Suppose I have a first order differential operator in matrix format:-
$Dx = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 \cdots \\ -1 & 0 & 1 & 0 & 0 & 0 & 0 \cdots \\ 0 & -1 & 0 & 1 & 0 & 0 & 0 \cdots \\ 0 & 0 & -1 & 0 & 1 & 0 & 0 \cdots \\ 0 & 0 & 0 & -1 & 0 & 1 & 0 \cdots \\ \vdots & \vdots & \vdots & & \ddots & &\ddots & \end{bmatrix}$
as we can see it is skew-symmetric and its eigenvalues are pure imaginary or zero(odd order) and its eigenvectors are complex conjugates. Now I am in a situation that I need a symmetric matrix to have real eigenvalues and eigenvectors for this first order equation. I need to know "is it possible or not ?" please let me know.
Edited:
Actually in finite difference time domain method, we use a differential equation like, $du/dt+Au=b$, $A =$ curl matrix
and $b=$ external input to the system
, to solve unknown vectors. My quest is to expand the unknown vector in its eigenvectors. In this case, it seems a little bit different because the eigenvectors are not real anymore. That's why i was wondering about the real symmetric matrix.