Question

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

C=\( \begin{pmatrix}1&1&0&0\\ -1&-1&1&0\\ 0&1&1&0\\ -1&-1&1&1\end{pmatrix} \)  

Answer 1

Solution

C= \( \begin{pmatrix}1&1&0&0\\ -1&-1&1&0\\ 0&1&1&0\\ -1&-1&1&1\end{pmatrix} \)  The characteristics polynomial.

\( \implies P(\lambda)=(-1)^n(\lambda^n-S_1\lambda^{n-1}+S_2\lambda^{n-2}+...+(-1)^4S_n) \)

\( \bullet S_1=tr(C)=1-1+1+1=2 \)

\( \bullet S_2=\begin{vmatrix}1&0\\ 1&1\end{vmatrix}+\begin{vmatrix}-1&0\\ -1&1\end{vmatrix}+\begin{vmatrix}-1&-1\\ 1&1\end{vmatrix}+\begin{vmatrix}1&1\\ -1&1\end{vmatrix}+\begin{vmatrix}1&0\\ 0&1\end{vmatrix}+\begin{vmatrix}1&0\\ -1&1\end{vmatrix}=2 \)

\( \bullet S_3=\begin{vmatrix}-1&1&0\\ 1&1&0\\ -1&1&1\end{vmatrix}+\begin{vmatrix}1&1&0\\ -1&-1&0\\ -1&-1&1\end{vmatrix}+\begin{vmatrix}1&0&0\\ 0&1&0\\ -1&1&1\end{vmatrix}+\begin{vmatrix}1&1&0\\ -1&-1&1\\ 0&1&1\end{vmatrix}=-2 \)

\( \bullet S_4=\begin{vmatrix}1&1&0&0\\ -1&-1&1&0\\ 0&1&1&0\\ -1&-1&1&1\end{vmatrix}=-1 \)

Therefore,\( P(\lambda)=\lambda^4-2\lambda^3+2\lambda^2+2\lambda-1=(\lambda-1)^3(\lambda+1) \)

The Jordan Conical form and \( m(\lambda) \)

For \( \lambda_1=1\implies \bigg(C-\lambda_1 I\bigg)x=0\implies \bigg(C-I\bigg)x=0 \)

\( \implies C-I=\begin{pmatrix}0&1&0&0\\ -1&-2&1&0\\ 0&1&0&0\\ -1&-1&1&0\end{pmatrix}\sim \begin{pmatrix}-1&-1&1&0\\ -1&-2&1&0\\ 0&1&0&0\\ 0&0&0&0\end{pmatrix}\sim \begin{pmatrix}1&1&-1&0\\ 0&1&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} \)

Let \( x_3=\alpha ,x_4=\beta,x_1=x_3=\alpha \)

\( \implies E_{\lambda_1}=E_{-1}=\left\{\begin{pmatrix}\alpha\\ 0\\ \alpha\\ \beta\end{pmatrix}\right\}=span\left\{\begin{pmatrix}1\\ 0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 0\\ 1\end{pmatrix}\right\} \)

\( \implies gm(\lambda_1)=gm(1)=dimE_1=2 \)

For \( \lambda_2=-1\implies \bigg(C-\lambda\bigg)x=0\implies \bigg(C+I\bigg)x=0 \)

\( \implies C+I=\begin{pmatrix}2&1&0&0\\ -1&0&1&0\\ 0&1&2&0\\ -1&-1&1&2\end{pmatrix}\sim \begin{pmatrix}1&1&-1&-2\\ 0&1&2&0\\ 0&0&1&1\\ 0&0&0&1\end{pmatrix} \)

Let \( x_3=-x_4,x_2=2x_4,x_1=-x_4 \)

\( \implies E_{\lambda_2}=E_{-1}=\left\{\begin{pmatrix}-x_4\\ 2x_4\\ -x_4\\ x_4\end{pmatrix},x_4\in \mathbb{R}^4,x_4\in \mathbb{R} \right\}=span\left\{\begin{pmatrix}-1\\ 2\\ -1\\ 1\end{pmatrix}\right\} \)

\( \implies gm(\lambda_2)=gm(-1)=dim_{-1}=1 \)

The possible Jordan Canonical form is  \( J=\begin{pmatrix}-1&0&0&0\\ 0&1&0&0\\ 0&0&1&1\\ 0&0&0&1\end{pmatrix} \)

Therefore, minimal polynomial is  \( m(\lambda)=\bigg(\lambda-1\bigg)^2\bigg(\lambda+1\bigg) \)

 

Answer

Therefore. \( P(\lambda)=\lambda^4-2\lambda^3+2\lambda^2+2\lambda-1=(\lambda-1)^3(\lambda+1) \)

 minimal polynomial is \( m(\lambda)=\bigg(\lambda-1\bigg)^2\bigg(\lambda+1\bigg) \)