TITLE: Over which fields are symmetric matrices diagonalizable ? 
QUESTION [37 upvotes]: The question is motivated by this one real symmetric matrix has real eigenvalues - elementary proof: 
Are there other fields $F$ than $\mathbb{R}$ (maybe some valued fields or real closed fields) with the property that every symmetric matrix in $M_n(F)$ is diagonalizable ?

REPLY [27 votes]: This is a generalization of the idea of Will Sawin.
The Stufe of such field should be infinite. In fact if $-1$ is a sums of
squares, i.e., $-1=a_1^2+\cdots+a_{n-1}^2$, then 
$$A:=
\begin{pmatrix}
1 & a_1&\cdots  & a_{n-1} \\
a_1 & a_1^2 &\cdots & a_1a_{n-1} \\
\vdots & \vdots&\ddots  &\vdots \\
a_{n-1} & a_{n-1}a_1&\cdots  & a_{n-1}^2\\
\end{pmatrix}
$$ would be a symmetric matrix with $A^2=0$ and is not diagonalizable.
So the base field should be a formally
real field.
A complete characterization is given in the following article (a
necessary and sufficient condition is that such field should be an
intersection of real closed fields):
MR1237224
D. Mornhinweg, D. B. Shapiro and K. G. Valente, The Principal Axis Theorem Over Arbitrary Fields
(The American Mathematical Monthly, Vol. 100, No. 8 (Oct., 1993), pp. 749-754).