Find all Jordan canonical forms of 2 x 2, 3 x 3 and 4 x 4 matrices over C. Can someone walk me through understanding this question/solution to it? On the one hand, I'm thinking something simple like (for the $2 \times 2$ case):
\begin{bmatrix}
    \alpha & 1 \\
    0 & \beta \\
\end{bmatrix}
and
\begin{bmatrix}
    \alpha & 0 \\
    0 & \beta \\
\end{bmatrix}
Since there's no other way to be in JCF for a $2 \times 2$ matrix. On the other hand, this seems to naive. 
 A: In the following, $\alpha,$ $\beta,$ $\gamma,$ $\delta$ do not necessarily represent distinct eigenvalues. Jordan blocks can be permuted, so the second three-by-three and last four-by-four matrices below each have two permutations, and the second four-by-four matrix has three permutations. I assume that you are interested only in the structure of the Jordan blocks; otherwise, there are more permutations to consider by permuting the order of the eigenvalues.
$$\begin{bmatrix}
\alpha & 0\\
0 & \beta
\end{bmatrix}
\begin{bmatrix}
\alpha & 1\\
0 & \alpha
\end{bmatrix}$$
$$\begin{bmatrix}
\alpha & 0 & 0\\
0 & \beta & 0\\
0 & 0 & \gamma\\
\end{bmatrix}
\begin{bmatrix}
\alpha & 0 & 0\\
0 & \beta & 1\\
0 & 0 & \beta\\
\end{bmatrix}
\begin{bmatrix}
\alpha & 1 & 0\\
0 & \alpha & 1\\
0 & 0 & \alpha\\
\end{bmatrix}$$
$$\begin{bmatrix}
\alpha & 0 & 0 & 0\\
0 & \beta & 0 & 0\\
0 & 0 & \gamma & 0\\
0 & 0 & 0 & \delta\\
\end{bmatrix}
\begin{bmatrix}
\alpha & 0 & 0 & 0\\
0 & \beta & 0 & 0\\
0 & 0 & \gamma & 1\\
0 & 0 & 0 & \gamma\\
\end{bmatrix}
\begin{bmatrix}
\alpha & 1 & 0 & 0\\
0 & \alpha & 0 & 0\\
0 & 0 & \beta & 1\\
0 & 0 & 0 & \beta\\
\end{bmatrix}
\begin{bmatrix}
\alpha & 0 & 0 & 0\\
0 & \beta & 1 & 0\\
0 & 0 & \beta & 1\\
0 & 0 & 0 & \beta\\
\end{bmatrix}$$
