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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.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 B:= (Ax – ux, Ax – vx) = (Ax – ux)*(Ax – vx) | where (, ) denotes the scalar product in C" and µ,v are real constants with µ < v. Show that if no eigenvalue lies between µ and v, then B > 0.
Transcribed Image Text: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 B:= (Ax – ux, Ax – vx) = (Ax – ux)*(Ax – vx) | where (, ) denotes the scalar product in C" and µ,v are real constants with µ < v. Show that if no eigenvalue lies between µ and v, then B > 0.
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