6.Let A be an n x n hermitian matrix, i.e. A = A*. Assumethat all n eigenvalues are different. Then the normalized eigenvectors { v; : j =1,...,n} form an orthonormal basis in C". ConsiderB:= (Ax – ux, Ax – vx) = (Ax – ux)*(Ax – vx)|where (, ) denotes the scalar product in C" and µ,v are real constants withµ < v. Show that if no eigenvalue lies between µ and v, then B > 0.

Answered: Image /qna-images/answer/ebbe61cb-a87e-43ab-b4c7-0161e2e94752.jpg

6. Let A be an n x n hermitian matrix, i.e. A = A*. Assume that all n eigenvalues are different. Then the normalized eigenvectors { v; : j = 1,...,n} form an orthonormal basis in C". Consider В %3 (Ах — их, Ах - их) %3D (Ах — их)"(Ах — vх) | where (,) denotes the scalar product in C" and u,v are real constants with µ < v. Show that if no eigenvalue lies between µ and v, then B > 0.

6. Let A be an n x n hermitian matrix, i.e. A = A*. Assume that all n eigenvalues are different. Then the normalized eigenvectors { v; : j = 1,...,n} form an orthonormal basis in C". Consider В %3 (Ах — их, Ах - их) %3D (Ах — их)"(Ах — vх) | where (,) denotes the scalar product in C" and u,v are real constants with µ < v. Show that if no eigenvalue lies between µ and v, then B > 0.

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter7: Distance And Approximation

Section7.1: Inner Product Spaces

Problem 11AEXP

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