# 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}

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You must have a Jordan block associated to the eigenvalue 2 of size 2. Otherwise the minimal polynomial would be $(x-1)(x-2)$. – Brandon Carter Jan 6 '13 at 0:15
I see. So the last matrix is knocked off. Good. – Frank White Jan 6 '13 at 0:18