Find the permutation matrix

Question

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.

Answer 1

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 $MJM^{-1}=J^t$.

Answer 2

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 $MJ\stackrel{?}=J^tM$.)