# about first order differential equation

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.

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So let's take an example, $x_1'=x_2$, $x_2'=-x_1$. As you say, the matrix is skew-symmetric, the eigenvalues $\pm i$ are purely imaginary, the eigenvectors are complex conjugates. This is intrinsic to this system, and there is no way around it. You can't make the matrix symmetric, or make the eigenvalues and eigenvectors real.
And yet, you can express the solutions in purely real terms, as obviously a solution is given by $x_1=\sin t$, $x_2=\cos t$.