MATH 581D FINAL EXAM Autumn December 12, 2016

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1 MATH 58D FINAL EXAM Autumn 206 December 2, 206 NAME: SIGNATURE: Instructions: there are 6 problems on the final. Aim for solving 4 problems, but do as much as you can. Partial credit will be given on all 6. If you have questions about the formulation of any of the problems, ask me. Problem Number of points Points obtained Total

2 Problem. Let V and V 2 be subspaces of R n such that R n = V V 2 (the direct sum of V and V 2 ), but V and V 2 are not orthogonal. Let P be the orthogonal proection onto V and P 2 the orthogonal proection onto V 2. Show that a) I P P 2 is invertible (hint: P P 2 x = x implies x = 0.) b) Q = (I P P 2 ) P (I P P 2 ) is the proection onto V in the direction of V 2 (recall what that means in terms of the range and nullspace of Q). Points: if attempted, 3 for a), 6 for b) Solution. a) Assume P P 2 x = x for some x. Then by orthogonality P 2 x 2 2 = x 2 2 x P 2 x 2 2 x 2 2 and P (P 2 x) 2 2 = P 2 x 2 2 P 2 x P P 2 x 2 2 P 2 x 2 2, so P P 2 x 2 x 2. To have equality, the way P P 2 x = x requires, one must have both that x P 2 x = 0, which means that x V 2, and that P P 2 x = P 2 x, so P 2 x V. Since P 2 x V 2 also, P 2 x = 0. However, since x V 2, this can only happen if x = 0. P P 2 x = 0 for all x is not true. See the picture below. Since V V 2, if v V 2, P v is not necessarily 0. b) We must check three things: that Q is a proection (Q 2 = Q), that its range is V, and that its nullspace is V 2. First note that since P 2 = P. Q 2 = (I P P 2 ) P (I P P 2 )(I P P 2 ) P (I P P 2 ) = (I P P 2 ) P 2 (I P P 2 ) = (I P P 2 ) P (I P P 2 ),

3 Let now x = y + z, with y V and z V 2 be the unique decomposition for x. Then Qx = (I P P 2 ) P (I P P 2 )(y + z) = (I P P 2 ) P (y + z P P 2 y P z) (since P 2 z = z) = (I P P 2 ) (y + P z P P 2 y P z) (since P 2 = P and y V ) = (I P P 2 ) (y P P 2 y) = (I P P 2 ) (I P P 2 )y = y. Since x was chosen arbitrarily (and hence so are y and z), the above proves both that the range of Q is V and that the nullspace of Q is V 2.

4 Problem 2. ) Prove that M n (C n ) is a vector space over C of dimension n 2. 2) We call a complex-valued (over C) function f of an n n complex matrix Lipschitz if it has the property that f(a) f(b) c A B for some matrix norm and some constant c R. Explain why the choice of norm doesn t matter in this definition. [ ] 0 3) Use the matrix A = (for ɛ appropriately small) to illustrate the fact that ɛ 0 the largest eigenvalue of a (non-symmetric) matrix is NOT a Lipschitz function of the matrix. Points: for ), 4 for 2), 5 for 3) Solution. a) Matrix addition and multiplication by a scalar work element-by-element, so they are associative, commutative, and distributive. The all-zero matrix is the zero element for addition, and each matrix s inverse with respect to addition is obtained by changing the sign of each element. The scalar is the inverse element with respect to multiplication by a scalar. Alternately, you could note that the matrices E i, for i i, n, made out of all zeros except at the (i, ) entry, which is, are a basis. b) Since the set of all matrices is a finite-dimensional vector space (or dimension n 2 ), as we have learned, all norms on it are equivalent. Hence, if a function f is Lipschitz with constant c in norm α and the equivalence constant between norm α and norm β is M ( X α M X β for all X M n ), then f is Lipschitz in β with constant cm. c) Assume that the largest eigenvalue were a Lipschitz function, and let C be a constant which works with the Frobenius norm F (meaning λ max [ (A) ] λ max (B) C A 0 B F. But taking A as given in the problem and B = we see that, while 0 0 λ max (A) = ɛ, A B F = ɛ. Hence by assumption of Lipschitz property, ɛ Cɛ, and since C is independent of ɛ this will clearly fail for ɛ small enough. Contradiction.

