This question is a more precise version of [this question.][1]

Let's assume we have the matrix 

$$\left(
\begin{array}{ccccc}
 0 & a & 0 & 0 & 0 \\
 f & 0 & b & 0 & 0 \\
 0 & e & 0 & c & 0 \\
 0 & 0 & d & 0 & r \\
 0 & 0 & 0 & g & 0 \\
\end{array}
\right)$$

If we square it, we get the matrix 

$$\left(
\begin{array}{ccccc}
 a f & 0 & a b & 0 & 0 \\
 0 & a f+b e & 0 & b c & 0 \\
 e f & 0 & b e+c d & 0 & c r \\
 0 & d e & 0 & c d+g r & 0 \\
 0 & 0 & d g & 0 & g r \\
\end{array}
\right)$$

We see that this matrix decomposes into two submatrices

$$C_1:=\left(\begin{array}{cccc}
 a f  & a b & 0 \\
 e f  & b e+c d & c r \\
 0  & d g & g r \\
\end{array}\right)$$
and 
$$C_2:=\left(\begin{array}{cc}
 a f+b e & b c \\
 d e & c d+g r \\
\end{array}\right)$$
Now, one can check explicitly that the two submatrices are isospectral apart  from one eigenvalue zero. I wonder if there is an abstract argument why this is so? 

It would for instance follow if we can write the matrices as $C_1 = AB$ and $C_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.


  [1]: https://mathoverflow.net/questions/446033/eigenvalues-two-fold-degenerate