How do I find out that the following two matrices are similar? How do I find out that the following two matrices are similar?
$N =
\begin{pmatrix} 
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 
\end{pmatrix}$
and $M=
\begin{pmatrix} 
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 
\end{pmatrix}$
I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried
$P=
\begin{pmatrix} 
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 
\end{pmatrix}$
such that $PN = \begin{pmatrix} 
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 
\end{pmatrix}$ but then $PNP^{-1} \neq M$.
My linear algebra is a bit rusty. Is there a more elaborate way to do this?
 A: The two matrices are made of Jordan blocks; in $2\times2$ block format, they are
$$
N=\begin{bmatrix} J & 0 \\ 0 & 0 \end{bmatrix}
\qquad
M=\begin{bmatrix} 0 & 0 \\ 0 & J \end{bmatrix}
$$
You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
$$
M=\begin{bmatrix} 0 & I_2 \\ I_2 & 0 \end{bmatrix} N
\begin{bmatrix} 0 & I_2 \\ I_2 & 0 \end{bmatrix}
$$
A: Let $(e_1,e_2,e_3,e_3)$ be the standard basis of $\mathbb{R}^4$. You have:


*

*$N.e_1=0$;

*$N.e_2=e_1$;

*$N.e_3=0$;

*$N.e_4=0$.


You also have:


*

*$M.e_3=0$;

*$M.e_4=e_3$;

*$M.e_1=0$;

*$M.e_2=0$.


So, if you see $M$ as a linear map from $\mathbb{R}^4$ into itself, the matrix of $M$ with respect to the basis $(e_3,e_4,e_1,e_2)$ is the matrix $N$. Therefore, $N$ and $M$ are similar.
Or you can take$$P=\begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix},$$which is basically the same thing.
A: Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus,  $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.
