Math 250 Problem Set 7 Solutions

Math 250 Problem Set 7 Solutions - Math 250 quad Fall 2006...

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Math 250 quad Fall 2006 Selected Solutions HH 3.7.10: Computing Operator Norm Let A be an ` × k matrix. Let A T denote the transpose of A . (a) Show that A T A is symmetric. (b) Show that all eigenvalues of A T A are non-negative, and that they are all positive if and only if the kernel of A is zero (i.e., A if and only if A is one-to-one). (c) Show that || A || = max λ λ , where λ ranges over all eigenvalues of A T A . Response. For part (a), we appeal to the easily verified fact that the transpose map A A T is antimultiplicative: ( AB ) T = B T A T . Given this, we can compute that ( A T A ) T = ( A T )( A T ) T = A T A . This means that A T A equals its own transpose, i.e., it is symmetric. (b) Suppose that λ is an eigenvalue of A T A , with corresponding eigenvector v . (We assume that v 6 = 0. A number is not called an eigenvalue unless there is a non-zero eigenvector attached to it.) Then ( A T A ) v = λ v . Take the inner product of this equation with v , and use the fact that v u = u T v . Then, taking into account also that matrix multiplication is associative, we see that ( A T A ) v v = v T ( A T A v ) = ( v T A T )( A v ) = ( A v ) T ( A v ) = A ( v ) A ( v ) . On the other hand if A T A ( v ) = λ v , then ( A T A )( v ) v = λ v v , so we can write λ = A v A v v v , and this is clearly non-negative. It is positive unless A v = 0, that is, unless A annihilates the non-zero vector v . (c) We can rewrite the expression A v A v v v = | A v | 2 | v | 2 = ± | A v | | v | ² 2 . In the final form, we recognize this as the square of the stretching factor for v by the linear transformation A . The operator norm || A || is by definition the largest possible value over v of this stretching factor. As is discussed in the note on “Operator Norm”, this may also be taken as the maximum over the unit sphere S 1 = { v : | v | = 1 } of the function | A v | . Clearly this is the same as the square root of the maximum value over S 1 of the function | A v | 2 = ( A v ) ( A v ) = ( A T A ) v v = Q A T A ( v ). We can think of finding this maximum value as a constrained maximization problem: maximize Q A T A subject to the condition Q I ( v ) = v v = 1. We can apply the doctrine of Lagrange multipliers to this problem. This says that at the maximum
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This note was uploaded on 07/18/2008 for the course MATH 250 taught by Professor Rogerhowe during the Fall '06 term at Yale.

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Math 250 Problem Set 7 Solutions - Math 250 quad Fall 2006...

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