5 Problem 3.. Let J λ denote an n n Jordan block, where λ C \ R. Calculate log(j λ ), defined appropriately through power series (in the entry-by-entry format log(j λ ) i,k =..., and using the canonical branch cut given above). Argue that this is valid for any λ C\R. 2. Let A M n be a matrix with eigenvalues λ,..., λ k of (algebraic) multiplicities m,..., m k, and let ϕ be an entire function (analytical everywhere on C). Prove or disprove: ϕ(a) is diagonalizable iff A is diagonalizable. 3. Find the Jordan form (NOT the Jordan decomposition, ust the form) for 2 0 A = Points: 4 for., 3 for 2., 3 for 3. Solution.. Denote by N the nilpotent corresponding to J λ, i.e., J λ = λi + N. Then N n+ = 0, for all 0 integer. The power series for the logarithm at 0 is log( + x) = = for log x at to log x = ( ) (x ) = = We will define (for λ C \ R appropriately small) log(j λ ) = = = = (( λ)i N)i, = ( ( λ) I + = k= ( λ) I + = = = (log λ)i + = ( x). ( ) )( λ) k ( ) k N k k min{,n } k= ( ) k ( ( n λ) I + k=( ) k k n k= k! (log(k) (λ))n k. ( ) x, so we switch that ( λ) k ( ) k N k,, ( ) )( λ) k N k k

6 In the above, log (k) (λ) is the kth complex derivative of log, evaluated at λ. This definition extends to the entire λ in the range, since over C \ R, log x is a holomorphic function, and we can define it for matrices via the resolvent, as we have seen in class, as log A = log ξ R A (ξ) dξ, 2πi γ with R A (ξ) = (A ξi) for every ξ, and γ being any closed simple curve which includes in its interior σ(a); in this case, λ. Note that the entry-by-entry format for log(j λ ) is 0, if i > k, log(j λ ) i,k = log(λ), if i = k, (k i)! (log(k i) (λ)), if i < k, since N has the property of being all 0 except on its th upper diagonal, which has all ones. 2. It should be clear that if A is diagonal, so is ϕ(a) (due to the Jordan canonical form decomposition and the fact that all Jordan blocks of a diagonal matrix have dimension ). However, the other way around is NOT true. Similarly to the calculation we did for log(j λ ), we can get that 0, if i > k, ϕ(j λ ) i,k = ϕ(λ), if i = k, (k i)! (ϕ(k i) (λ)), if i < k Again, in the above, ϕ (k) (λ) is the kth complex derivative of ϕ, evaluated at λ. [ ] 0 So if we pick a 2 2 Jordan block J 0 = and pick ϕ(z) = e 0 0 z z, then ϕ(j 0 ) = I 2. This shows that ϕ(a) being diagonalizable does not mean that A must be. 3. A quick calculation shows that the eigenvalues are 2, 2, and that 0 0 A 2I =, and this matrix has rank 2, hence there is only one eigenvector for eigenvalue 2, and the Jordan form of A looks like

7 Problem 4. Prove Hadamard s inequality: Given a matrix A M m n n det A A(:, ) 2, = where A(:, ) denotes the th column of A. (Hint: think in terms of the QR decomposition.) Solution. Recall that in the QR decomposition, the entry r ii is calculated as i r ii = v 2, where v = a i a i, q q, where a i is the ith column of A while q i is the ith column of Q. Also recall that q i = v/r ii. Due to the orthonormal property of q,..., q i, it follows that = v 2 = a i 2 a i, q 2 a i 2. i= On the other hand, since A = QR and thus det A = det Q det R, as det Q =, det A = det R = n i= r ii (as R is upper triangular). Thus, n n n det A = r ii a i 2 = A(:, ) 2. i= i= =

