Homework 9. Ha Pham. December 6, 2008

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1 Homework 9 Ha Pham December 6, 2008 Problem (Ch7 - Problem 30). Suppose S L(V ). Prove that S is an isometry if and only if all the singular values of S equal. Proof. S S is self-adjoint operator with non-negative eigenvalues, ie a positive operator. By theorem 7.27-Axler, S S has a unique positive square root denoted by S S. If S is an isometry then S S = Id by Theorem Axler. Hence we have S S = I, all of whose eigenvalues are then exactly those of Id, whis comprises of only. By definition, the singular values of S are the eigenvalues of S S, thus are equal. On the other hand, suppose all the singular values of S equal to, ie all the eigenvalues of S S are. S S is self-adjoint thus diagonalizable. This together with the fact it only has eigenvalues forces S S = Id S S = Id. By theorem 7.36-Axler, this is equivalent to S being an isometry. Problem 2 (Ch 7 - Ex 3). Suppose T, T 2 L(V ). Prove that T and T 2 have the same singular values if and only if there exist isometries S, S 2 L(V ) such that T = S T 2 S 2. Proof. T, T 2 have the same singular values decomposition s,..., s n. By The Singular- Value Decomposition theorem (thm Axler), there exists orthonormal bases (e,..., e n ) and (ẽ i,..., ẽ n ) and isometries S and S such that for all v V, we have T v = s v, e Se s n v, e n Se n T 2 v = s v, ẽ Sẽ s n v, ẽ n Sẽ n Guesswork : from the requirement T = S T 2 S 2, we would like to find isometries S, S 2 so that S 2 v, ẽ i S Sẽi = v, e i Se i ( )

2 Hence it is reasonable to define S as the linear transformation S Sẽi = Se i Note that (ẽ,..., e n ) is an orthonormal basis and S is an isometry so ( Sẽ,..., Sẽ n ) is also an orthonormal basis. Similarly for (Se,..., Se n ). Hence S maps an orthonormal basis to an orthonormal basis. By theorem 7.36 in Axler, this shows that S is an isometry. For S 2, we define S 2 as the linear transformation such that S 2 ẽ i, ẽ j = ẽ i, e j This gives the values of S 2 ẽ i for each i since we have specified the coefficients when S 2 ẽ i is written as a linear combination of the basis elements (ẽ,..., ẽ n ). It remains to verify that S 2 is also an isometry. We need to check S 2 ẽ i, S 2 ẽ j = ẽ j, ẽ j = δ ij LHS is equal to S 2 ẽ i, ẽ k S 2 ẽ j, ẽ l δ kl = ẽ i, e k ẽ j, e k k,l k = ẽ i, ẽ j = δ ij Hence we have found isometries S, S 2 with the required property : S T 2 S 2 ẽ i = j s j S 2 ẽ i, ẽ j S Sẽj = j s j ẽ i, e j Se j = T ẽ i Since this true for each basis elements, this equality holds for all v V hence T = S T 2 S 2 Problem 3 (Ch 7 - Ex 32). Suppose T L(V ) has singular-value decomposition given by T v = s v, e f s n v, e n f n 2

3 . Prove that for every v V T v = s v, f e s n v, f n e n 2. Prove that if T is invertible, then for every v V Proof. T v = v, f e v, f n s. (e,..., e n ) and (f,..., f n ) are orthonormal basis. Hence from the expression s n e n T v = s v, e f s n v, e n f n the matrix representation of T : (V, (e,..., e n )) (V, (f,..., f n )) is the diagonal matrix with entries s,..., s n s s M = s n Hence the matrix reprensentation of T : (V, (f,..., f n )) (V, (e,..., e n )) is the transpose of the above matrix M which means exactly for all v V T v = s v, f e s n v, f n e n 2. The matrix representation of T : (V, (f,..., f n )) (V, (e,..., e n )) is then the inverse of matrix M which is s M 0 s = s n which means that for all v V T v = v, f e v, f n s s n e n 3

4 Problem 4 (Chapter 7 - ex 33). Suppose T L(V ). Let ŝ denote the smallest singular value of T, and let s denote the largest singular value of T. Prove that for every v V ŝ v T v s v Proof. We have the singular value decomposition for T. T v = s v, e f s n v, e n f n where s,..., s n are singular values of T and (e,..., e n ) and (f,..., f n ) are orthonormal bases of V. Since (f,..., f n ) is an orthonormal basis, we have T v 2 = (s v, e ) (s n v, e n ) 2 s 2 [ ( v, e ) ( v, e n ) 2] = s 2 v 2 Similarly for the other inequality we have T v 2 = (s v, e ) (s n v, e n ) 2 ŝ 2 [ ( v, e ) ( v, e n ) 2] = ŝ 2 v 2 ŝ v T v s v Problem 5 (Chapter 7 - ex 34). Suppose T, T L(V ). Let s, s, s denote the largest singular values of T, T, T + T correspondingly. Prove that s s + s. Proof. For all v V by triangle s inequality and previous excercise we have (T + T )(v) T v + T v s v + s v = (s + s ) v Now we can choose v to be e i in the orthonormal basis in the singular decomposition so that (T + T )e i = s i e i, e i f i and s i = s. Thus the inequality above becomes s s + s 4

5 Problem 6 (Extra problem). Proof. If an operator T is invertible then the smallest singular value ŝ of T is s T, where s T is the largest singular value for T. If T is noninvertible then its smallest singular value is 0.. Consider the singular values of the product of two matrices. (a) Denote s, s, s to be the largest singular values for A, B, AB correspondingly. We want to show that s s s. By ex 33, we have ABv = A(Bv) s Bv s s v Now pick v to be the e i in the orthonormal basis in the singular decomposition so that (AB)e i = s i e i, e i f i and s i = s. Thus the inequality above becomes s s s. (b) Denote ŝ, ŝ, ŝ to be the smallest singular values for A, B, AB correspondingly. We want to show s s s. From a previous hw we have that AB is invertible if and only if A and B are both invertible. If AB is not invertible then either A or B is not invertible. For an invertible operator one of its singular value has to be 0 hence the smallest singular value is also 0. Hence both sides of the inequality are 0. If AB is invertible, so are both A and B. From the work shown above for the largest singular value we have Hence in either case, we have ŝ ŝ ŝ s T s T s T ŝ ŝ ŝ ŝ ŝ ŝ (c) The inequality established for the largest singular value does not hold for second largest. Consider the following 2 2 matrices. ( ) ( ) In this case, we have the second largest singular value of the product is 2 is greater than the product of the second largest singular values of each matrix, which is. (d) The inequality that was established for smallest singular values does not hold in the second smallestcase. Take for example ( ) ( )

6 The second smallest singular value of the product is 2 while the product of the second smallest singular values for each of the matrix is 4. Hence it is not true that 2 > Sum of matrices (a) The inequality established for the largest singular value does not hold in the case of second largest singular values. Consider the counter example with the matrices ( ) ( ) The second largest value for the sum is while the sum of second largest value is 0. Hence it is not true that (b) For the smallest singular values, we do not ŝ ŝ + ŝ. Consider the counterexample. ( ) ( ) The smallest singular value of the sum is 0 while the sum of the smallest singular values is 2 hence it is false that 0 + (c) For second smallest, consider the examples Here the second smallest singular value of the sum which is 4 is greater than the sum of the second smallest which is While for the case The second smallest singular value of the sum which is 2 is leass the sum of the second smallest which is 2. 6