# Find the permutation matrix

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.

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

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

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

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