How are the eigenvalues and eigenvectors of $A$ and $B$ related if $B=\left[\begin{smallmatrix}O&A\\A&O\end{smallmatrix}\right]$?

Question

Given matrix $$A=\begin{bmatrix}a&b&b\\b&a&b\\b&b&a \end{bmatrix}$$ and matrix $$B=\begin{bmatrix}0&0&0&a&b&b\\0&0&0&b&a&b\\0&0&0&b&b&a\\a&b&b&0&0&0\\b&a&b&0&0&0\\b&b&a&0&0&0 \end{bmatrix}$$ I want to calculate the eigenvalues and eigenvectors of $B$, and I know the eigenvalues of $A$ are $(a-b)$ and $(a+2b)$, and the corresponding eigenvectors are $(-1,1,0)$, $(-1,0,1)$, $(1,1,1)$. I can calculate $B$ in the normal way but it seems that there should be some shortcuts, so is there any relation between $A$ and $B$?

Thanks in advance.

Answer 1

The larger matrix $B$ is the Kronecker product of $S=\binom{0~1}{1~0}$ and $A$. Since $S$ is diagonalisable with eigenvalues $-1,1$, then the eigenvalues of $B$ are $\pm\lambda_i$ where $\lambda_i$ runs through the eigenvalues of $A$.

Answer 2

Suppose $B=\begin{bmatrix}0&A\\ A&0 \end{bmatrix}$ has an eigenpair $(\lambda$, $\begin{bmatrix}x\\ y \end{bmatrix})$ (note that $x$ and $y$ are row vectors). We have $ Ax=\lambda y $ and $Ay = \lambda x$. Therefore $A^2 x = A (Ax) = \lambda A y = \lambda^2 x$. We conclude that $(\lambda^2,x)$ is an eigenpair of $A^2$.

In your case $A$ is of full rank. Therefore all eigenpairs of $A^2$ can be generated by those of $A$ (I'll leave it to you), i.e., either $(\lambda,x)$ or $(-\lambda,x)$ is an eigenpair of $A$.

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In brief, you get that if $(\lambda$, $\begin{bmatrix}x\\ y \end{bmatrix})$ is an eigenpair of $B$, then either $(\lambda,x)$ or $(-\lambda,x)$ is an eigenpair of $A$.