I have a rotationally symmetric matrix of arbitrary size, for example,
\begin{equation} 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} \end{equation}
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
\begin{equation}
J =
\begin{pmatrix}
0 & 0 & ... & 0 & 1 \\
0 & 0 & ... & 1 & 0 \\
& & ... & & \\
1 & 0 & ... & 0 & 0
\end{pmatrix},
\end{equation}
then I can represent $A$ as
\begin{equation}
A =
\begin{pmatrix}
M & MJ \\
JM & JMJ
\end{pmatrix},
\end{equation}
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
\begin{equation}
U = \frac{1}{\sqrt{2}}
\begin{pmatrix}
I & -J \\
J & I
\end{pmatrix},
\end{equation}
then
\begin{equation}
U^TAU =
\begin{pmatrix}
0 & 0 \\
0 & 4JMJ
\end{pmatrix}.
\end{equation}
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.
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 \begin{equation} U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & -J \\ J & I \end{pmatrix}, \end{equation} then \begin{equation} U^TAU = \begin{pmatrix} 0 & 0 \\ 0 & 4JMJ \end{pmatrix}. \end{equation} 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?