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Let:

$$J=\begin{bmatrix} \lambda&1&0\\ 0&\lambda&1\\ 0&0&\lambda \end{bmatrix}$$

Find a permutation matrix $M$ such that $$M J M^{-1} = J^{t}$$

I know that $J$ is a Jordan form matrix, but I don't even have an idea as to how to approach the problem.

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  • $\begingroup$ I presume you mean let $J = $ that matrix. $\endgroup$ – Robert Israel Jun 23 at 23:12
  • $\begingroup$ Forget $M$ for the moment. Can you come up with a sequence of row and column swaps that transpose $J$? $\endgroup$ – amd Jun 23 at 23:12
  • $\begingroup$ Oh, forgot about that, thanks for notyfing me ! $\endgroup$ – Guilherme takata Jun 24 at 17:06
  • $\begingroup$ @amd I can come up with such sequence but does that help in any way in finding the matrix M ? $\endgroup$ – Guilherme takata Jun 24 at 17:43
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There are only five permutation matrices to try.

I found $M=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}$ works, by trial and error. (I checked $MJ\stackrel{?}=J^tM$.)

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Let $\{e_1,e_2,e_3\}$ be standard basis. Then $J.e_1=\lambda e_1$, $J.e_2=e_1+\lambda e_2$, and $J.e_3=e_2+\lambda e_3$. In other words, $J.e_3=\lambda e_3+e_2$, $J.e_2=\lambda e_2+e_1$, and $J.e_1=e_1$. So, consider the permutation

  • $e_1\rightarrow e_3$
  • $e_2\rightarrow e_2$;
  • $e_3\rightarrow e_1$.

The matrix of the linear map $v\mapsto J.v$ with respect to the basis which consists of $e_3$, $e_2$, and $e_1$ is then$$\begin{bmatrix}\lambda&0&0\\1&\lambda&0\\0&1&\lambda\end{bmatrix}=J^T.$$So, let $M$ be the inverse of the change-of-bases matrix between $\{e_1,e_2,e_3\}$ and $\{e_3,e_2,e_1\}$, which is$$\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}.$$Then you will have $MJM^{-1}=J^t$.

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  • $\begingroup$ I dont quite get the reasoning behind this solution, I see that I need to solve the equation: $MJ=J^{t}M$ but i dont know ho to proceed now $\endgroup$ – Guilherme takata Jun 24 at 17:40
  • $\begingroup$ No, you certainly don't need to solve that equation, since what I did proves that $MJM^{-1}=J^T$. $\endgroup$ – José Carlos Santos Jun 24 at 17:50
  • $\begingroup$ Still why do you considered said permutation, I sitll dont see the reasoning behind this $\endgroup$ – Guilherme takata Jun 24 at 17:52
  • $\begingroup$ Do you agree that $J.e_1=\lambda e_1$? $\endgroup$ – José Carlos Santos Jun 24 at 17:59
  • $\begingroup$ Yes but i dont see how multiplying by the standart basis helps in any way $\endgroup$ – Guilherme takata Jun 24 at 18:00

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