: translate a vector from to b b0 B C • Solution: do this in two steps:

T B⇒S · v | {z } ﬁrst translate from B to S...

−1 T S⇒C · T B⇒S · v = (T C⇒S ) · T B⇒S · v | {z } ...then translate from S to C

A. Kissinger Version: autumn 2017 Matrix Calculations 21 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Transforming bases, example

• For bases: 100 100 −1 1 B = { , }C = { , } 0 1 2 2 • ...we need to ﬁnd a0 and b0 such that a0 a = b0 b C B • Translating both sides to the standard basis gives: −1 1 a0 100 100 a · = · 2 2 b0 0 1 b • This we can solve using the matrix-inverse: a0 −1 1−1 100 100 a = · · b0 2 2 0 1 b

A. Kissinger Version: autumn 2017 Matrix Calculations 22 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Transforming bases, example For: a0 −1 1−1 100 100 a = · · b0 2 2 0 1 b | {z } | {z } | {z } |{z} in basis C T S⇒C T B⇒S in basis B we compute

−1 ! −1 1 100 100 − 1 1 100 100 −200 −199 · = 2 4 · = 1 2 2 0 1 1 1 0 1 4 200 201 2 4

which gives:

a0 −200 −199 a = 1 · b0 4 200 201 b | {z } | {z } |{z} in basis C T B⇒C in basis B

A. Kissinger Version: autumn 2017 Matrix Calculations 23 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Basis transformation theorem

Theorem n Let S be the standard basis for R and let B = {v 1,..., v n} and C = {w 1,..., w n} be other bases.

1 Then there is an invertible n × n basis transformation matrix T B⇒C such that:

a0  a  a0  a  .1 .1 .1 .1  .  = T B⇒C ·  .  with  .  =  .  a0 a a0 a n n n C n B 2 T B⇒S is the matrix which has the vectors in B as columns, and −1 T B⇒C := (T C⇒S ) · T B⇒S

−1 3 T C⇒B = (T B⇒C)

A. Kissinger Version: autumn 2017 Matrix Calculations 24 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Matrices in other bases

• Since vectors can be written with respect to diﬀerent bases, so too can matrices. • For example, let g be the linear map deﬁned by: 1 0 0 1 g( ) = g( ) = 0 1 1 0 S S S S • Then, naturally, we would represent g using the matrix: 0 1 1 0 S • Because indeed: 0 1 1 0 0 1 0 1 · = and · = 1 0 0 1 1 0 1 0 (the columns say where each of the vectors in S go, written in the basis S)

A. Kissinger Version: autumn 2017 Matrix Calculations 25 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen On the other hand...

• Lets look at what g does to another basis:

1 1 B = { , } 1 −1

• First (1, 1) ∈ B:

1 1 1 0 g( ) = g( ) = g( + ) = ... 0 1 0 1 B

• Then, by linearity:

1 0 0 1 1 1 ... = g( ) + g( ) = + = = 0 1 1 0 1 0 B

A. Kissinger Version: autumn 2017 Matrix Calculations 26 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen On the other hand...

1 1 B = { , } 1 −1

• Similarly (1, −1) ∈ B:

0 1 1 0 g( ) = g( ) = g( − ) = ... 1 −1 0 1 B

• Then, by linearity:

1 0 0 1 −1 0 ... = g( )−g( ) = − = = 0 1 1 0 1 −1 B

A. Kissinger Version: autumn 2017 Matrix Calculations 27 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen A new matrix

• From this: 1 1 0 0 g( ) = g( ) = 0 0 1 −1 B B B B • It follows that we should instead use this matrix to represent g: 1 0 0 −1 B • Because indeed: 1 0 1 1 1 0 0 0 · = and · = 0 −1 0 0 0 −1 1 −1

(the columns say where each of the vectors in B go, written in the basis B)

A. Kissinger Version: autumn 2017 Matrix Calculations 28 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen A new matrix

• So on diﬀerent bases, g acts in a totally diﬀerent way!

1 0 0 1 g( ) = g( ) = 0 1 1 0 S S S S

1 1 0 0 g( ) = g( ) = 0 0 1 −1 B B B B

• ...and hence gets a totally diﬀerent matrix:

0 1 1 0 vs. 1 0 0 −1 S B

A. Kissinger Version: autumn 2017 Matrix Calculations 29 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Transforming bases, part II

Theorem n Assume again we have two bases B, C for R . n n If a linear map f : R → R has matrix A w.r.t. to basis B, then, w.r.t. to basis C, f has matrix A0 : 0 A = T B⇒C · A · T C⇒B

Thus, via T B⇒C and T C⇒B one tranforms B-matrices into C-matrices. In particular, a matrix can be translated from the standard basis to basis B via:

0 A = T S⇒B · A · T B⇒S

A. Kissinger Version: autumn 2017 Matrix Calculations 30 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Example basis transformation, part I

2 • Consider the standard basis S = {(1, 0), (0, 1)} for R , and as alternative basis B = {(−1, 1), (0, 2)} 2 2 • Let the linear map f : R → R , w.r.t. the standard basis S, be given by the matrix: 1 −1 A = 2 3

• What is the representation A0 of f w.r.t. basis B? −1 0 • Since S is the standard basis, T = contains the B⇒S 1 2 B-vectors as its columns

A. Kissinger Version: autumn 2017 Matrix Calculations 31 / 32 Determinants Change of basis Matrices and basis transformations Radboud University Nijmegen Example basis transformation, part II

• The basis transformation matrix T S⇒B in the other direction is obtained as matrix inverse: −1 0−1 2 0 −2 0 T = T −1 = = 1 = 1 S⇒B B⇒S 1 2 −2−0 −1 −1 2 1 1

• Hence: 0 A = T S⇒B · A · T B⇒S −2 0 1 −1 −1 0 = 1 · · 2 1 1 2 3 1 2 −2 2 −1 0 = 1 · 2 3 2 1 2 4 4 = 1 2 −1 4 2 2 = 1 − 2 2

A. Kissinger Version: autumn 2017 Matrix Calculations 32 / 32