Quick way to find eigenvalues of anti-diagonal matrix

Question

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 $\lambda$.

Answer 1

For ease of formatting and explanation, I'll be doing everything for the $5 \times 5$ example. However, the same trick works for any $n \times n$ antidiagonal matrix (though slightly differently for even $n$).

Suppose $$ A = \begin{pmatrix}0&0&0&0&a_{15}\\0&0&0&a_{24}&0\\0&0&a_{33}&0&0\\0&a_{42}&0&0&0\\a‌​_{51}&0&0&0&0 \end{pmatrix} $$

Here's a neat trick: we note that $$ A^2 = \pmatrix{ a_{15}a_{51}&&&&\\ &a_{24}a_{42}&&&\\ &&(a_{33})^2&&\\ &&&a_{24}a_{42}&\\ &&&&a_{15}a_{51}\\ } $$ So, the eigenvalues of $A^2$ are precisely $\{a_{15}a_{51}, a_{24}a_{42}, (a_{33})^2\}$.

Now, note that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ must be an eigenvalue of $A^2$. This gives you six candidates for the eigenvalues of $A$.

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In fact, with more thorough analysis, we can guarantee that the eigenvalues will be precisely $\lambda = \pm \sqrt{a_{i,(n+1-i)}a_{(n+1-i),i}}$ for $i = 1,\dots,\lfloor n/2\rfloor$ and, for odd $n$, $\lambda = a_{(n+1)/2,(n+1)/2}$.

Proof that this is the case: Let $e_1,\dots,e_n$ denote the standard basis vectors. Let $S_{ij}$ denote the span of the vectors $e_i$ and $e_j$.

Note that $A$ is invariant over $S_{i(n-i)}$ for $i = 1,\dots,\lfloor n/2\rfloor$. We may then consider the restriction $A_{i(n-i)}: S_{i(n-i)} \to S_{i(n-i)}$, which can be represented by the matrix $$ \pmatrix{0 & a_{i(n-i)}\\a_{(n-i)i} & 0} $$ It suffices to find the eigenvalues of this transformation.

For the case of an odd $n$, it is sufficient to note that $a_{(n+1)/2,(n+1)/2}$ lies on the diagonal with zeros in its row and column.

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Another explanation: denote the matrix $S = \pmatrix{e_1 & e_{n} & e_2 & e_{n-1} & \cdots}$

Noting that $S$ is orthogonal (i.e. $S^{-1} = S^{T}$), we find that $$ SAS^{-1} = \pmatrix{ 0&a_{1,n}\\ a_{n,1}&0\\ &&0&a_{2,n-1}\\ &&a_{n-1,2}&0\\ &&&&\ddots } $$ This matrix is similar, and therefore has the same eigenvalues. However, it is also block diagonal.

Answer 2

Suppose $A$ has even size, say, $2m \times 2m$. Then by reordering the basis we can produce a block diagonal matrix with the same eigenvalues as the original.

$$ \left( \begin{array}{cc} 0 & a_{1,2m} \\ a_{2m,1} & 0 \\ \end{array} \right) \oplus \cdots \oplus \left( \begin{array}{cc} 0 & a_{m,m+1} \\ a_{m+1,m} & 0 \\ \end{array} \right). $$ The characteristic polynomial is

$$\det(\lambda I - A) = \det\left(\lambda I - \left( \begin{array}{cc} 0 & a_{1,2m} \\ a_{2m,1} & 0 \\ \end{array} \right)\right) \cdots \det\left(\lambda I - \left( \begin{array}{cc} 0 & a_{m,m+1} \\ a_{m+1,m} & 0 \\ \end{array} \right)\right) = (\lambda^2 - a_{1,2m}a_{2m,1})\cdots(\lambda^2 - a_{m,m+1}a_{m+1,m}) . $$ and so the eigenvalues are the roots of these factor polynomials, namely both square roots of each of the products $a_{1,2m}a_{2m,1}, \ldots, a_{m,m+1} a_{m+1,m}$.

If $A$ has odd size, say, $(2m + 1) \times (2m + 1)$ when we change basis we can send the middle ($(m+1)$st) element to the end, in which case the characteristic polynomial takes a similar form as in the even case (with indices $> m$ shifted up one), but with an additional factor of $\lambda - a_{m+1, m+1}$, so the additional eigenvalue in this case is just the middle entry of the matrix.

Answer 3

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 $$ (x^2-a_{1n}a_{n1})(x^2-a_{2,n-1}a_{n-1,2})\cdots\left(x^2-a_{\frac n2,\frac n2+1}a_{\frac n2+1,\frac n2}\right). $$