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  • Prove that $J_{n}(0)$ and $(J_{n}(0))^t$ are similar ($J_n(0)$ is a $n \times n$ Jordanian block which belongs to the eigenvalue $0$).

  • Use your answer and Jordanian form to prove that every matrix $A \in M_{nxn}^C$ is similar to $A^t$.

My problem is with the first question. Let's take for example the following matrices from a Jordanian form:

$$ B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix} $$

$$ B^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$

Who's ranks are 1 and 2 respectively. Similar matrices should have the same rank, therefore $B$ and $B^2$ aren't similar and the first question can't be proven right. Since I didn't prove the first question, I guess I can't continue to the second one.

I probably missed something, any help would be appriciated, thanks!

Irregular User
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Eytan Manor
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    $t$ here means transpose – Ben Grossmann Apr 30 '16 at 13:32
  • Ohh that's right, I don't know why I missed that. Thank you! – Eytan Manor Apr 30 '16 at 15:37
  • The matrix $B^2$ is not in Jordan normal form (the only nonzero entries should be directly above the main diagonal, which the top-right entry $1$ is not). – Marc van Leeuwen May 07 '16 at 10:42
  • For comparison with the duplicate question, the first point here is answered in the last (8th) point of my answer there: it suffices to conjugate a Jordan block by the order-reversing permutation matrix (the one with entries $1$ on the (bottom-let to top-right) anti-diagonal to get the transpose of a Jordan block. The remainder of the points there are needed to get from there to the second point of this question (the general case, at least if the field is $\Bbb C$), which I presume is supposed to be the "easy" part. – Marc van Leeuwen May 07 '16 at 10:52

1 Answers1

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The minimal polynomial of $J_n(0)^t$ is $x^n$. This is its characteristic polynomial as well. This tells you that the Jordan canonical of $J_n(0)^t$ is $J_n(0)$. But then they are similar.

Ken Duna
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