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 $\lambda$.
Linear Algebra – How to Quickly Find Eigenvalues of Anti-Diagonal Matrix
eigenvalues-eigenvectorslinear algebramatrices
Best Answer
For ease of formatting and explanation, I'll be doing everything for the $5 \times 5$ example. However, the same trick works for any $n \times n$ antisymmetric 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 $A$.
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.
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.