Question

Find the characteristics and the minimal polynomial of the following matrices over $\mathbb{R}$, then deduce the their corresponding Jordan Canonical Form J.

A=$\begin{pmatrix}1&-1&-1\\ 0&0&-1\\ 0&1&2\end{pmatrix}\:$

Answer 1

Solution

A=$\begin{pmatrix}1&-1&-1\\ 0&0&-1\\ 0&1&2\end{pmatrix}\:$ The characteristics polynomial.

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

$\bullet S_1=tr(A)=1+0+2=3$

$\bullet S_2=\begin{vmatrix}1&-1\\ 1&2\end{vmatrix}+\begin{vmatrix}1&-1\\ 0&2\end{vmatrix}+\begin{vmatrix}1&-1\\ 0&0\end{vmatrix}=3$

$\bullet S_3=\begin{vmatrix}1&-1&-1\\ 0&0&-1\\ 0&1&2\end{vmatrix}=1$

$\implies P(\lambda )=-\bigg(\lambda^3-3\lambda^2+3\lambda-1\bigg)=-\bigg(\lambda-1\bigg)^3$

$we\hspace{2mm} take P(\lambda)=0\iff -\bigg(\lambda-1\bigg)^3=0\implies \lambda =1\hspace{2mm}Thus,spact(A)=1$$Let \hspace{2mm}x=\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix} \in E_1\hspace{2mm},Then\hspace{2mm}\bigg(A-I\bigg)x=0\hspace{2mm},for\hspace{2mm} x \neq 0$

$\implies \begin{pmatrix}0&-1&-1\\ 0&-1&-1\\ 0&0&0\end{pmatrix} \sim \begin{pmatrix}0&1&1\\ 0&0&0\\ 0&0&0\end{pmatrix} \hspace{2mm}Let\hspace{2mm} x_3=S\hspace{2mm}and\hspace{2mm}x_1=t\in\mathbb{R}$

$\implies x_2+x_3=0\iff x_2=-x_3=-S$

 

$\implies x=\begin{pmatrix}t\\ -S\\ S\end{pmatrix}=t\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}+S\begin{pmatrix}0\\ -1\\ 1\end{pmatrix}\hspace{2mm}Thus,\hspace{2mm}E_1=span\left\{\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ -1\\ 1\end{pmatrix}\right\}$

$\iff m(\lambda)=\bigg(\lambda-1\bigg)^2\implies gm(1)=dim(E_1)=2$

Therefore. there are 2 jordan block and index(1)=2

$$

Answer :

Therefore. spact(A)=1 and  there are 2 jordan block and index(1)=2.