Spectral Theorem version for $ \mathbb{Q}$ or field with characteristic $0$ in general?

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I have wondered if the spectral theorem for finite $n\times n$ matrices over $\mathbb{R}$ can also hold on $\mathbb{Q}$. Since $\mathbb{Q}\subset\mathbb{R}$ obviously any rational symmetric matrix has to be diagonalizable. However we can get irrational eigenvalues like in this example:
$A = \left(\begin{array} &0 &1 \\ 2 &0 \end{array}\right)$ where $A$ has characteristic Polynomial $X^2-2$, so eigenvalues are $\pm \sqrt{2}\not\in\mathbb{Q}$
So my question is: are there any precise conditions for a symmetric rational matrix to only have rational eigenvalues? Such that every rational symmetric matrix which also fulfills this condition is diagonalizable over $\mathbb{Q}$
And how about other fields of characteristic $0$? I am currently working a lot with $p$-adic numbers and I am especially wondering about them.

It seems to me that really the only field I can think of where a general spectral theorem holds are the reals. What is so special about the real numbers in that case? Is it that they are "only" a degree 2 extension away from their algebraic closure? What makes the reals so special in this case and is there ANY other field in which a general spectral theorem holds?