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 ?

  • $\begingroup$ You may also be interested in: mathoverflow.net/questions/105870/… $\endgroup$
    – Qfwfq
    Jan 11, 2013 at 22:57
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    $\begingroup$ I forget the exact reference or construction, but every square complex matrix is similar to a complex symmetric one, so already the case $F={\mathbb C}$ fails $\endgroup$
    – Yemon Choi
    Jan 11, 2013 at 23:00
  • $\begingroup$ (Which doesn't say anything against your question; just thought it might be worth pointing out) $\endgroup$
    – Yemon Choi
    Jan 11, 2013 at 23:00
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    $\begingroup$ That's the reason why in $\mathbb{C}$ one has to consider the adjoint and not the transpose. $\endgroup$
    – tomasz
    Jan 11, 2013 at 23:06
  • $\begingroup$ A related question has now been answered at math.stackexchange.com/questions/3307062/… . $\endgroup$ Jul 29, 2019 at 18:44

2 Answers 2


This is a countable family of first-order statements, so it holds for every real-closed field, since it holds over $\mathbb R$.

From a square matrix, we immediately derive that such a field must satisfy the property that the sum of two perfect squares is a perfect square. Indeed, the matrix:

$ \left(\begin{array}{cc} a & b \\ b & -a \end{array}\right)$

has characteristic polynomial $x^2-a^2-b^2$, so it is diagonalizable as long as $a^2+b^2$ is a pefect square.

Moreover, $-1$ is not a perfect square, or else the matrix:

$ \left(\begin{array}{cc} i & 1 \\ 1 & -i \end{array}\right)$

would be diagonalizable, thus zero, an obvious contradiction.

So the semigroup generated by the perfect squares consists of just the perfect squares, which are not all the elements of the field, so the field can be ordered.

However, the field need not be real-closed. Consider the field $\mathbb R((x))$. Take a matrix over that field. Without loss of generality, we can take it to be a matrix over $\mathbb R[[x]]$. Looking at it mod $x$, it is a symmetric matrix over $\mathbb R$, so we can diagonalize it using an orthogonal matrix. If its eigenvalues mod $x$ are all distinct, we are done, because we can find roots of its characteristic polynomial in $\mathbb R[[x]]$ by Hensel's lemma. If they are all the same, say $\lambda$ we can reduce: subtract $\lambda I$, divide by $x$ and diagonalize again. The only remaining case is if some are the same and some are distinct. If we can handle that case, then we can diagonalize any matrix.

Lemma: Let $M$ be a symmetric matrix over $\mathbb R[[x]]$ such that some eigenvalues are distinct mod $x$. There exists an orthogonal matrix $A$ such that $AMA^{-1}$ is block diagonal, with the blocks symmetric.

Proof: Consider the scheme of such orthogonal matrices. Each connected component of this scheme corresponds to a partition of the eigenvalues into blocks. Choose one. Since we can diagonalize the matrix with an orthogonal matrix mod $x$, there is certainly a mod $x$ point on this component. We want to lift this to a point on the whole ring. We can do this if the scheme is smooth over $\mathbb R[[x]]$.

Assuming the blocks have distinct eigenvalues, the variety of ways to do this looks, over an algebraically closed field, like $O(n_1) \times O(n_2) \times.. \times O(n_k)$ where $n_1,\dots,n_k$ are the sizes of the blocks. This is because the only way to keep a diagonal matrix block diagonal is to hit it with one of those. So as long as the blocks are chosen such that the eigenvalues in different blocks are distinct and remain so on reduction mod $x$, the variety is smooth over $\mathbb R((x))$ and smooth over $\mathbb R$, and has the same dimension over both, so is smooth over $\mathbb R[[x]]$. (This bit might not be entirely correct.) Thus there is a lift and the matrix can be put in this form.

Then we do an induction on dimension. The only way we would be unable to put a matrix in a form where two of its eigenvalues are distinct mod $x$ is if its eigenvalues are all the same, in which case,since $\mathbb R((x))$ is contained in a real closed field, it's a scalar matrix and we're done.

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    $\begingroup$ A trivial comment on this very nice answer: a field for which the sum of two squares is a square is called Pythagorean. $\endgroup$
    – LSpice
    Feb 24, 2016 at 20:16

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).

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    $\begingroup$ Nice complete characterization! Not to be obsessed with credit, but I posted that idea first and then ccnotboy copied and pasted it. Also, "$-1$ is not a perfect square" and "the sum of two perfect squares is a perfect square" imply "$-1$ is not a sum of perfect squares" by induction. $\endgroup$
    – Will Sawin
    Jan 12, 2013 at 15:24
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    $\begingroup$ Will, (in the future) to be more convincing, I suggest the second sentence in your comment above start instead with "To clarify the history of this thread of ideas," or words to the effect of matching your perception of the chain of events. The sentence would then have less in common with nonsequiturs and things like "I don't need to tell you this, but..." Gerhard "The Costs Of Unwanted Advice" Paseman, 2013.01.12 $\endgroup$ Jan 12, 2013 at 17:23

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