Question

Let A be a square matrix defined by $A = \begin{pmatrix}-2&-1&-5\\ 2&2&3\\ 4&2&2\end{pmatrix}$

(a) Find the characteristic polynomial of A.

(b) Find the eigenvalues and eigenspaces of A.

(c) Show that A is not diagonalizable, but it is triangularizable, then triangularize A.

(d) Find the three real sequences $(a)_n, (b)_n ,(c)_n$ satisfying.

$\begin{cases} a_{n+1}=-2a_n-b_n-5c_n \hspace{2mm},a_0=1 & \quad \\ b_{n+1}=2a_n+2b_n+3c_n \hspace{2mm}, b_0=0 & \quad \\ c_{n+1}=4a_n+2b_n+6c_n \hspace{2mm},c_0=1 & \quad \end{cases}$

 

Answer 1

Solution

(a) Find the characteristic polynomial of A.

we have $A = \begin{pmatrix}-2&-1&-5\\ 2&2&3\\ 4&2&2\end{pmatrix}$

$\implies P(\lambda)=\lambda^3+6\lambda^2-12\lambda-8=-\bigg(\lambda-2\bigg)^3$

Thus, $P(\lambda)=-\bigg(\lambda-2\bigg)^3$

(b) Find the eigenvalues and eigenspaces of A.

$\lambda \in sp(A) \iff P(\lambda)=0\implies \lambda =2 \hspace{2mm}with \hspace{2mm} am(2)=3$

$A-2I=\begin{pmatrix}-4&-1&-5\\ 2&0&3\\ 4&2&4\end{pmatrix}\sim \begin{pmatrix}2&0&3\\ 0&1&-1\\ 0&0&0\end{pmatrix}$

Let $x_3=t\implies x_2=t\implies x_1=-\frac{3}{2}t$

Thus, $E_2=span\left\{\begin{pmatrix}-\frac{3}{2}\\ 1\\ 1\end{pmatrix}\right\}$

(c) Show that A is not diagonalizable, but it is triangularizable, then triangularize A.

since $P(\lambda)$ is splitted and $am(\lambda) \neq gm(\lambda)$ so A is not diagonalizable but it's trianglarizable 

$\bullet \hspace{2mm}Triangularizable$

$\implies A=PJP^{-1}\hspace{2mm}, J=\begin{pmatrix}2&1&0\\ 0&2&1\\ 0&0&2\end{pmatrix}$

Then $\bigg(A-2I\bigg)^2=\begin{pmatrix}-6&-6&-3\\ 4&4&2\\ 4&4&2\end{pmatrix}\hspace{2mm},\bigg(A-2I\bigg)^3=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$

Let $V_3\in G_{\lambda}-E_{\lambda}^2 \hspace{2mm},G_{\lambda}=\left\{\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}\right\}$

$\implies V_3=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix} \hspace{2mm},V_2=\bigg(A-2I\bigg)V_3=\begin{pmatrix}-4\\ 2\\ 4\end{pmatrix}\hspace{3mm},v_1=\bigg(A-2I\bigg)V_2=\begin{pmatrix}-6\\ 4\\ 4\end{pmatrix}$

Thus, $A=PJP^{-1}\hspace{2mm}where\hspace{2mm}J=\begin{pmatrix}2&1&0\\ 0&2&1\\ 0&0&0\end{pmatrix},P=\begin{pmatrix}-6&-4&1\\ 4&2&0\\ 4&4&0\end{pmatrix}$

(d) Find the three real sequences $(a)_n, (b)_n ,(c)_n$ satisfying.

$\begin{cases} a_{n+1}=-2a_n-b_n-5c_n \hspace{2mm},a_0=1 & \quad \\ b_{n+1}=2a_n+2b_n+3c_n \hspace{2mm}, b_0=0 & \quad \\ c_{n+1}=4a_n+2b_n+6c_n \hspace{2mm},c_0=1 & \quad \end{cases}$

Let $X_{n+1}=\begin{pmatrix}a_{n+1}\\ b_{n+1}\\ c_{n+1}\end{pmatrix} \implies X_{n+1}=AX_n=A^{n+1}X_0 \hspace{2mm}or\hspace{2mm} X_n=A^nX_0$

where  $A^n=PJ^nP^{-1}\hspace{2mm}with\hspace{2mm} J=2I+M \hspace{2mm},M=\begin{pmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix} \hspace{2mm}and\hspace{2mm}M^3=0$

Then $J^n=\bigg(2I\bigg)^n+n\bigg(2I\bigg)^{n-1}M+\frac{n(n-1)}{2}M^2\times 2^{n-2}$

Thus, $X_n=P\bigg(2^nI+2^nnM+\frac{n(n-1)}{2}\times 2^{n-2}M^2\bigg)P^{-1}X_0$

 

Answer

Therefore. 

a). $P(\lambda)=-\bigg(\lambda-2\bigg)^3$

b). $E_2=span\left\{\begin{pmatrix}-\frac{3}{2}\\ 1\\ 1\end{pmatrix}\right\}$

c).$A=PJP^{-1}\hspace{2mm}where\hspace{2mm}J=\begin{pmatrix}2&1&0\\ 0&2&1\\ 0&0&0\end{pmatrix},P=\begin{pmatrix}-6&-4&1\\ 4&2&0\\ 4&4&0\end{pmatrix}$

d). $X_n=P\bigg(2^nI+2^nnM+\frac{n(n-1)}{2}\times 2^{n-2}M^2\bigg)P^{-1}X_0$