Question

Let A be a square matrix defined by $A =\begin{pmatrix}-1&3&-1\\ -3&5&-1\\ -3&3&1\end{pmatrix}$

(a) Find the characteristic polynomial of A.

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

(c) Write $A^n$ in term of n.

Answer 1

Solution

(a) Find the characteristic polynomial of A.

we have $A =\begin{pmatrix}-1&3&-1\\ -3&5&-1\\ -3&3&1\end{pmatrix}$

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

$\bullet \hspace{2mm}S_1=tr(A)=-1+5+1=5$

$\bullet \hspace{2mm}S_2=\begin{vmatrix}5&-1\\ 3&1\end{vmatrix}+\begin{vmatrix}-1&-1\\ -3&1\end{vmatrix}+\begin{vmatrix}-1&3\\ \:-3&5\end{vmatrix}=8-4+4=8$

$\bullet \hspace{2mm}S_3=|A|=\bigg(-5+9+9\bigg)-\bigg(15+3-9\bigg)=13-9=4$

so, $P(\lambda)=-\bigg(\lambda^3-5\lambda^2+8\lambda-4\bigg)=-\lambda\bigg(\lambda-1\bigg)\bigg(\lambda-2\bigg)^2$

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

since,

$P(\lambda)=0\implies \lambda\bigg(\lambda-1\bigg)\bigg(\lambda-2\bigg)^2 \iff \lambda =1,2,2\hspace{2mm}$

$\implies spect(A)=\left\{1,2\right\}$

Then $am(1)=1,am(2)=2$

$\bullet \hspace{2mm}For\hspace{2mm} \lambda =2$ $x=\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}\in E_2\hspace{2mm}$ then $x\neq 0$ and $\bigg(A-2I\bigg)x=0$$\implies A-2I=\begin{pmatrix}-3&3&-1\\ -3&3&-1\\ -3&3&-1\end{pmatrix}\sim \begin{pmatrix}-3&3&-1\\ 0&0&0\\ 0&0&0\end{pmatrix}$

Let $x_1=S,x_2=t\implies x_3=-3S+3t$

$\implies x=\begin{pmatrix}S\\ t\\ -3S+3t\end{pmatrix}=t\begin{pmatrix}0\\ 1\\ 3\end{pmatrix}+S\begin{pmatrix}1\\ 0\\ -3\end{pmatrix},S,t\in \mathbb{R}$

So, $E_2=span\left\{\begin{pmatrix}0\\ \:1\\ \:3\end{pmatrix},\begin{pmatrix}1\\ \:0\\ \:-3\end{pmatrix}\right\}\implies gm(2)=dim(E_2)=am(2)=2$

since. $\forall \lambda\in Spect(A)\hspace{2mm},gm(\lambda)=am(\lambda)$ and

$P(\lambda)=-\bigg(\lambda-1\bigg)\bigg(\lambda-2\bigg)^2\hspace{2mm}$ is  split

$\bullet \hspace{2mm}For \hspace{2mm}\lambda=1 : \hspace{2mm}Let \hspace{2mm}x=\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}\in E_1 \hspace{2mm},Then \hspace {2mm} x\neq 0 \hspace{2mm}and \hspace{2mm} \bigg(A-I\bigg)x=0$

$\implies A-I=\begin{pmatrix}-2&3&-1\\ -3&4&-1\\ -3&3&0\end{pmatrix}\sim \begin{pmatrix}1&-1&0\\ -3&4&-1\\ -2&3&-1\end{pmatrix}$

                      $\sim \begin{pmatrix}1&-1&0\\ 0&1&-1\\ 0&1&-1\end{pmatrix}\sim \begin{pmatrix}1&-1&0\\ 0&1&-1\\ 0&0&0\end{pmatrix}$

Let $x_3=t\implies x_2=t,x_1=t\iff x=\begin{pmatrix}t\\ t\\ t\end{pmatrix}=t\begin{pmatrix}1\\ 1\\ 1\end{pmatrix},t\in \mathbb{R}$

Then $gm(1)=dimE_1=am(1)=1\hspace{2mm}$so A is diagonalizable

Thus $A\sim D\iff A=PDP^{-1}\iff AP=PD$

where \( D=\begin{pmatrix}1&0&0\\ 0&2&0\\ 0&0&2\end{pmatrix},P=\begin{pmatrix}1&1&0\\ 1&0&1\\ 1&-3&3\end{pmatrix}$\implies Index(1),Index(2)=2 \)

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

we have $m(x)=\bigg(x-1\bigg)\bigg(x-2\bigg)$

$\implies x^n=m(x)q(x)+ax+b$

$\iff \begin{cases} a+b=1 & \quad \\ 2a+b=2^n & \quad \end{cases} \implies \begin{cases} a=2^n-1 & \quad \\ b=1-a=2-2^n & \quad \end{cases}$

Then $A^n=aA+bI=\bigg(2^n-1\bigg)A+\bigg(2-2^n\bigg)I$

               $=\bigg(2^n-1\bigg)\begin{pmatrix}-1&3&-1\\ -3&5&-1\\ -3&3&1\end{pmatrix}+\bigg(2-2^n\bigg)\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

Threfore. $A^n=\begin{pmatrix}3-2^{n+1}&3\left(2^n-1\right)&1-2^n\\ -3\left(2^n-1\right)&2^{n+2}-3&1-2^n\\ -3\left(2^n-1\right)&3\left(2^n-1\right)&1\end{pmatrix}$

Answer

Therefore.

a). $P(\lambda)=-\bigg(\lambda^3-5\lambda^2+8\lambda-4\bigg)=-\lambda\bigg(\lambda-1\bigg)\bigg(\lambda-2\bigg)^2$

b). \( D=\begin{pmatrix}1&0&0\\ 0&2&0\\ 0&0&2\end{pmatrix},P=\begin{pmatrix}1&1&0\\ 1&0&1\\ 1&-3&3\end{pmatrix}$\implies Index(1),Index(2)=2 \)

c). $A^n=\begin{pmatrix}3-2^{n+1}&3\left(2^n-1\right)&1-2^n\\ -3\left(2^n-1\right)&2^{n+2}-3&1-2^n\\ -3\left(2^n-1\right)&3\left(2^n-1\right)&1\end{pmatrix}$