Question

Let A be a square matrix defined by \( A =\begin{pmatrix}-1&-2&-1&3\\ -6&-5&1&6\\ -6&-4&0&6\\ -6&-7&1&8\end{pmatrix} \) and its characteristics polynomial \( P(\lambda)=\bigg(\lambda+1\bigg)^2\bigg(\lambda-2\bigg)^2 \)

(a) Find the minimal polynomial of A.

(b) Deduce that A is not diagonalizable, but it is triangularizable, then triangularize A.

(c) Write \( A^n \) in terms of n.

Answer 1

Solution

(a) Find the minimal polynomial of A.

we have \( A =\begin{pmatrix}-1&-2&-1&3\\ -6&-5&1&6\\ -6&-4&0&6\\ -6&-7&1&8\end{pmatrix} \) and \( P(\lambda)=\bigg(\lambda+1\bigg)^2\bigg(\lambda-2\bigg)^2 \)

\( \implies m_A({\lambda})=\bigg(\lambda+1\bigg)^{\alpha_1}\bigg(\lambda-2\bigg)^{\alpha_2} \hspace{2mm} \) where  \( \alpha_1,\alpha_2 \)   are index of  

\( \lambda_1,\lambda_2 \) respectivdy.

\( A+I=\begin{pmatrix}0&-2&-1&3\\ -6&-5&1&6\\ -6&-4&0&6\\ -6&-7&1&8\end{pmatrix}\sim \begin{pmatrix}0&-2&-1&3\\ -6&-4&1&6\\ 0&0&0&0\\ 0&-3&0&3\end{pmatrix}\sim \begin{pmatrix}0&-2&-1&3\\ -6&0&3&0\\ 0&0&0&0\\ 0&-3&0&3\end{pmatrix} \)

Let \( x_4=t\implies x_2=t, x_3=S\implies x_1=-2x_2+3x_3 ,x_1=\frac{t}{2} \)

Then \( E_{\lambda_1}=span\left\{\begin{pmatrix}\frac{1}{2}\\ 1\\ 1\\ 1\end{pmatrix}\right\} \iff gm(\lambda_1)=1\neq am(\lambda_1)\implies \alpha_1=2 \)

\( \implies A-2I=\begin{pmatrix}-3&-2&-1&3\\ -6&-7&1&6\\ -6&-4&-2&6\\ -6&-7&1&6\end{pmatrix}\sim \begin{pmatrix}-3&-2&-1&3\\ -6&-7&1&6\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}\sim \begin{pmatrix}-3&-2&1&3\\ 0&-3&3&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} \)

Let \( x_3=t\hspace{2mm},x_2=t\hspace{2mm}and\hspace{2mm} x_4=S\implies x_1=S-\frac{t}{3} \)

Then  \( E_{\lambda_2}=span\left\{\begin{pmatrix}1\\ 0\\ 0\\ 1\end{pmatrix},\begin{pmatrix}-\frac{1}{3}\\ 1\\ 1\\ 0\end{pmatrix}\right\}\implies gm(\lambda_2)=am(\lambda_2)=2\implies \alpha_2=2 \)

Thus, \( m_A(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda+1\bigg)^2 \)

(b) Deduce that A is not diagonalizable, but it is triangularizable, then triangularize A.

since \( am(\lambda_1)\neq gm(\lambda_1) \)

so A is not diagonalizable. But it's triangularizable

\( \bullet \hspace{2mm} Tringularizable \hspace{2mm}that\hspace{2mm} is \hspace{2mm} A=PJP^{-1} \)

where \( J=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&1\\ 0&0&0&-1\end{pmatrix}\hspace{2mm}, \)

\( \bigg(A+I\bigg)^2=\begin{pmatrix}0&-9&0&-9\\ -18&-18&9&18\\ -18&-18&8&18\\ -18&-27&9&27\end{pmatrix}\sim \begin{pmatrix}0&1&0&1\\ 2&2&-1&-2\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} \)

Let  \( x_4=t\implies x_2=-t\hspace{2mm},x_3=S\implies x_1=\frac{1}{2}S+2t \)

Thus, \( E_(\lambda_1)^2=span\left\{\begin{pmatrix}\frac{1}{2}\\ 0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}2\\ -1\\ 0\\ 1\end{pmatrix}\right\} \)

Let  \( V_2\in E_{\lambda_1}^2-E_{\lambda_1}\implies V_2=\begin{pmatrix}\frac{1}{2}\\ 0\\ 1\\ 0\end{pmatrix} \)

\( \implies V_1=\bigg(A+I\bigg)V_2=\begin{pmatrix}0&-2&-1&3\\ -6&-4&1&6\\ -6&-4&1&6\\ -6&-7&1&9\end{pmatrix}\sim \begin{pmatrix}\frac{1}{2}\\ \:0\\ \:1\\ \:0\end{pmatrix}=\begin{pmatrix}-1\\ -2\\ -2\\ -2\end{pmatrix} \) 

Then, \( P\bigg(u_1,u_2,V_1,V_2\bigg)=\begin{pmatrix}1&-\frac{1}{3}&-1&\frac{1}{2}\\ 0&1&-2&0\\ 0&1&-2&1\\ 1&0&-2&0\end{pmatrix} \)

Thus, \( A=PJP^{-1}\hspace{2mm},J=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix},P=\begin{pmatrix}1&-\frac{1}{3}&-1&\frac{1}{2}\\ 0&1&-2&0\\ 0&1&-2&1\\ 1&0&-2&0\end{pmatrix} \)

(c) Write \( A^n \)in terms of n.

\( \implies A=PJP^{-1}\iff A^n=PJ^nP^{-1}\hspace{2mm},J=D+M \)

since  \( D=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix} \)  and  \( M=\begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}\hspace{2mm} where \hspace{2mm}M^2=0 \)

so,\( J^n=D^n+nD^{n-1}M=\begin{pmatrix}2^n&0&0&0\\ 0&2^n&0&0\\ 0&0&\left(-1\right)^n&n\left(-1\right)^{n-1}\\ 0&0&0&\left(-1\right)^{n\:}\end{pmatrix} \)

 

Answer

Therefore

a). \( m_A(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda+1\bigg)^2 \)

b). \( A=PJP^{-1}\hspace{2mm},J=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix},P=\begin{pmatrix}1&-\frac{1}{3}&-1&\frac{1}{2}\\ 0&1&-2&0\\ 0&1&-2&1\\ 1&0&-2&0\end{pmatrix} \)

c). \( A^n=P\begin{pmatrix}2^n&0&0&0\\ \:0&2^n&0&0\\ \:0&0&\left(-1\right)^n&n\left(-1\right)^{n-1}\\ \:0&0&0&\left(-1\right)^{n\:}\end{pmatrix}P^{-1} \)