# Over which fields (besides $\mathbb{R}$) is every symmetric matrix potentially diagonalizable?

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?

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

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.)
• Very nice! ${}{}$ Jul 29, 2019 at 4:19
• 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 ... Jul 29, 2019 at 18:41
• ... 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. Jul 29, 2019 at 18:43