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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

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

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  • $\begingroup$ I think it's quite easy to see that (in the even-dimensional case) the characteristic polynomial is even, so the eigenvalues of $A$ come in pairs $\pm \lambda$. (In odd dimensions, it is odd, and the conclusions are the same.) $\endgroup$ May 3, 2023 at 0:42
  • $\begingroup$ how do you see that it is even? $\endgroup$ May 3, 2023 at 0:43
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    $\begingroup$ The coefficient of $\lambda^k$ is the sum of determinants of smaller matrices, and these are singular when $k$ is odd. (For example, in the $4\times 4$ case, the coefficient of $\lambda$ comes from the $3\times 3$ submatrices obtained by deleting the $k$th row and column.) $\endgroup$ May 3, 2023 at 0:49
  • $\begingroup$ Induction on the size of the matrix should work too. $\endgroup$ May 3, 2023 at 0:51
  • $\begingroup$ see also mathoverflow.net/q/446048/11260 $\endgroup$ May 3, 2023 at 14:04

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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.

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    $\begingroup$ OP wrote, "I don't want an explicit computation," so would probably be happier with a more theoretical argument showing the eigenvalues (before squaring) come in $\pm$ pairs. Such an argument is given in the comments by user Christian Remling. $\endgroup$ May 3, 2023 at 2:18

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