Question

In: Advanced Math

Determine all possible Jordan canonical forms for a linear operator L whose characteristic polynomial is

Determine all possible Jordan canonical forms for a linear operator L whose characteristic polynomial is 

\( P(\lambda)=\bigg(\lambda-3\bigg)^3\bigg(\lambda-4\bigg)^2 \)

Solutions

Expert Solution

solution

we have the characteristic polynomial is 

\( P(\lambda)=\bigg(\lambda-3\bigg)^3\bigg(\lambda-4\bigg)^2 \)

\( \implies am(3)=3,am(4)=2\hspace{2mm}so,\hspace{2mm}gm(3)=1,2,3 \)

and \( gm(4)=1,2 \)

By the order of decreasing of dimension of blocks matrix corresponding to the same eigenvalues.

The possible Jordan matrices are : 

\( J_1=\begin{pmatrix}3&0&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix}\: \) \( J_2=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix}\: \)

\( J_3=\begin{pmatrix}3&1&0&0&0\\ 0&3&1&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix} \)  \( J_4=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&1&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)

\( J_5=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)  \( J_6=\:\:\begin{pmatrix}3&0&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)


Answer

The possible Jordan matrices are : 

\( J_1=\begin{pmatrix}3&0&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix}\: \)   \( J_2=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix}\: \)

\( J_3=\begin{pmatrix}3&1&0&0&0\\ 0&3&1&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix} \)  \( J_4=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&1&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)

\( J_5=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)  \( J_6=\:\:\begin{pmatrix}3&0&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)

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