Some Useful Results in the Theory of Matrix Functions

Transcription

1 Some Useful Results in the Theory of Matrix Functions Nick Higham School of Mathematics The University of Manchester Matrix Analysis and Applications, CIRM Luminy, October 2007 MIMS Nick Higham Theory of Matrix Functions 1 / 14

2 Sherman Morrison Woodbury Formula If U, V C n p and I p + V A 1 U is nonsingular then (A + UV ) 1 = A 1 A 1 U(I p + V A 1 U) 1 V A 1. Question Why does this formula hold? MIMS Nick Higham Theory of Matrix Functions 2 / 14

3 Sherman Morrison Woodbury Formula If U, V C n p and I p + V A 1 U is nonsingular then (A + UV ) 1 = A 1 A 1 U(I p + V A 1 U) 1 V A 1. Question Why does this formula hold? Obtained using A + UV = A(I + A 1 U V ) and (I m + AB) 1 = I A(I n + BA) 1 B { A C m n B C n m MIMS Nick Higham Theory of Matrix Functions 2 / 14

4 Identity (I + AB) 1 = I A(I + BA) 1 B. I = I + AB (I + AB)A(I + BA) 1 B = I + AB A(I + BA)(I + BA) 1 B = I + AB AB = I MIMS Nick Higham Theory of Matrix Functions 3 / 14

5 Identity (I + AB) 1 = I A(I + BA) 1 B. I = I + AB (I + AB)A(I + BA) 1 B = I + AB A(I + BA)(I + BA) 1 B = I + AB AB = I Key equation: (I + AB)A = A(I + BA), or MIMS Nick Higham Theory of Matrix Functions 3 / 14

6 Identity (I + AB) 1 = I A(I + BA) 1 B. I = I + AB (I + AB)A(I + BA) 1 B = I + AB A(I + BA)(I + BA) 1 B = I + AB AB = I Key equation: (I + AB)A = A(I + BA), or World s Most Fundamental Matrix Equation (AB)A = A(BA). MIMS Nick Higham Theory of Matrix Functions 3 / 14

7 Application of WMFME (AB)A = A(BA) (AB) 2 A = ABA(BA) = A(BA) 2. In general, for any poly p, p(ab)a = Ap(BA). Does the same hold for arbitrary f? MIMS Nick Higham Theory of Matrix Functions 4 / 14

8 f(ab) and f(ba) Lemma Let A C m n and B C n m and let f(ab) and f(ba) be defined. Then Af(BA) = f(ab)a. MIMS Nick Higham Theory of Matrix Functions 5 / 14

9 f(ab) and f(ba) Lemma Let A C m n and B C n m and let f(ab) and f(ba) be defined. Then Af(BA) = f(ab)a. Proof. There is a single polynomial p such that f(ab) = p(ab) and f(ba) = p(ba). Hence Af(BA) = Ap(BA) = p(ab)a = f(ab)a. MIMS Nick Higham Theory of Matrix Functions 5 / 14

10 Special Case Take f(t) = t 1/2. When AB (and hence also BA) has no eigenvalues on R, A(BA) 1/2 = (AB) 1/2 A. MIMS Nick Higham Theory of Matrix Functions 6 / 14

11 Special Case Take f(t) = t 1/2. When AB (and hence also BA) has no eigenvalues on R, Useful, but A(BA) 1/2 = (AB) 1/2 A. Af(BA) = f(ab)a cannot be solved for f(ba) in terms of f(ab). MIMS Nick Higham Theory of Matrix Functions 6 / 14

12 New Theorem Theorem (Harris 1993; H 2008) Let A C m n and B C n m, with m n, and assume BA is nonsingular. Then f(αi m + AB) = f(α)i m + A(BA) 1( f(αi n + BA) f(α)i n ) B. MIMS Nick Higham Theory of Matrix Functions 7 / 14

13 New Theorem Theorem (Harris 1993; H 2008) Let A C m n and B C n m, with m n, and assume BA is nonsingular. Then f(αi m + AB) = f(α)i m + A(BA) 1( f(αi n + BA) f(α)i n ) B. Proof. Let g(t) = t 1 (f(α + t) f(α)). Then f(αi + X) = f(α)i + Xg(X). Hence, using the lemma, f(αi m + AB) = f(α)i m + ABg(AB) = f(α)i m + Ag(BA)B = f(α)i m + A(BA) 1( f(αi n + BA) f(α)i n ) B. MIMS Nick Higham Theory of Matrix Functions 7 / 14

14 Example: Rank 2 Perturbation of I Consider f(αi n + uv + xy ), where u, v, x, y C n. Write [ ] v uv + xy = [ u x ] AB. y Then [ ] v C := BA = u v x y u y C 2 2. x f(αi n + uv + xy ) = f(α)i n + [ u x ] C 1( f(αi 2 + C) f(α)i 2 ) [ v y ] MIMS Nick Higham Theory of Matrix Functions 8 / 14

15 Example: Rank 2 Perturbation of I Consider f(αi n + uv + xy ), where u, v, x, y C n. Write [ ] v uv + xy = [ u x ] AB. y Then [ ] v C := BA = u v x y u y C 2 2. x f(αi n + uv + xy ) = f(α)i n + [ u x ] C 1( f(αi 2 + C) f(α)i 2 ) [ v For A C 2 2, f(a) = f(λ 1 )I + f[λ 1,λ 2 ](A λ 1 I). y ] MIMS Nick Higham Theory of Matrix Functions 8 / 14

