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

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    $\begingroup$ It depends what you define JCF to be. Many insist that for a given eigenvalue the block sizes decrease. $\endgroup$ – ancientmathematician Aug 5 '18 at 17:08
  • $\begingroup$ @ancientmathematician In that case, Frank White's first and forth matrices still yield 3 and 4 forms because the blocks for a given eigenvalue are the same size. The second matrix, however, now yields only 4 forms, and the third matrix yields only 6 forms for a total of 17 forms. $\endgroup$ – Maurice P Aug 6 '18 at 15:11
  • $\begingroup$ I agree with all the calculations.. $\endgroup$ – ancientmathematician Aug 6 '18 at 15:51

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