Question

Determine all possible Jordan canonical forms J for a matrix of order 6 whose minimal polynomial is 

\( m(\lambda)=\bigg(\lambda-1\bigg)^3\bigg(\lambda-3\bigg)^2 \)

Answer 1

Solution

we have the minimal polynomial is 

\( m(\lambda)=\bigg(\lambda-1\bigg)^3\bigg(\lambda-3\bigg)^2 \)

we take \( m(\lambda)=0\implies \bigg(\lambda-1\bigg)^3\bigg(\lambda-3\bigg)^2 =0 \)

since \( \lambda_1=1 \hspace{2mm}and\hspace{2mm}\lambda_2=3 \)

\( \implies Index(1)=3 \) (means the bigest Jordan block is \( 3\times3 \))\( \implies Index(3)=2 \)  (means the bigest Jordan block is \( 2\times2 \))

Therefore. all possible Jordan Canonical form is :

\( J_{1\:}=\begin{pmatrix}1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&3&1&0\\ 0&0&0&0&3&0\\ 0&0&0&0&0&3\end{pmatrix},J_2=\begin{pmatrix}1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&3&1\\ 0&0&0&0&0&3\end{pmatrix} \)

 

Answer

Therefore. all possible Jordan Canonical form is :

\( J_{1\:}=\begin{pmatrix}1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&3&1&0\\ 0&0&0&0&3&0\\ 0&0&0&0&0&3\end{pmatrix},J_2=\begin{pmatrix}1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&3&1\\ 0&0&0&0&0&3\end{pmatrix} \)