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We know that for a symmetric matrix $P$ on a $n$-dimensional real linear space $V$, it always can be diagonalized. The reason lies in the proposition that

$V$ can be expressed as the sum of all eigenspaces of $P$.

However,in a general number field $\mathbb{F}$,the above process may not be achievable, as the eigenvalues do not necessarily lie in $\mathbb{F}$,hence,l try to weaken the conclusion : For any number field $\mathbb{F}$, does the symmetric matrix $P$ always remain similar to the following matrix?
$\begin{bmatrix} A& 0\\0 &B\end{bmatrix},A\in M_{r} (\mathbb{F} ), B\in M_{n-r} (\mathbb{F} ),\forall 0 \le r\le n$

user1405622
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  • this question is similar to https://math.stackexchange.com/questions/4378117/can-a-conjugate-symmetric-matrix-over-an-arbitrary-field-be-diagonalized/ . As there, you are missing the point -- it is not so much that the "eigenvalues do not necessarily lie in $\mathbb F$" it is that you don't have a compatible inner product. Consider the $2\times 2$ matrix of all $1$'s over $\mathbb F_2$ for a counterexample. Also look into the case of $\mathbb F:= \mathbb C$-- complex symmetric matrices can have a single Jordan Block. – user8675309 Jun 24 '25 at 17:08
  • for the complex case ref , ref e.g. https://math.stackexchange.com/questions/5075803/proving-every-complex-matrix-is-similar-to-a-symmetric-matrix-via-jordan-form – user8675309 Jun 24 '25 at 17:15

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