TITLE: Eigenvalues of block matrix $\begin{bmatrix}0 &A\\A&0\end{bmatrix}$
QUESTION [9 upvotes]: 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?

REPLY [13 votes]: 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 $-\lambda$.