Eigenvalues of block matrix $\begin{bmatrix}0 &A\\A&0\end{bmatrix}$

Question

If the eigenvalues of a matrix $A$ are $\lambda_1, \lambda_2, \dots, \lambda_n$, what are the eigenvalues of the following matrix?

$$\begin{bmatrix}0 &A\\A&0\end{bmatrix}$$

From some numerical examples, I have found that the eigenvalues of this block matrix are just $\lambda_1, \lambda_2, \dots, \lambda_n$ and $-\lambda_1, -\lambda_2, \dots, -\lambda_n$, but I am not sure how to go about proving it. I have tried taking $ \det(A - \lambda I) = 0 $, but I could not find a way to work with the results. Where is a good place to start?

Answer 1

Hint: using the last formula on this list, note that $$ \det\left( \begin{bmatrix}0 &A\\A&0\end{bmatrix} - \lambda I_{2n} \right)= \det\left(-\lambda I_n-A\right)\det\left(-\lambda I_n+A\right) $$

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Edit: Another approach, since this question still seems to be getting attention.

One could find the eigenvalues of this block-matrix using the properties of the Kronecker product. In particular, we have $$ M = \pmatrix{0 & A\\A & 0} = \pmatrix{0&1\\1&0} \otimes A. $$ Because the eigenvalues of $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ are given by $\mu_1 = 1, \mu_2 = -1$, we can conclude that the eigenvalues of $M$ are given by all products $\lambda_i \mu_j$ for $i = 1,\dots,n$ and $j = 1,2$.

Answer 2

If $x=(x_1, \ldots, x_n)^T$ is an eigenvector of $A$ so that $Ax = \lambda x$, consider the vector $x' = (x_1, \ldots, x_n, x_1, \ldots x_n)^T$. What can be said about the product of your matrix with $x'$?

Answer 3

Let $$B=\begin{bmatrix} 0 & A \\ A & 0 \\ \end{bmatrix}.$$ We observe that $$\begin{bmatrix} 0 & A \\ A & 0 \\ \end{bmatrix}\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \end{bmatrix}=\lambda\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \end{bmatrix}$$ implies that $A\mathbf{v}=\lambda\mathbf{u}$ and $A\mathbf{u}=\lambda\mathbf{v}$. This implies $\mathbf{u}$ satisfies $$A^2\mathbf{u}=\lambda^2 \mathbf{u}$$ and $\mathbf{v}$ satisfies $$A^2\mathbf{v}=\lambda^2 \mathbf{v}.$$

So the eigenvectors of $B$ must have the form $$\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \\ \end{bmatrix}$$ where $\mathbf{u}$ and $\mathbf{v}$ are both eigenvectors of $A^2$ such that $\mathbf{u}$ and $\mathbf{v}$ belong to the same eigenspace. This eigenvector of $B$ has corresponding eigenvalue either $\pm \lambda$.

We conclude that the eigenvalues of $B$ belong to $$\{\pm \lambda: \lambda \text{ is an eigenvalue of } A\}.$$

Finally, we check that these eigenvalues are achieved. If $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$\begin{bmatrix} \mathbf{v} \\ \mathbf{v} \end{bmatrix}$$ is an eigenvector of $B$ with eigenvalue $\lambda$ and $$\begin{bmatrix} -\mathbf{v} \\ \mathbf{v} \end{bmatrix}$$ is an eigenvector of $B$ with eigenvalue $-\lambda$.