Question

I need help with Q12) please

and eigenvectors of the row-echelon matrix VWV) 37dldl IV 31076 IW NO LOHS 1 U = 2 -4 0 2 1 0 0 3 0 0 0 3 --3 3 5 d the eigenvalues and eigenvectors of the following matrices. a) A= 1 3 0 2 2 0 0 0 6 3 0 b) B= 0 -4 0 6 0 -1 3 Problems 8.2 : Eigenvectors, bases, and diagonalisation 11. [R] For each of the matrices in Questions 7, 9 and 10, decide if the matrix is diagonalisable, and if it is find an invertible matrix M and a diagonal matrix D such that D = M-AM. 12. [H] Show that if I is an eigenvalue of A then I is also an eigenvalue of the matrix A' = B-AB, where B is any invertible matrix. Also show that if v is an eigenvector of A for eigenvalue then B-lv is an eigenvector of A' for eigenvalue .. © 2019 School of Mathematics and Statistics, UNSW Sydney

Answer 1

Solution:

$\small 12.\: \left [ H \right ]$

Consider $\small A'=B^{-1}AB$

$\small A'-\lambda I=B^{-1}AB-\lambda I$

Because $\small B^{-1}\left ( \lambda I \right )B=\lambda B^{-1}B=\lambda I$

$\small \therefore A'-\lambda I=B^{-1}AB-B^{-1}\left ( \lambda I \right )B$

$\small \therefore A'-\lambda I=B^{-1}\left ( A-\lambda I \right )B$

$\small \therefore \mathrm{det}\left ( A'-\lambda I \right )=\mathrm{det}\left ( B^{-1}\left ( A-\lambda I \right )B \right )$

$\small \therefore \mathrm{det}\left ( A'-\lambda I \right )=\mathrm{det}\left ( B^{-1} \right )\cdot \mathrm{det} \left ( A-\lambda I \right )\cdot \mathrm{det}\left ( B \right )$

$\small \therefore \mathrm{det}\left ( A'-\lambda I \right )=\frac{1}{\mathrm{det}\left ( B \right )}\cdot \mathrm{det} \left ( A-\lambda I \right )\cdot \mathrm{det}\left ( B \right )$

$\small \therefore \mathrm{det}\left ( A'-\lambda I \right )= \mathrm{det} \left ( A-\lambda I \right )$

Thus, the matrices  $\small A$ and $\small A'$ have the same characteristic polynomial and hence, the same eigenvalues.

$\small v$ is an eigenvector of $\small A$ for eigenvalue $\small \lambda$

$\small \therefore Av=\lambda v$

Consider $\small A'\left ( B^{-1}v \right )=B^{-1}AB\left ( B^{-1} v\right )$

$\small \therefore A'\left ( B^{-1}v \right )=B^{-1}A\left ( BB^{-1} \right )v$

$\small \therefore A'\left ( B^{-1}v \right )=B^{-1}A\left ( I \right )v$

$\small \therefore A'\left ( B^{-1}v \right )=B^{-1}Av$

$\small \therefore A'\left ( B^{-1}v \right )=B^{-1}\lambda v$

$\small \therefore A'\left ( B^{-1}v \right )=\lambda B^{-1} v$

This shows that $\small B^{-1}v$ is an eigenvector of  $\small A'$ for eigenvalue $\small \lambda .$