UCSD ECE269 Handout #8 Prof Young-Han Kim Monday, March 9, 208 Final Examination (Total: 30 points) There are 5 problems, each with multiple parts Your answer should be as clear and succinct as possible You may include a simple justification to your answer in the space provided, but please note that an erroneous argument could count against an otherwise correct answer True or false (5 points for each correct answer, -5 points for each wrong answer, 0 point for each blank) Fill in each blank with true or false A statement is considered true if it holds for all matrices satisfying the conditions stated, and false otherwise For example, A nonsingular matrix is invertible A strictly tall matrix is onto False True Here the matrix dimensions are such that each expression makes sense, but they are otherwise unspecified (a) A diagonalizable matrix A is nonsingular False As a counterexample, the matrix A = 0 is diagonalizable but singular (b) A nonsingular matrix A is diagonalizable False [ ] As a counterexample, the matrix A = is nonsingular but is not diagonalizable 0 (c) A positive square matrix A is positive definite False [ ] 3 5 As a counterexample, the matrix A = is positive, but for x = 5 3 2 Thus, A is not positive definite [ 2 2 ], x T Ax = (d) A positive definite and Hermitian matrix A is invertible True For a positive definite Hermitian matrix A = UΛU H, A is given by UΛ U H (e) A square matrix A with real and positive eigenvalues is positive definite False [ ] 0 As a counterexample, consider A = The eigenvalue of A is (with multiplicity 0 [ ] two) but for x =, x T Ax = 8 Hence A is not positive definite
(f) For a square matrix A, rank(a) rank(e A ) True Using properties of the matrix exponential from Homework #5, e A e A = e A A = e 0 = I Thus, e A is invertible, which implies that e A is full-rank and hence rank(e A ) rank(a) (g) For a square matrix A, e A I e A True Using the definition of e A, e A I = A + A2 2! + From properties of the spectral norm it follows that e A I = A + A2 2! A 2 + A + A2 + A + + = e A 2! 2! (h) If A is nilpotent, then I A is invertible True The only unique eigenvalue of a nilpotent matrix A is 0 (cf Question 4, Homework #6) and thus the only unique eigenvalue of I A is This implies that I A has no zero eigenvalue, and is thus invertible (i) If A C n n such that A <, then I A is invertible True For I A to be non-invertible, A has to have as an eigenvalue eigenvalue of A with the associated eigenvector v Then, Assume is an A = max x 0 Ax x Av v = v v =, which contradicts the fact that A < Thus, A cannot have an eigenvalue of, implying that I A is invertible (j) Every eigenvalue λ of a complex square matrix A satisfies λ A True Consider an eigenvalue λ of A with associated eigenvector v By definition of A, A = max x 0 Ax x Av v = λ v v = λ Thus for all eigenvalues λ of A, λ A 2
2 Perpendicular distances (30 points) For v, w R n with w =, define the line L in R n that goes through v in the direction w as L = {v + wt : t R} (a) Let z R n Which point x on the line L is the closest to z? Your answer should be in terms of v, w, and z x = v + ww T (z v) = v + w, z v w This can can be formulated as a least-squares problem whose solution is min t wt (z v) 2, t = (w T w) w T (z v) = w 2 wt (z v) = w T (z v), which leads to x = v + wt = v + ww T (z v) (b) Find the distance from z to the closest point x on L in terms of v, w, and z d(l, z) = (z v) T (I ww T ) (z v) We have d(l, z) 2 = z x 2 = z v ww T (z v) 2 = z v 2 (z v) T ww T (z v) = (z v) T ( I ww T ) (z v) 3
