TITLE: Quick way to find eigenvalues of anti-diagonal matrix
QUESTION [28 upvotes]: If $A \in M_n(\mathbb{R})$ is an anti-diagonal $n \times n$ matrix, is there a quick way to find its eigenvalues in a way similar to finding the eigenvalues of a diagonal matrix? The standard way for finding the eigenvalues for any $n \times n$ is usually straightforward, but may sometimes lead to painstaking computational time. Just wondering if there was a quicker way in doing so for any anti-diagonal matrix without having to resort to the standard method of finding the determinant of $A - \lambda I$, where $I$ is the $n \times n$ identity matrix, and setting it equal to $0$ and solving for $\lambda$.

REPLY [5 votes]: One can actually find the eigenvalues of $A$ directly without reordering the indices (note: I'm not saying that this is the quickest or preferred way to find those eigenvalues -- this is not). The eigenvalues of $A$ are just the roots of the characteristic polynomial $\det(xI-A)$. When $n$ is odd, let $m=\frac{n+1}2$ and write
$$
A=\left[\begin{array}{c|c|c}0&0&A_{13}\\ \hline 0&a_{mm}&0\\ \hline A_{31}&0&0\end{array}\right]\tag{$\ast$}
$$
where $A_{13}$ and $A_{31}$ are two anti diagonal matrices of size $m-1$. By Laplace expansion along the middle row, we get $\det(xI-A)=(x-a_{mm})\det\left[\begin{array}{c|c}xI&-A_{13}\\ \hline -A_{31}&xI\end{array}\right]$. Using the formula $\det\pmatrix{X&Y\\ Z&W}=\det(XW-YZ)$ when $Z$ and $W$ commute, we further obtain
\begin{align*}
\det(xI-A)&=(x-a_{mm})\det(x^2I-A_{13}A_{31})\\
&= (x-a_{mm})(x^2-a_{1n}a_{n1})(x^2-a_{2,n-1}a_{n-1,2})\cdots\left(x^2-a_{m-1,m+1}a_{m+1,m-1}\right)
\end{align*}
and finding its roots is a trivial matter.
When $n$ is even, the central element in $(\ast)$ vanishes and we may skip the Laplace expansion step. The characteristic polynomial of $A$ is then
$$
(x^2-a_{1n}a_{n1})(x^2-a_{2,n-1}a_{n-1,2})\cdots\left(x^2-a_{\frac n2,\frac n2+1}a_{\frac n2+1,\frac n2}\right).
$$