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

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