Derivatives of Matrix Functions and their Norms.


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1 Derivatives of Matrix Functions and their Norms. Priyanka Grover Indian Statistical Institute, Delhi India February 24, / 41
2 Notations H : ndimensional complex Hilbert space. L (H) : the space of bounded linear operators on H. A L (H) : identified with n n matrix. S k : the group of permutations on k symbols. L k (X; Y) : the space of continuous klinear mappings of X X into Y. (X, Y are Banach spaces.) 2 / 41
3 Derivative φ : X Y φ is called (Fréchet) differentiable at u if there exists a linear transformation Dφ(u) : X Y such that for all v φ(u + v) φ(u) Dφ(u)(v) = o( v ). (1) The linear operator Dφ(u) is called the derivative of φ at u. If φ is differentiable at u, then for every v X, Dφ(u)(v) = d dt φ(u + tv). t=0 3 / 41
4 Dφ : X L (X; Y) φ is twice differentiable at u if Dφ is differentiable at u. The derivative is the second derivative of φ at u, written as D 2 φ(u). D 2 φ(u) L (X; L (X; Y)). L (X; L (X; Y)) is identified with L 2 (X; Y), the space of continuous bilinear mappings of X X into Y. The action of D 2 φ(u) on (v 1, v 2 ) is given by D 2 φ(u)(v 1, v 2 ) = 2 φ(u + t 1 v 1 + t 2 v 2 ). t 1 t 2 t1 =t 2 =0 4 / 41
5 One can similarly define D k φ at any u. D k φ(u) L k (X; Y). Its action at (v 1,..., v k ) is given by D k φ(u)(v 1,..., v k ) = k t 1 t k t1 = =t k =0 φ(u + t 1 v t k v k ). 5 / 41
6 Taylor s theorem : Let φ : X Y be a (p + 1)times differentiable map. For u X and for small h, φ(u + h) φ(u) p k=1 1 k! Dk φ(u)(h,..., h) = O( h p+1 ). From here, p 1 φ(u + h) φ(u) k! Dk φ(u) h k + O( h p+1 ). k=1 6 / 41
7 Mean Value theorem : Let φ : X Y be a differentiable map. Let u, v X and let L be the line segment joining them. Then φ(u) φ(v) u v sup Dφ(w). w L 7 / 41
8 Tensor power Similar to binomial theorem: m (A + X) = j i 0 j 1 + +j k =m ( j 1 A) ( j 2 X) ( j 3 A) ( j k X). For every X M(n), the expression for D m (A)(X) is the coefficient of t in m (A + tx). D m (A)(X) = X A... A + A X... A A A... X. 8 / 41
9 Norm: For all A M(n), D m (A) = m A m 1. Proof. : Triangle inequality and A X = A X. Equality at X = A/ A. 9 / 41
10 Antisymmetric tensor power m H is the range of the projection P m, defined as P m (x 1 x m ) = 1 sgn(σ) x σ(1) x σ(m). m! σ S m Inner product : x 1 x m = m! P m (x 1 x m ) x 1 x m, y 1 y m = det( x i, y j ). m H is invariant under m A. The operator m A is the restriction of m A to this subspace. It follows D m (A) m A m / 41
11 Theorem (Bhatia, Friedland; 1981) Let s 1 s 2 s n 0 be singular values of A. Then, for 1 m n, D m (A) = m p=1 j=1 j =p m s j = p m 1 (s 1,..., s m ), where p m 1 denotes the (m 1)th elementary symmetric polynomial in m variables. 11 / 41
12 Perturbation bound: By mean value theorem, Corollary For any two elements A, B of L (H), m (B) m (A) m M m 1 B A, where M = max( A, B ). 12 / 41
13 Application Determinant det A = n A. For any two elements A, B of L (H), det(b) det(a) n M n 1 B A, where M = max( A, B ). 13 / 41
14 Eigenvalues Definition (Distance between eigenvalues) For A, B L (H), let Eig A = {α 1,..., α n } and Eig B = {β 1,..., β n } denote their respective eigenvalues counted with multiplicity. A distance between these ntuples can be defined as Theorem For all A, B L (H). d(eiga, EigB) = min max σ S k 1 i n α i β σ(i). d(eiga, EigB) C M 1 1/n B A 1/n, where M = max( A, B ). 14 / 41
15 Symmetric tensor power m H, space of symmetric tensors, is the range of the projection Q m, defined as Q m (x 1 x m ) = 1 x σ(1) x σ(m). m! σ S m Inner product: x 1 x m = m! Q m (x 1 x m ) x 1 x m, y 1 y m = per( x i, y j ). Permanent: The permanent of A = (a ij ), written as per A, is defined by per A = a 1σ(1) a 2σ(2) a nσ(n), where the summation extends over all the permutations of 1,2,...,n. σ 15 / 41
16 m H is invariant under m A. The operator m A is the restriction of m A to this subspace. D m (A) m A m 1. D m A =? 16 / 41
17 Theorem (Bhatia; 1984) D m (A) = m A m 1. Using mean value theorem, Corollary For every A, B L (H), we have m A m B m M m 1 A B, where M = max( A, B ). 17 / 41
