A question in simillarity of matrices I have a 4x4 matrix 
$$ A = 
\left[ 
\begin{matrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1
\end{matrix}
\right] $$
and i want to check if there is any similar matrix of matrix $A$ of the form:
$$ B = 
\left[ 
\begin{matrix}
0 & 0 & x & y \\
0 & 0 & z & w \\
a & b & 0 & 0 \\
c & d & 0 & 0
\end{matrix}
\right] $$
Is there any way to come in a conclusion if there such of this form similarity ? 
thanks.
 A: note that two similar matrix have the same caracteristic polinomial $P(t)$, if we calcul this polynomial for each matrix we tack an have coefficient -1 at $t^3$ and othere $0$ as coefficient at $t^3$ so  the similarity is impossible
A: Working over $\mathbb C$, let's write $B$ as a block matrix
$$ B = \begin{bmatrix}0 & X \\ E & 0 \end{bmatrix} $$
Then
$$ B^2 = \begin{bmatrix}XE & 0 \\ 0 & EX\end{bmatrix} $$
and since $XE$ and $EX$ have the same eigenvalues, we can do a case analysis on the number of these eigenvalues:


*

*If there's only one eigenvalue for $EX$ and $XE$, then $B^2$ has only one eigenvalue. However, Wolfram Alpha tells us that $A^2$ has three different eigenvalues, so $A$ cannot be similar to such a $B$.

*On the other hand, if $EX$ has two eigenvalues, it is diagonalizable, and so is $XE$, so $B^2$ is diagonalizable. But Wolfram Alpha just showed us that $A^2$ isn't diagonalizable, so $A$ is not similar to this kind of $B$ either.
Since $A$ is not similar to any $B$-shaped matrix over $\mathbb C$, this is in particular not the case over $\mathbb R$ either.
