18.06 (Spring 14) Problem Set 9

This problem set is due Thursday, May 1, 2014 by 4pm in E17-131. The problems are out of the 4th edition of the textbook. This homework has 7 questions worth 70 points in total. Please WRITE NEATLY. You may discuss with others (and your TA), but you must turn in your own writing.

1. Problem 6.6.17 P.362. True or false, with a good reason:

(a) A symmetric matrix can’t be similar to a nonsymmetric matrix. (b) An invertible matrix can’t be similar to a singular matrix. (c) A can’t be similar to −A unless A = 0. (d) A can’t be similar to A + I.

solution

(a) False. Let 0 0 1 −1 1 1 A = ,P = ,P −1 = 0 1 0 1 0 1 then we have 1 1 0 0 1 −1 0 1 P −1AP = = , 0 1 0 1 0 1 0 1

which is not symmetric. (b) True. Let A, P be nonsingular matrices. Suppose −1P AP is singular, v 6= 0 is in the null space of P −1AP . We have P −1AP v = 0. Multiply by P on the left yields AP v = 0. Since P is nonsingular, v 6= 0, we have P v 6= 0 and is in the nullspace of A. We get a contradiction. (c) False. Let Let

−1 0 1 0 0 1 A = , −A = P = P −1 = 0 1 0 −1 1 0

then we have P −1AP = −A. The key is that A and −A could have the same set of eigenvalues. (d) True. Let v be a eigenvector of A with λ the corresponding eigenvalue(Both of them may be complex). Suppose there is a nonsingular P such that P −1AP = A + I. We have

AP −1v + P −1v = (A + I)(P −1v) = P −1AP (P −1v) = P −1Av = λP −1v

Therefore, we have

AP −1v = λP −1Av − P −1v = (λ − 1)P −1v

1 Since P −1v 6= 0, λ − 1 is also an eigenvalue of A. Use the same reasoning, we get λ − k is an eigenvalue of A for any positive integer k. This contradict the fact that A can only have finite eigenvalues. Remark: The above argument works fine for matrices over R or C. If the field we are working on is finite, then the set {λ − k} may be finite. The argument above does not work and the statement may be false. However, it’s beyond the scope of 18.06.

2. Problem 6.6.18 P.362. If B is invertible, prove that AB is similar to BA. They have the same eigenvalues.

solution

Since B is invertible, B−1 exists and is invertible. (B−1)−1 = B. Therefore, (B−1)−1AB(B−1) = BA. AB and BA are similar. From P.355, we can conclude they have the same eigenvalues.

T 3. Problem 6.7.1 P.371. Find the eigenvalues and unit eigenvectors v1, v2 of A A. Then ﬁnd u1 = Av1/σ1:

1 2 10 20 5 15 A = ,AT A = , AAT = . 3 6 20 40 15 45

T Verify that u1 is a unit eigenvector of AA . Complete the matrices U, Σ, V . 1 2 σ 0 T = u u 1 v v . 3 6 1 2 0 0 1 2

solution

T 2 The characteristic polynomial for A A is λ − 50λ = 0. The eigenvalues are λ1 = 50, 1 T 1 T λ2 = 0. The corresponding unit eigenvectors are v1 = √ (1, 2) , v2 = √ (2, −1) . √ 5 5 We have σ = 50. u = Av /σ = √1 (1, 3)T . Choose u from the null space of AT , 1 1 1 1 10 2 we have u = √1 (3, −1)T . Therefore, we have 2 10 √ 1 1 −3 50 0 1 1 2 U = √ , Σ = ,V = √ . 10 3 1 0 0 5 2 −1

4. Problem 6.7.14 P.372. Suppose A is invertible (with σ1 > σ2 > 0). Change A by as small a matrix as possible to produce a singular matrix A0. Hint: U and V do not change: σ1 0 T From A = u1 u2 v1 v2 , ﬁnd the nearest A0. 0 σ2

2 solution

The nearest matrix is obtained by letting σ2 to be 0, σ 0 T which is A = u u 1 v v . 0 1 2 0 0 1 2 0 0 T T The distance from A to A0 is u1 u2 v1 v2 = u2v2 . 0 σ2 5. Problem 10.2.3 P.506. Solve Az = 0 to ﬁnd a vector in the nullspace of

i 1 i A = . 1 i i

Show that z is orthogonal to the columns of AH . Show that z is not orthogonal to the columns of AT . The good row space is no longer C(AT ). Now it is C(AH ).

solution

Use the elimination to get the upper triangular matrix

i 1 i U = , 0 2i i − 1

and the reduced row echelon form 1 1 0 2 (1 + i) R = 1 . 0 1 2 (1 + i) The free variable is the third variable. A basis of the homogeneous solution is given 1 1 T H T T by z = (− 2 (1 + i), − 2 (1 + i), 1) . The columns of A are (−i, 1, −i) , (1, −i, −i) . Since the inner product of any u, v is deﬁned as uH v. We can conclude that the null space is orthogonal to C(AH ). Consider the inner product with the ﬁrst row. 1 1 1 − i i − 1 (− (1 − i), − (1 − i), 1)(i, 1, i) = + + i = i 2 2 2 2 Therefore, z is not orthogonal to C(AT ).

6. Problem 10.2.6 P.507. True or false (give a reason if true or a counterexample if false):

(a) If A is a real matrix then A + iI is invertible. (b) If A is a Hermitian matrix then A + iI is invertible. (c) If A is a unitary matrix then A + iI is invertible.

3 solution

(a) False. Let 0 1 A = . −1 0 We have i 1 A + iI = . −1 i Let v = (i, 1)T . We have (A + iI)v = 0. Therefore, A + iI is not invertible. (b) True. If A is Hermitian, then A is unitarily diagonizable and all the eigenvalues are real. Therefore, all the eigenvalues of A+iI has imaginary part 1. Therefore, A + iI cannot have 0 as its eigenvalue and therefore invertible. (c) False. The counterexample in part (a) is also unitary. It can also serve as a counterexample for part (c).

7. Problem 10.2.19 P.508. The functions e−ix and eix are orthogonal on the interval R 2π 0 ≤ x ≤ 2π because their inner product is 0 = 0.

solution Z 2π Z 2π Z 2π −ix ix −ix ¯ix −ix −ix −2ix i −2ix 2π he , e i = e e dx = e e dx = e dx = e |0 = 0. 0 0 0 2