Answer to Question #216168 in Linear Algebra for Chris

Question

Given that matrix A has 1has the eigenvalue with the corresponding eigenvector (1,-3) and 5 as the second eigenvalue with (1,1) as the corresponding eigenvector. (I) find the matrix A (Ii) D the diagonal matrix such that P^-1 AP=D

Answer 1

Let

$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$

$\begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix}=0$

$(a-\lambda)(d-\lambda)-bc=0$

$\lambda^2-(a+d)\lambda+ad-bc=0$

$\lambda_1=1, v_1=\begin{pmatrix} 1 \\ -3 \end{pmatrix}$

$\lambda_2=5, v_1=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

By Vieta Theorem

$a+d=1+5$

$ad-bc=1(5)$

$Av_i=\lambda_iv_i, i=1,2$

$\begin{pmatrix} a-1 & b \\ c & d-1 \end{pmatrix}\begin{pmatrix} 1 \\ -3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$a-1-3b=0$ $c-3d+3=0$

$a=3b+1$ $c=3d-3$

$\begin{pmatrix} a-5 & b \\ c & d-5 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$a-5+b=0$ $c+d-5=0$

$a=-b+5$ $c=-d+5$

$3b+1=-b+5=>b=1, a=4$

$3d-3=-d+5=>d=2, c=3$

$A=\begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix}$

$P=\begin{pmatrix} 1 & 1 \\ -3 & 1 \end{pmatrix}, P^{-1}=\begin{pmatrix} 0.25 & -0.25 \\ 0.75 & 0.25 \end{pmatrix}$

$D=P^{-1}AP$

$D=\begin{pmatrix} 1 & 0 \\ 0 & 5 \end{pmatrix}$