18 Permanent per A is one of the diagonal entries of n A : the (I, I)entry for I = (1, 2,..., n). Since, each entry of the matrix is dominated by its norm, we get Theorem For any A, B M(n), per A per B n M n 1 A B, where M = max( A, B ). 18 / 41
19 Higher order derivatives? We study higher order derivatives of these functions. kth order perturbation bounds follow by Taylor s theorem. 19 / 41
20 Tensor power Higher order derivative: For 1 k m, D k m (A)(X 1,..., X k ) = ( j 1 A) X σ(1) ( j k A) X σ(k) ( j k+1 A). σ S k j i 0 j 1 + +j k+1 =m k Norm: D k m (A) = m! (m k)! A m k. Perturbation bound: m (A + X) m A m m k A m k X k k=1 = ( A + X ) m A m. 20 / 41
21 Determinant Derivative : Jacobi s formula : D det A(X) = tr(adj(a)x), where adj(a) stands for the adjugate (the classical adjoint) of A. The following are three equivalent descriptions of Jacobi s formula. 1 adj(a) can be identified as an operator on n 1 H. Call this operator n 1 A. D det A(X) = tr(( n 1 A)X). 21 / 41
22 2 For 1 i, j n, let A(i j) be the (n 1) (n 1) matrix obtained from A by deleting its ith row and jth column. Then, D det A(X) = n ( 1) i+j det A(i j)x ij. i,j=1 3 For 1 j n, let A(j; X) be the matrix obtained from A by replacing the jth column of A by the jth column of X and keeping the rest of the columns unchanged. Then, D det A(X) = n det A(j; X). j=1 22 / 41
23 Higher order derivatives : Let Q k,n denote the set of multiindices I = (i 1,..., i k ) in which 1 i 1 < < i k n. For I, J Q k,n, A[I J ]: the k k submatrix obtained from A by picking its entries from the rows I and columns J. A(I J ): the (n k) (n k) submatrix obtained from A by deleting rows I and columns J. Given operators X 1,..., X k on H, consider the operator 1 X σ(1) X σ(2) X σ(k), k! σ S k on the space k H. This leaves the subspace k H invariant. The restriction to this subspace is denoted by X 1 X k. 23 / 41
24 I denotes the sum i i k. The transpose of the matrix with entries ( 1) I + J det A(I J ) can be identified as an operator on the space n k H. We call it n k A (it is unitarily similar to the transpose of n k A.) Theorem (Bhatia, Jain; 2009) 1 For 1 k n, D k det A(X 1,..., X k ) = k! tr[( n k A)(X 1 X k )]. When k = 1 1 D det A(X) = tr(( n 1 A)X). 24 / 41
25 Theorem 2 For 1 k n, we have D k det A(X 1,..., X k ) = ( 1) I + J det A(I J ) det Y σ [I J ], [J ] I,J Q k,n σ S k where Y σ [J ] : n n matrix whose j pth column is the j p th column of X σ(p) for 1 p k, and the remaining n k columns are zero. When k = 1, 2 n D det A(X) = ( 1) i+j det A(i j)x ij. i,j=1 25 / 41
26 Theorem 3 A(J ; X 1,..., X k ): matrix obtained from A by replacing the j p th column of A by the j p th column of X p for 1 p k, and keeping the rest of the columns unchanged. Then for 1 k n, we have D k det A(X 1,..., X k ) = det A(J ; X σ(1),..., X σ(k) ). J Q k,n When k = 1, σ S k 3 n D det A(X) = det A(j; X). j=1 26 / 41
27 Norm : Theorem Let s 1 s 2 s n 0 be singular values of A. Then, for 1 k n, we have D k det A = k! p n k (s 1,..., s n ). Perturbation bound : Corollary Let A and B be n n matrices. Then n det(a) det(b) p n k (s 1,..., s n ) A B k. k=1 27 / 41
28 Antisymmetric Tensor Power Theorem Let A M(n). Then for 1 k m n, D k m (A)(X 1,..., X k ) = k! det A[γ δ] (X 1 X k ) (m) (γ, δ). γ,δ Q m k,n γ α if {γ 1,..., γ k } {α 1,..., α m }. α γ denotes (γ 1,..., γ m k ) Q m k,n where {γ 1,..., γ m k } = {α 1,..., α m } \ {γ 1,..., γ k }. π α := the permutation on {1, 2,..., n} π α (α i ) = i for all i = 1,..., m. On α = (1,..., n) α, π α (α ) = m + j for all j = 1,..., n m. j (X 1 X k ) (m) (γ, δ) : n n m m matrix whose indexing set is Q m,n and for α, β Q m,n, the (α, β)entry is ( 1) π α(γ) + π β (δ) (α γ, β δ)entry of X 1 X k if γ α and δ β and is 0 otherwise. 28 / 41
29 Norm : Theorem Let A M(n). Then for 1 k m n, D k m A = k! p m k (s 1 (A),..., s m (A)), where s 1 (A) s n (A) are singular values of A. Perturbation bound : Corollary For n n matrices A and X, m (A + X) m A m p m k (s 1 (A),..., s m (A)) X k k=1 m m A m k X k k k=1 = ( A + X ) m A m. 29 / 41
