Question

Determine the value of a so that $\lambda = 2$ is an eigenvalue of 

$A=\begin{pmatrix}1&-1&0\\ a&1&1\\ 0&1+a&3\end{pmatrix}$

Then show that A is diagonallizable and diagonalize it. 

Answer 1

Solution

we have $A-2I=\begin{pmatrix}-1&-1&0\\ a&-1&1\\ 0&1+a&3\end{pmatrix}$

Then if $\lambda =2\in SP(A)$

$\implies |A-2I|=\begin{vmatrix}-1&-1&0\\ \:a&-1&1\\ \:0&1+a&3\end{vmatrix}=0$ Thus, $a=-1$

we have $A=\begin{pmatrix}1&-1&0\\ -1&1&1\\ 0&0&3\end{pmatrix}$

$\implies P(\lambda)=|A-\lambda I|=\begin{vmatrix}1-\lambda &-1&0\\ \:-1&1-\lambda \:&1\\ \:0&0&3-\lambda \:\end{vmatrix}$

                    $=\bigg(3-\lambda\bigg)\bigg[\bigg(1-\lambda\bigg)^2-1\bigg]$

$\implies P(\lambda)=\bigg(3-\lambda\bigg)\bigg(1-2\lambda+\lambda^2-1\bigg)$

                     $=\lambda\bigg(\lambda-2\bigg)\bigg(3-\lambda\bigg)$

since, $A\in M_3(\mathbb{R})$ and $P(\lambda)=\lambda\bigg(\lambda-2\bigg)\bigg(3-\lambda\bigg)$

$\color{red}Note : Theorem21$

(1) The characteristic polynomial $P_L(\lambda)$ splits over $\mathbb{K}$

(2) Every eigenvalue $\lambda$ of L $am(\lambda)=gm(\lambda)$

By Theorem 21$\implies am(\lambda)=gm(\lambda)=1 ,\forall \lambda \iff SP(\lambda)$

$\bullet$ Find  $E_0$

$\implies A-(0)I= \begin{pmatrix}1&-1&0\\ -1&1&1\\ 0&0&3\end{pmatrix}\sim \begin{pmatrix}0&0&1\\ 1&-1&0\\ 0&0&0\end{pmatrix}\hspace{2mm}$

Then $E_0=span\left\{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}\right\}$

$\bullet$ Find $E_2$

$\implies A-2I=\begin{pmatrix}-1&-1&0\\ -1&-1&1\\ 0&0&1\end{pmatrix}\sim \begin{pmatrix}-1&-1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}$

Then $E_2=span\left\{\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}\right\}$

$\bullet$ Find $E_3$ 

$\implies A-3I=\begin{pmatrix}-2&-1&0\\ -1&-2&1\\ 0&0&0\end{pmatrix}\sim \begin{pmatrix}2&1&0\\ 0&3&1\\ 0&0&0\end{pmatrix}$

Then $E_3=span\left\{\begin{pmatrix}-1\\ 2\\ 3\end{pmatrix}\right\}$

since. $dimE_0+dimE_2+dimE_3=3=\mathbb{R}$

A is diagonalizable in $\mathbb{R}$

Therefore. $A=PDP^{-1}$ where $D=\begin{pmatrix}0&0&0\\ 0&2&0\\ 0&0&3\end{pmatrix},P=\begin{pmatrix}1&1&-1\\ 1&-1&2\\ 0&0&3\end{pmatrix}$

Answer

Therefore.

$A=PDP^{-1}$ where  $D=\begin{pmatrix}0&0&0\\ 0&2&0\\ 0&0&3\end{pmatrix}$ and $P=\begin{pmatrix}1&1&-1\\ 1&-1&2\\ 0&0&3\end{pmatrix}$