8 Problem 5. Let A M m n (C), m n, with A = invertible, and A 2 is (m n) n. [ A A 2 ]. Assume A is n n and (i) Show that A is full-rank. (ii) Let UΣV be the SVD of A. Partition U = [ U ] U 2 U 2 U 22 with U being n n and U 22 being (m n) (m n). Show that U is invertible. (Hint: use block matrix multiplication and express A in terms of submatrices of U, Σ, and V.) (iii) Use the expression for U found in (ii) to conclude that A 2 A 2, where we recall that 2 is the spectral norm and A is the pseudoinverse of A. Points: if attempted, 3 for (i), 3 for (ii), 3 for (iii) Solution. (i) Suppose A were not full-rank, then there would be some column A(:, i) which is a linear combination of the others. But then, the restriction of A to A would have the same column as a linear combination of the others. Contradiction, so A must be full-rank. (ii) We write [ ] [ ] [ ] A U U A = = 2 Σ V A 2 U 2 U 22 0 yields A = U ΣV. As A is invertible, all matrices on the right, including U, must also be. (iii) From the above, A = V Σ U. Since V is unitary and doesn t change the 2-norm of the matrices it multiplies, A 2 = Σ U 2. Recall that the matrix U is a (non-strict) contraction (as we learned, e.g., from the CS Decomposition). Thus U must be an expansion, i.e., U x 2 x 2 for all x. As it is also invertible, there is some x such that U x = e n, and for this x we get Σ U x 2 x 2 Thus, for this particular x, Σ U x 2 U x 2 U Σ e n 2 =. x 2 x 2 e n 2 σ n A 2 A x 2. x 2 σ n [ ] Σ Given that A = V U 0, it follows that A 2 = σ n, where σ... σ n are the singular values of A. Hence A 2 A 2.

9 Problem 6. Let ρ(a) be the spectral radius of the matrix A M n (C), and suppose ρ(a) <. Recall that a linear transformation L is a contraction in the vector norm if Lx x for all x C n.. Give an example of A with ρ(a) < which is not a contraction in the Euclidean norm Show that there exists a norm on C n such that A is a contraction in that norm. 3. Show that the canonical norm example in the above is inner-product-induced. Points: if attempted, 2 for., 3 for 2., 4 for 3. Solution. [ ] [ ] [ ] /2 3/ A = ; or A = ; both of them map 0 / norm; the first has ρ(a) = /2, the second has ρ(a) = 0. into vectors of higher 2. If ρ(a) <, let ɛ > 0 be small enough so that ρ(a) + ɛ <. We know that for any such ɛ there exists a norm so that A ρ(a) + ɛ, so for this norm A < and hence the norm is a contraction. 3. This one is too hard to do on a final, unless you knew very well the example. Sorry guys. Recall that the example for the norm in (ii) shown in class was to find a matrix S such that T = S AS is upper triangular and such that every entry T (i, ) with i has T (i, ) < ɛ. Then, the norm we chose in class was X = S XS, and this satisfies the fact that A < ρ(a) + ɛ. But if we pick instead ɛ/n in the above, and X = S XS 2, then it will still be true that A < ρ(a) + ɛ: T x 2 2 i ( t i x ) 2 i t ii 2 x i 2 + n(n )ɛ 2 /n 2 < ρ(a) 2 + ɛ 2 < (ρ(a) + ɛ) 2. So the 2-norm also works and has a clear inner product flavor. Now note that X = S XS 2 on matrices is a norm induced by the vector norm x = S x 2 : S S XSx 2 XS 2 = sup x 0 x 2 = sup y 0 XSy Sy Xy = sup y 0 y = X.

10 In turn, x = S x 2 is induced by the inner product x, S S x. This finishes the proof.

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