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

$$a+d=1+5$$

$$ad-bc=1(5)$$

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

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

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

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

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

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

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

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

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

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

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

$$D=P^{-1}AP$$

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