Consider the matrix $$A:=\left(
\begin{array}{cccc}
 0 & a & 0 & 0 \\
 f & 0 & b & 0 \\
 0 & e & 0 & c \\
 0 & 0 & d & 0 \\
\end{array}
\right)$$


I noticed that if I square this matrix then the eigenvalues of $A^2$ are two-fold degenerate. Does anyone see how this follows? I don't want an explicit computation but rather an argument that generalizes to arbitrary matrix sizes.


The same phenomenon follows (aside from an eigenvalue 0, since the matrix size is odd) if the matrix is continued analogously, i.e. if I consider


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



Eigenvalues two-fold degenerate
For the eigenvalues of the matrix powers the following identity holds:



> 
> If $A$ is a square ($d \times d$) matrix with associated eigenvalues $\lambda\_1,\dots,\lambda\_d$, then the eigenvalues of $A^n$ are
> $$\lambda\_1^n,\dots,\lambda\_d^n$$.
> 
> 
> 


This can be shown by considering the eigenvalue/eigenvector equality for $A$


$$A\;\mathbf{e}\_i=\lambda\_i \mathbf{e}\_i,$$


where $\mathbf{e}\_1, \mathbf{e}\_2, \dots, \mathbf{e}\_d$ are the eigenvectors, and take the product of the power of $A$ with an eigenvector and progressively replace products of $A$ with the eigenvector with $\lambda\_i \mathbf{e}\_i$


\begin{align}
A^d \mathbf{e}\_i&=A^{d-1} A\mathbf{e}\_i \\&=A^{d-1}\lambda\_i \mathbf{e}\_i\\&=\lambda\_i A^{d-2} A\mathbf{e}\_i \\&= \dots\\&=\lambda\_{i}^{d}\mathbf{e}\_i.\end{align}


In your case the eigenvalues of $A$ are


$\lambda\_1 \approx -0.707107 \sqrt{-\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $


$\lambda\_2 \approx 0.707107 \sqrt{-\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $


$\lambda\_3 \approx-0.707107 \sqrt{\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $


$\lambda\_4 \approx 0.707107 \sqrt{\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $


so the pairs $(\lambda\_1, \lambda\_2 )$ and $(\lambda\_3, \lambda\_4 )$ have same magnitudes and opposite signs, which results in them being equal pairwise once you square them.


