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

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

From some numerical examples I have found that the eigenvalues 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?

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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 $-\lambda$.

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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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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 $x'$?

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