Now consider two lines and where w = w 2 = L = {v + w t : t R} L 2 = {v 2 + w 2 t 2 : t 2 R}, (c) Find the distance between the two lines, that is, the smallest distance between a point x on L and a point x 2 on L 2, in terms of the vectors v, v 2, w, w 2, and any matrices involving them (Hint: You can consider the case w = ±w 2 separately) { (v2 v ) T (I W (W T W ) W T ) (v 2 v ), w ±w 2, d(l, L 2 ) = (v 2 v ) ( ) T I w w T (v2 v ), w = ±w 2, where W := [w w 2 ] The square of the distance can be found as the optimal valuefor the following leastsquares problem: min t R 2 W t b 2, where t := [t W := [ w w 2 ], t 2 ] T, and b := v 2 v Therefore, we have d(l, L 2 ) 2 = b Proj R(W ) (b) 2 = b W W + b 2 = b T ( I W W +) b Now, if w ±w 2, then W is tall and full-rank, therefore W W + = W (W T W ) W T, whence d(l, L 2 ) 2 = b T ( I W (W T W ) W T ) b If w = ±w 2, then W + = (/2)W T, so that W W + = (/2)W W T, whence d(l, L 2 ) 2 = b T ( I (/2)W W T ) b 4
3 Matrix its (20 points) (a) Let Find the it A = n An = [ ] /3 2/3 2/3 /3 n An [ ] /2 /2 /2 /2 For any primitive matrix A with Perron Frobenius eigenvalue λ, n ( A λ ) n vw T where v and w are respectively the right and left eigenvectors associated with λ, normalized so that v T w = In this case, since A is the transition [ ] matrix of an irreducible and aperiodic Markov chain, λ =, v is the all-ones vector, and w is the stationary distribution [ /2 /2 ] Then, ( ) A n [ ] [ ] [/2 ] /2 /2 = n n An = /2 = /2 /2 (b) For 0 < α, β, α β <, let α β α β B = α β α β β α β α Find n Bn /3 /3 /3 n Bn = /3 /3 /3 /3 /3 /3 Following the reasoning in Part (a), B is also the transition matrix of an irreducible, aperiodic Markov chain with a state space of size 3 By symmetry, it is straightforward to verify that [ /3 /3 /3 ] is the stationary distribution Hence, in this case λ =, v = and w = [ /3 /3 /3 ] Then, n Bn = [ /3 /3 /3 ] /3 /3 /3 = /3 /3 /3 /3 /3 /3 5
4 Jordan canonical form (20 points) Suppose that a matrix A R n n has distinct eigenvalues λ,, λ n Let [ ] A I B = 0 A (a) Find the eigenvalues of B (including multiplicity) B has eigenvalues λ,, λ n, each with multiplicity 2 Since A has distinct eigenvalues, A is diagonalizable and we can write A = T ΛT, where Λ := diag(λ,, λ n ) We therfore have [ ] A I B = 0 A [ ] [ ] [ ] T 0 Λ I T 0 = 0 T 0 Λ 0 T [ ] [ ] [ ] T 0 Λ I T 0 = 0 T 0 Λ 0 T Therefore, B is similar to the upper-triangular matrix [ ] Λ I 0 Λ and therefore has the eigenvalues λ,, λ n, each with multiplicity 2 (b) Find the Jordan blocks of B J J 2 J =, Jn with [ ] λi J i := 0 λ i We have [ ] [ ] [ ] T 0 Λ I T 0 B = 0 T 0 Λ 0 T 6
Letting v,, v n be the columns of T, this relation can also be written as B = T J T, where and [ ] v 0 v T = 2 0 v n 0 0 v 0 v 2 0 v n J J 2 J =, Jn with [ ] λi J i := 0 λ i 7
5 Matrix polynomial (0 points) Is there a real square matrix A of any size such that A 2 + 2A + 5I = 0? If yes, find one such matrix If not, prove that no such A exists Yes, real square matrices satisfying A 2 + 5A + I exist One example is A = [ ] 2 2 By the Cayley Hamilton theorem, any matrix satisfies its own characteristic equation Consider all 2 2 matrices For a 2 2 matrix A to satisfy A 2 + 2A + 5I = 0, it needs to have tr(a) = 2 and det(a) = 5 One such possible matrix is [ ] 2 A = 2 It can be verified [ that ] for this A, A 2 +2A+5I = 0 Note that this is not a unique solution as 4 an example, is also a valid solution 8