# Eigenvalues of a rotationally symmetric matrix

I have a rotationally symmetric matrix of arbitrary size, for example,

$$A = \begin{pmatrix} a & b & c & b & a \\ b & d & e & d & b \\ c & e & f & e & c \\ b & d & e & d & b \\ a & b & c & b & a \end{pmatrix}$$

I am trying to find the eigenvectors & eigenvalues of the matrix, but am really struggling. I realise that if A is size $2n \times 2n$, and if $J$ is an exchange matrix of size $n \times n$, i.e $$J = \begin{pmatrix} 0 & 0 & ... & 0 & 1 \\ 0 & 0 & ... & 1 & 0 \\ & & ... & & \\ 1 & 0 & ... & 0 & 0 \end{pmatrix},$$ then I can represent $A$ as $$A = \begin{pmatrix} M & MJ \\ JM & JMJ \end{pmatrix},$$ for a symmetric matrix $M$ of size $n \times n$. That is as far as I got. there is a lot of dependence. There is a matrix $U$ that block diagonalises $A$; if $$U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & -J \\ J & I \end{pmatrix},$$ then $$U^TAU = \begin{pmatrix} 0 & 0 \\ 0 & 4JMJ \end{pmatrix}.$$ I would then find the $n$ eigenvectors of $JMJ$?
I'm new to matrix differential equations so I'm not sure how to interpret the dependence. Perhaps it's a silly question?

Any help would be greatly appreciated as it is driving me nuts!

Thank you very much for your help, Katie.

• Don't you have lots of dependent row/columns here? – A.Γ. Oct 9 '15 at 7:51
• The title says determinant, the post says you are looking for eigenvalues...? – Martin Sleziak Oct 9 '15 at 13:34

## 2 Answers

The matrix structure that you’ve described as “rotationally symmetric” is often referred to as a bisymmetric matrix. Bisymmetric matrices are symmetric centrosymmetric, where centrosymmetric matrices are those matrices $A$ which commute with exchange matrix $J$. There’s a fair amount of literature that has been written on such structures, albeit with considerable replication of elementary results and usually not much new to say without further constraints on the entries. Your initial investigation, involving a $2 \times 2$ block decomposition of $A$ involving the exchange matrix, is headed in the right general direction. Many papers on centrosymmetric matrices exploit the fact that one can simultaneously diagonalize $A$ and $J$ (since they commute) and $J$’s diagonalization involves two natural blocks corresponding to its eigenvalues $-1$ and $1$.

Assume $A$ is $n \times n$. I would say that the most relevant result for your question, which deals with unconstrained bisymmetric matrices, is that an eigenvector $x$ of $A$ will be either symmetric (i.e., $x = Jx$) or skew-symmetric ($x=-Jx$), and moreover that $\left\lceil {n/2} \right\rceil$ of them will be symmetric while $\left\lfloor {n/2} \right\rfloor$ of them will be skew-symmetric. Using this fact, it is possible to generate the eigenvectors of $A$ by solving two eigenproblems of roughly half the size of the original. For details on the eigenproblem construction, I’ll refer you to Theorem 2 of the paper Eigenvalues and eigenvectors of symmetric centrosymmetric matrices by Cantoni and Butler in Linear Algebra and its Applications, 13 (1976) where it is fully laid out (see pages 280-281 per the journal pagination).

Two brief notes regarding Cantoni and Butler’s result on the quantification of $A$’s symmetric and skew-symmetric eigenvectors. First, this result is implicit in the earlier paper of Alan Andrew, Eigenvectors of certain matrices in Linear Algebra and its Applications, 7 (1973). Cantoni and Butler’s explicit construction, however, is probably more of what you’re looking for here. Second, there’s a natural generalization of this result that deals with real symmetric matrices $A$ that commute with a fixed involutory matrix (so, not just limited to matrices that commute with the exchange matrix $J$). If you’re interested in learning about this, it’s given in Lemma 3.11 and Remark 3.12 on page 890 of the paper of Tao and Yasuda, A spectral characterization of generalized real symmetric centrosymmetric and generalized real symmetric skew-centrosymmetric matrices, SIAM J. Matrix Analysis and Applications, 23 (2002).

Thanks for your response A.G. I can't seem to respond to your comment as I'm an unregistered user (a guest). Yes, there is a lot of dependence. There is a matrix U that block diagonalises A; if $$U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & -J \\ J & I \end{pmatrix},$$ then $$U^TAU = \begin{pmatrix} 0 & 0 \\ 0 & 4JMJ \end{pmatrix}.$$ I would then find the n eigenvectors of JMJ? I'm new to matrix differential equations so I'm not sure how to interpret the dependence. Perhaps it's a silly question?

• You cannot comment until you get more reputation. In the meantime, I would suggest you edit your original post instead of posting this answer. Furthermore, to ping someone directly, you can say @Katie – NoseKnowsAll Oct 9 '15 at 13:09
• @Katie You've ended up creating two accounts. Please follow the instructions at "I accidentally created two accounts; how do I merge them?". – epimorphic Oct 9 '15 at 18:37