Possible Jordan Canonical Forms Given Minimal Polynomial I was supposed to find all possible Jordan canonical forms of a $5\times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester.  I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2\times 2$ while the largest Jordan block with eigenvalue 1 is $1\times 1$.  Did I miss any matrices or interrupt the minimal polynomial incorrectly?
\begin{pmatrix}
  2 &1  &0  &0  &0\\
  0 &2  &0  &0  &0\\
  0 &0  &2  &1  &0\\
  0 &0  &0  &2  &0\\
  0 &0  &0  &0  &1
\end{pmatrix}
\begin{pmatrix}
  2 &1  &0  &0  &0\\
  0 &2  &0  &0  &0\\
  0 &0  &2  &0  &0\\
  0 &0  &0  &2  &0\\
  0 &0  &0  &0  &1
\end{pmatrix}
\begin{pmatrix}
  2 &1  &0  &0  &0\\
  0 &2  &0  &0  &0\\
  0 &0  &2  &0  &0\\
  0 &0  &0  &1  &0\\
  0 &0  &0  &0  &1
\end{pmatrix}
\begin{pmatrix}
  2 &1  &0  &0  &0\\
  0 &2  &0  &0  &0\\
  0 &0  &1  &0  &0\\
  0 &0  &0  &1  &0\\
  0 &0  &0  &0  &1
\end{pmatrix}
\begin{pmatrix}
  2 &0  &0  &0  &0\\
  0 &2  &0  &0  &0\\
  0 &0  &1  &0  &0\\
  0 &0  &0  &1  &0\\
  0 &0  &0  &0  &1
\end{pmatrix}

Yes, you did.
Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1.  You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2.  Using those rules yields precisely your first four matrices.  Your fifth matrix is not correct.
Furthermore, you can permute the blocks.  Thus,


*

*your first matrix yields 3!/2! = 3 Jordan forms,

*your second and third matrices yield 4!/2! = 12 forms each, and

*your four matrix yields 4!/3! = 4 forms


for a total of 3 + 2 · 12 + 4 = 31 forms.
