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