Answer
See below
Work Step by Step
Given: $p(\lambda)=(-2-\lambda)^2(6-\lambda)^5$
Find the eigenvalues:
$\lambda_1=-2$ with multiplicity of 2
$\lambda_2=6$ with multiplicity of 5
The possibilities for $\lambda_1=-2$ with multiplicity of 2 are:
One block of $2 \times 2$ matrix; Two blocks of $1\times 1$
The possibility for $\lambda_1=6$ with multiplicity of 5 are:
One block of $5 \times 5$; One blocks of $4 \times 4$ and one block of $1 \times 1$; One block of $3 \times 3$ and one block of $2 \times 2$; Two blocks of $2 \times 2$ and one block of $1 \times 1$; One block of $2 \times 2$ and three blocks of $1 \times 1$; Five blocks of $1 \times 1$.
The possibilities for $\lambda_1=0$ with multiplicity of 2 are:
One block of $2 \times 2$ matrix; Two blocks of $1\times 1$
The set S of matrices include:
$\begin{bmatrix}
6 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 6 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
$\begin{bmatrix}
6 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 6 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
$\begin{bmatrix}
6 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 6 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
$\begin{bmatrix}
6 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 6 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
$\begin{bmatrix}
6 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 6 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
$\begin{bmatrix}
6 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 0& 0 & 0 & 0\\
0 & 0 & 0 & 6 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
$\begin{bmatrix}
6 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 6 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 6 & 0& 0 & 0 & 0\\
0 & 0 & 0 & 6 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 6 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & -2
\end{bmatrix}$
There are $7 \times 2=14$ possible Jordan canonical forms.
