Answer
See below
Work Step by Step
There are 3 possible Jordan canonical forms:
$J_1=\begin{bmatrix}
4 & 0 & 0 & 0 & 0 \\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 5 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $1$
$J_2=\begin{bmatrix}
4 & 1 & 0 & 0 & 0 \\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 4 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $2$
$J_3=\begin{bmatrix}
4 & 1 & 0 & 0 & 0 \\
0 & 4 & 0 & 0 & 0\\
0 & 0 & 4 & 1 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 3 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $2$
$J_4=\begin{bmatrix}
4 & 1 & 0 & 0 & 0 \\
0 & 4 & 1 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 3 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $3$
$J_5=\begin{bmatrix}
4 & 1 & 0 & 0 & 0 \\
0 & 4 & 1 & 0 & 0\\
0 & 0 & 4 & 0 & 0\\
0 & 0 & 0 & 4 & 1\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 2 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $3$
$J_6=\begin{bmatrix}
4 & 1 & 0 & 0 & 0 \\
0 & 4 & 1 & 0 & 0\\
0 & 0 & 4 & 1 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 2 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $4$
$J_1=\begin{bmatrix}
4 & 1 & 0 & 0 & 0 \\
0 & 4 & 1 & 0 & 0\\
0 & 0 & 4 & 1 & 0\\
0 & 0 & 0 & 4 & 1\\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$
There are 1 linearly independent eigenvectors of a matrix with this Jordan canonical form, and the maximum length of a cycle is $5$