In: Advanced Math
Determine all possible Jordan canonical forms for a linear operator L whose characteristic polynomial is
\( P(\lambda)=\bigg(\lambda-3\bigg)^3\bigg(\lambda-4\bigg)^2 \)
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 :
\( J_1=\begin{pmatrix}3&0&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix}\: \) \( J_2=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix}\: \)
\( J_3=\begin{pmatrix}3&1&0&0&0\\ 0&3&1&0&0\\ 0&0&3&0&0\\ 0&0&0&4&0\\ 0&0&0&0&4\end{pmatrix} \) \( J_4=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&1&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)
\( J_5=\:\begin{pmatrix}3&1&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \) \( J_6=\:\:\begin{pmatrix}3&0&0&0&0\\ 0&3&0&0&0\\ 0&0&3&0&0\\ 0&0&0&4&1\\ 0&0&0&0&4\end{pmatrix} \)