Over which fields (besides the well-known $\mathbb{R}$) is every symmetric matrix potentially diagonalizable? A matrix is potentially diagonalizable in a field $F$ if it is diagonalizable in the algebraic closure of $F$.

It appears to me that the fields $\mathbb{F}_2$ and $\mathbb{C}$ do not have this property. What about other finite fields?

  • $\begingroup$ Nice question! I have found something wrong in this post. $\endgroup$
    – Bach
    Jul 29, 2019 at 4:49
  • $\begingroup$ Possible duplicate of this MO post: mathoverflow.net/q/118680/123740 $\endgroup$
    – Bach
    Jul 29, 2019 at 12:48

1 Answer 1


This is true iff $F$ is formally real, i.e. $-1$ is not a sum of squares in $F$. On the one hand, if $F$ is formally real, then by extending $F$ we may assume it is real-closed, and then the exact same argument as for $\mathbb{R}$ applies to matrices over $F$ (or to use a sledgehammer, the result for $F$ follows from the result for $\mathbb{R}$ since all real-closed fields are elementarily equivalent).

Alternatively, here's a more direct argument: if $F$ is formally real, then no element of $F^n$ is orthogonal to itself with respect to the dot product, and so for any subspace $V\subseteq F^n$, the orthogonal complement $V^\perp$ with respect to the dot product is a linear complement of $V$. If an $n\times n$ matrix $A$ is symmetric, then if a subspace $V$ is invariant under $A$, then so is $V^\perp$. This means the action of $A$ on $F^n$ is semisimple, or equivalently that the minimal polynomial of $A$ over $F$ is squarefree. Since $F$ has characteristic $0$, this minimal polynomial will remain squarefree over the algebraic closure of $F$, which means $A$ is still semisimple over the algebraic closure and thus is potentially diagonalizable.

Conversely, suppose $F$ is not formally real; say $a_1^2+\dots+a_n^2=-1$ in $F$. Let $u=(a_1,\dots,a_n,1)\in F^{n+1}$ and consider the linear map $A:F^{n+1}\to F^{n+1}$ given by $A(v)=\langle v,u\rangle u$, where $\langle\cdot,\cdot\rangle$ is the dot product. Then $A$ is symmetric: for any $v,w\in F^{n+1}$, $$\langle Av,w\rangle = \langle \langle v,u\rangle u,w\rangle=\langle v,u\rangle \langle u,w\rangle=\langle v,\langle w,u\rangle u\rangle=\langle v,Aw\rangle.$$ However, note that $\langle u,u\rangle=0$ so $A(u)=0$ and thus $A^2=0$ since the image of $A$ is spanned by $u$. Since $A\neq 0$ and $A^2=0$, $A$ is not potentially diagonalizable.

More generally, similar arguments show that if $V$ is a finite-dimensional vector space with a non-degenerate symmetric bilinear form $\langle \cdot,\cdot\rangle$, then every self-adjoint endomorphism of $V$ is semisimple iff there is no nonzero $v\in V$ such that $\langle v,v\rangle=0$. (If the base field is perfect, then semisimple is equivalent to potentially diagonalizable, but in general it is weaker.)

  • $\begingroup$ Very nice! ${}{}$ $\endgroup$ Jul 29, 2019 at 4:19
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    $\begingroup$ There is an even more elementary argument for "if $F$ is formally real, then the minimal polynomial of $A$ is squarefree". Indeed, let $F$ be formally real, and write the minimal polynomial of $A$ in the form $f^2 g$ where $f$ and $g$ are two polynomials. Our goal is to show that $f$ is constant. Well, $\left(f^2 g\right)\left(A\right) = 0$, which rewrites as $f\left(A\right)^2 g\left(A\right) = 0$. This further rewrites as $f\left(A\right)^T f\left(A\right) g\left(A\right) = 0$, because $f\left(A\right) = f\left(A\right)^T$ (since $A = A^T$). Multiplying this equality ... $\endgroup$ Jul 29, 2019 at 18:41
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    $\begingroup$ ... by $g\left(A\right)^T$ on the left, we obtain $g\left(A\right)^T f\left(A\right)^T f\left(A\right) g\left(A\right) = 0$. Set $B = f\left(A\right) g\left(A\right)$; then, this rewrites as $B^T B = 0$. But since $F$ is formally real, this yields $B = 0$ (because $\operatorname{Tr}\left(B^T B\right)$ is the sum of the squares of all entries of $B$). Since $B = f\left(A\right) g\left(A\right) = \left(fg\right)\left(A\right)$, this rewrites as $\left(fg\right)\left(A\right) = 0$. Hence, $fg$ is a multiple of $f^2 g$ (because $f^2 g$ is the minimal polynomial of $A$). Thus, $f$ is constant, qed. $\endgroup$ Jul 29, 2019 at 18:43

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