Question

Let A be a square matrix defined by$A =\begin{pmatrix}2&-3&1\\ 1&-2&1\\ 1&-3&2\end{pmatrix}$

(a) Find the characteristic polynomial of A.

(b) Show that A is diagonalizable then diagonalize it.

(c) Write $A^n$ \hspace{2mm} in term of n.

 

 

Answer 1

Solution

(a) Find the characteristic polynomial of A.

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

$\implies P(\lambda)=|A-\lambda I|=\bigg(-1\bigg)^3\bigg(\lambda^3-S_1\lambda^2+S_2\lambda-S_3\bigg)$

$\bullet \hspace{2mm}S_1=2-2+2=2$

$\bullet \hspace{2mm}S_2=\begin{vmatrix}-2&1\\ -3&2\end{vmatrix}+\begin{vmatrix}1&1\\ 1&2\end{vmatrix}+\begin{vmatrix}1&-2\\ 1&-2\end{vmatrix}$

         $=\bigg(-4+3\bigg)+\bigg(2+1\bigg)+\bigg(-3+2\bigg)=1$

$\bullet \hspace{2mm}S_3=\begin{vmatrix}2&-3&1\\ 1&-2&1\\ 1&-3&2\end{vmatrix}=2\bigg(-4+3\bigg)+3\bigg(2-1\bigg)+1\bigg(-3+2\bigg)=0$

Therefore, $P(\lambda)=-\lambda\bigg(\lambda-1\bigg)^2$

Show that A is diagonalizable then diagonalize it. 

we have $P(\lambda)=0\implies -\lambda\bigg(\lambda-1\bigg)^2=0$ $\iff \lambda_1=0,\lambda_2=1$

so, $spact(A)=\left\{0,1\right\}$

where, $am(0)=1,am(1)=2$  Then  $E_2=\left\{x\in \mathbb{R}^3|\bigg(A-I\bigg)x=0\right\}$

Let $x=\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}\hspace{2mm}$ Then $A=\begin{pmatrix}1&-3&1\\ 1&-3&1\\ 1&-3&1\end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}=0$

$\iff x_1-3x_2+x_0\implies x_3=-x_1+3x_2$

$\implies x=\begin{pmatrix}x_1\\ x_2\\ -x_1+3x_2\end{pmatrix}\hspace{2mm} =x_1\begin{pmatrix}1\\ 0\\ -1\end{pmatrix}+x_2\begin{pmatrix}0\\ 1\\ 3\end{pmatrix}$

$\iff E_2=\left\{\begin{pmatrix}1\\ \:0\\ \:-1\end{pmatrix},\begin{pmatrix}0\\ \:1\\ \:3\end{pmatrix}\right\}$

Then $gm(1)=dim(E_2)=2\implies E_1=\left\{x\in \mathbb{R}|Ax=0\right\}$

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

Let  $x_3=t\implies x_2=x_1=t$

$\implies x=\begin{pmatrix}t\\ t\\ t\end{pmatrix}=t\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}\hspace{2mm}\iff gm(0)=dim(E_1)=1$

so, $am(1)=gm(1)=2\hspace{2mm},am(0)=gm(0)=1$

Therefore,A is diagonallizable.

(c) Write$A^n$\hspace{2mm} in term of n.

we have $A^n=PDP^{-1}$ since $P=\begin{pmatrix}1&0&1\\ 0&1&1\\ -1&3&1\end{pmatrix} \hspace{2mm},D=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{pmatrix}$

$\implies A^n=\begin{pmatrix}1&0&1\\ 0&1&1\\ -1&3&1\end{pmatrix}\begin{pmatrix}1&0&0\\ \:0&1&0\\ \:0&0&0\end{pmatrix}^n\begin{pmatrix}1&0&1\\ \:0&1&1\\ \:-1&3&1\end{pmatrix}^{-1}$

Therefore. $A^n=\begin{pmatrix}\frac{1}{2}&\frac{3}{2}&-\frac{1}{2}\\ 0&1&0\\ -\frac{1}{2}&\frac{3}{2}&\frac{1}{2}\end{pmatrix}$

 

Answer

Therefore.

a). $P(\lambda)=-\lambda\bigg(\lambda-1\bigg)^2$

b). A is diagonallizable. $D=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{pmatrix}$

c). $A^n=\begin{pmatrix}\frac{1}{2}&\frac{3}{2}&-\frac{1}{2}\\ 0&1&0\\ -\frac{1}{2}&\frac{3}{2}&\frac{1}{2}\end{pmatrix}$