16 Function of Jordan block A = Z diag(j 1,...,J p )Z 1 f(a) = Z diag(f(j 1 ),...,f(j p ))Z 1. λ k 1 λ J k = k Cm k m k, λ k f(λ k ) f (λ k ).... f(j k ) = f(λ k ).. f (m k 1) )(λ k ) (m k 1)!... f (λ k ) f(λ k ).. MIMS Nick Higham Theory of Matrix Functions 9 / 14

17 Theorem Let A C n n with eigenvalues λ k. 1 If f (λ k ) 0 then for every J(λ k ) in A there is a Jordan block of the same size in f(a) for f(λ k ). 2 Let f (λ k ) = f (λ k ) = = f (l 1) (λ k ) = 0 but f (l) (λ k ) 0, where l 2, and consider J(λ k ) of size r in A. (i) If l r, J(λ k ) splits into r 1 1 Jordan blocks for f(λ k ) in f(a). (ii) If l r 1, J(λ k ) splits into Jordan blocks for f(λ k ) in f(a) as follows: l q Jordan blocks of size p, q Jordan blocks of size p + 1, where r = lp + q with 0 q l 1, p > 0. MIMS Nick Higham Theory of Matrix Functions 10 / 14

18 Theorem Let A C n n with eigenvalues λ k. 1 If f (λ k ) 0 then for every J(λ k ) in A there is a Jordan block of the same size in f(a) for f(λ k ). 2 Let f (λ k ) = f (λ k ) = = f (l 1) (λ k ) = 0 but f (l) (λ k ) 0, where l 2, and consider J(λ k ) of size r in A. (i) If l r, J(λ k ) splits into r 1 1 Jordan blocks for f(λ k ) in f(a). (ii) If l r 1, J(λ k ) splits into Jordan blocks for f(λ k ) in f(a) as follows: l q Jordan blocks of size p, q Jordan blocks of size p + 1, where r = lp + q with 0 q l 1, p > 0. MIMS Nick Higham Theory of Matrix Functions 10 / 14

19 Theorem Let A C n n with eigenvalues λ k. 1 If f (λ k ) 0 then for every J(λ k ) in A there is a Jordan block of the same size in f(a) for f(λ k ). 2 Let f (λ k ) = f (λ k ) = = f (l 1) (λ k ) = 0 but f (l) (λ k ) 0, where l 2, and consider J(λ k ) of size r in A. (i) If l r, J(λ k ) splits into r 1 1 Jordan blocks for f(λ k ) in f(a). (ii) If l r 1, J(λ k ) splits into Jordan blocks for f(λ k ) in f(a) as follows: l q Jordan blocks of size p, q Jordan blocks of size p + 1, where r = lp + q with 0 q l 1, p > 0. MIMS Nick Higham Theory of Matrix Functions 10 / 14

20 Application: Matrix Equation Does the equation 1 a a... a 1 a... a cosh(x) = a = A Cn n 1 have a solution for a 0 and n > 1? MIMS Nick Higham Theory of Matrix Functions 11 / 14

21 Application: Matrix Equation Does the equation 1 a a... a 1 a... a cosh(x) = a = A Cn n 1 have a solution for a 0 and n > 1? No! A has just one Jordan block. Λ(X) = {cosh 1 (1)} = {0}. But f (0) = sinh(0) = 0, so cosh(x) must have more than one Jordan block. MIMS Nick Higham Theory of Matrix Functions 11 / 14

22 Application: Matrix Square Root Show that if A C n n has a defective zero eigenvalue then A does not have a square root that is a polynomial in A. MIMS Nick Higham Theory of Matrix Functions 12 / 14

23 Application: Matrix Square Root Show that if A C n n has a defective zero eigenvalue then A does not have a square root that is a polynomial in A. Solution Key fact: squaring a matrix leaves intact the number and size of Jordan blocks for λ 0 but splits blocks for λ = 0. Hence defective zero eigenvalue of A X has Jordan block 3 3 for λ = 0. But then X p(a), because A = X 2 has more Jordan blocks than X. MIMS Nick Higham Theory of Matrix Functions 12 / 14

24 Function of Block Triangular Matrix (1) Theorem A = [ n 1 1 ] n 1 B c 1 0 λ where g(z) = f[z,λ]. f(a) = [ f(b) g(b)c 0 f(λ) If λ / Λ(B) then g(b) = (B λi) 1 (f(b) f(λ)i). ] Theorem f ([ ]) X E = 0 X [ ] f(x) L(X, E). 0 f(x) Recall, Fréchet derivative L: f(x + E) f(x) L(X, E) = o( E ). MIMS Nick Higham Theory of Matrix Functions 13 / 14

25 Function of Block Triangular Matrix (2) Theorem Let [ A11 A A = 12 0 A 22 ] [ ] A11 0, D =, N = 0 A 22 [ ] 0 A Then f(a) = f(d) + L(D, N) (i.e., o( ) term in Frechét definition is zero). MIMS Nick Higham Theory of Matrix Functions 14 / 14

26 Bibliography I Lawrence A. Harris. Computation of functions of certain operator matrices. Linear Algebra Appl., 194:31 34, Nicholas J. Higham. Functions of Matrices: Theory and Computation. Society for Industrial and Applied Mathematics, Philadelphia, PA, USA, Book to appear. MIMS Nick Higham Theory of Matrix Functions 14 / 14