30 Permanent We obtain three different expressions for all higher order derivatives of the permanent of a matrix. The permanental adjoint, denoted by padj (A), is the n n matrix whose (i, j)entry is per A(i j). Similar to the Jacobi formula : Theorem For each X M(n), Proof. D per (A)(X) = tr(padj(a) T X). D per A(X) is the coefficient of t in per (A + tx). per is a linear function of each of its columns. Combine the above two. 30 / 41
31 Let G k,n denote the set {(i 1,..., i k ) 1 i 1 i k n}. For k n, Q k,n is a subset of G k,n. If {e i } is an orthonormal basis of H, then for I = (i 1,..., i k ) G k,n, we define e I = e i1 e ik. Let P k be the canonical projection of k H onto the subspace generated by {e I : I Q k,n }. If we vary I, J in Q k,n, we get the submatrix P k ( k A)P k of k A. The matrix padj(a) T can be identified with a submatrix of an operator on the space n 1 H. We call this operator n 1 A. 1 D per A(X) = tr ((P n 1 ( n 1 A)P n 1 )X). 31 / 41
32 Given two elements I and J of G k,n, let A[I J ] denote the k k matrix whose (r, s)entry is the (i r, j s )entry of A. In general, A[I J ] is not a submatrix of A, unless I, J Q k,n. The Laplace expansion theorem for permanent says that for any I Q k,n, per A = per A[I J ] per A(I J ). J Q k,n In particular, for any i, 1 i n, per A = n a ij per (A(i j)). j=1 Using this, 32 / 41
33 2 n n D per (A)(X) = per A(i j)x ij. i=1 j=1 3 n D per (A)(X) = per A(j; X). j=1 33 / 41
34 Higher order derivatives : Again consider the following operator on k H: 1 X σ(1) X σ(2) X σ(k). k! σ S k It leaves the space k H invariant. We use the notation X 1 X 2 X k for the restriction of this operator to the subspace k H. The transpose of the matrix whose (I, J )entry is per A(I J ) can be identified as a submatrix of an operator on the space n k H. We call this operator n k A. Theorem 1 For 1 k n, D k per A(X 1,..., X k ) = k! tr [(P n k ( n k A)P n k )(P k (X 1 X k )P k )]. 34 / 41
35 Theorem 2 For 1 k n, D k per A(X 1,..., X k ) = In particular, per A(I J ) per Y σ [I J ]. [J ] I,J Q k,n σ S k D k per A(X,..., X) = k! I,J Q k,n per A(I J ) per X[I J ]. Y σ [J ] : n n matrix whose j pth column is the j p th column of X σ(p) for 1 p k, and the remaining n k columns are zero. 35 / 41
36 Theorem 3 For 1 k n, D k per A(X 1,..., X k ) = In particular, σ S k D k per A(X,..., X) = k! J Q k,n per A(J ; X σ(1), X σ(2),..., X σ(k) ). J Q k,n per A(J ; X,..., X). A(J ; X σ(1), X σ(2),..., X σ(k) ) : the matrix obtained from A by replacing the j p th column of A by the j p th column of X σ(p) and keeping the rest of the columns unchanged. 36 / 41
37 Theorem Let A be an n n matrix, we have n D k per A k! A n k. k Proof. D k per A = k! sup tr [(P n k ( n k A)P n k ) X 1 = = X k =1 k! P n k ( n k A) P n k 1 n k! P n k ( n k A) P n k k Use n k A = n k A = A n k. (P k (X 1 X k )P k )] 37 / 41
38 As a corollary, we obtain a perturbation bound for per. Corollary Let A and X be n n matrices. Then n n per (A + X) per A A n k X k. k k=1 Consider the simplest commutative case: A = I, X = xi. Then the expression on both the sides of the inequality is n n x k. k k=1 So no improvement on the bound is possible. 38 / 41
39 Symmetric tensor power γ α : {γ 1,..., γ k } {α 1,..., α m } such that α l occurs in α, say, d α times then α l cannot occur in γ for more than d α times. α γ denotes (γ 1,..., γ m k ) G m k,n where γ l {α 1,..., α m } and occurs in α γ exactly d α d γ times. (X 1 X k ) (m) (γ, δ): n+m 1 n+m 1 m m matrix whose indexing set is G m,n and for α, β G m,n, the (α, β)entry m(α γ)m(β δ) 1/2 is m(α)m(β) (α γ, β δ)entry of X 1 X k if γ α and δ β and zero otherwise. Theorem Let A M(n). Then for 1 k m n, D k m (A)(X 1,..., X k ) = k! per A[γ δ] (X 1 X k ) (m) (γ, δ). γ,δ G m k,n 39 / 41
40 Norm : Theorem Let A M(n). Then for 1 k m n, D k m A = m! (m k)! A m k. Perturbation bound : Corollary For n n matrices A and X, m (A + X) m A ( A + X ) m A m. 40 / 41
41 THANK YOU! 41 / 41
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