Question

Problem 1. (2 points) Find a 2 x 2 matrix A such that [-] and are eigenvectors of A with eigenvalues 9 and 3, respectively. A=

Answer 1

Solution:

Given matrix A has eigenvalues 9 and -3

$\lambda_1=9,~~ \lambda _2=-3$

Eigenvectors are

$v_1=\begin{bmatrix} 0\\ -5 \end{bmatrix},~~~~v_2=\begin{bmatrix} -2\\ -1 \end{bmatrix}$

From diagonalizable theorem

$A=S\sum S^{-1}\\\\S=\begin{bmatrix} 0 & -2\\ -5& -1 \end{bmatrix}\\\\\sum =\begin{bmatrix} 9 &0 \\ 0& -3 \end{bmatrix} \\\\S^{-1}=\left[ \begin{array}{cc} \frac{1}{10} & - \frac{1}{5} \\\\ - \frac{1}{2} & 0 \end{array} \right]$

$A=\begin{bmatrix} 0 & -2\\ -5& -1 \end{bmatrix}\cdot \begin{bmatrix} 9 &0 \\ 0& -3 \end{bmatrix} \cdot \left[ \begin{array}{cc} \frac{1}{10} & - \frac{1}{5} \\\\ - \frac{1}{2} & 0 \end{array} \right]\\\\$

$S\cdot \sum =\left[ \begin{array}{cc} 0 & -2 \\\\ -5 & -1 \end{array} \right] \cdot \left[ \begin{array}{cc} 9 & 0 \\\\ 0 & -3 \end{array} \right]=\left[ \begin{array}{cc} 0 & 6 \\\\ -45 & 3 \end{array} \right]$

$(S\cdot \sum )\cdot S^{-1}=\left[ \begin{array}{cc} 0 & 6 \\\\ -45 & 3 \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{1}{10} & - \frac{1}{5} \\\\ - \frac{1}{2} & 0 \end{array} \right]=\left[ \begin{array}{cc} -3 & 0 \\\\ -6 & 9 \end{array} \right]$

$A=\left[ \begin{array}{cc} -3 & 0 \\\\ -6 & 9 \end{array} \right]$