Answer
See below
Work Step by Step
Given: $p(\lambda)=(4-\lambda)^3(-1-\lambda)^2$
Find the eigenvalues:
$\lambda_1=4$ with multiplicity of 3
and $\lambda_2=-1$ with multiplicity of 2
The possibilities for $\lambda_1=4$ with multiplicity of 3 are:
One block of $3 \times 3$ matrix; One block of $2\times 2$ & two blocks of $1\times 1$; 3 blocks of $1\times 1$.
The possibility for $\lambda_1=-6$ with multiplicity of 1 are:
One block of $2 \times 2$; 2 blocks of $1 \times 1$.
The set S of matrices include:
$J_1=\begin{bmatrix}
4 & 1 & 0 & 0 & 0\\
0 & 4 & 1 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}$
$J_2=\begin{bmatrix}
4 & 1 & 0 & 0 & 0\\
0 & 4 & 1 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & -1 & 0\\
0 & 0 & 0 & 0 & -1
\end{bmatrix}$
$J_3=\begin{bmatrix}
4 & 1 & 0 & 0 & 0\\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}$
$J_4=\begin{bmatrix}
4 & 1 & 0 & 0 & 0\\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}$
$J_5=\begin{bmatrix}
4 & 0 & 0 & 0 & 0\\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}$
$J_6=\begin{bmatrix}
4 & 0 & 0 & 0 & 0\\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & -1
\end{bmatrix}$
There are $3 \times 2=6$ possible Jordan canonical forms.
