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Let $K$ be a field with an involution $*$, meaning $*:K\to K$ is a field homomorphism and $(x^*)^*=x$ for all $x\in K$. Let $M$ be a conjugate-symmetric $n\times n$ matrix with entries in $K$, meaning ${M_{ij}=M_{ji}}^*$ for all $1\leq i,j\leq n$.

Is $M$ diagonalizable? If so, must its eigenvalues be of the form $xx^*$? If not, what if we restrict to $*=\text{id}$ and/or char$(K)≠2$?

As might be clear, I'm interested in whether the standard real/complex spectral theorems in finite dimension generalize to arbitrary fields. Note that the above question reduces to some standard results for $(K,*)=(\mathbb R,\text{id})$ or $(K,*)=(\mathbb C,\overline{\phantom{x}}\,)$, for which the answers are "yes."

WillG
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  • yes, the greedy algorithm at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr can be adapted, rather than the original $D_{n+1} = E_{n+1}^T D_n E_{n+1},$ revise to $D_{n+1} = (E_{n+1}^\ast)^T D_n E_{n+1},$ – Will Jagy Feb 09 '22 at 18:41
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    note that my comment is talking about $ (P^\ast)^T H P = D.$ In Horn and Johnson they call this $\ast$coungruence. – Will Jagy Feb 09 '22 at 19:43
  • @WillJagy So is the difference between your comments and the answers below stemming from a difference in the meaning of "diagonalizing"? I.e., for the answers below, the intended meaning is $A^{-1}MA=D$, and for your comments, it's something else? – WillG Feb 09 '22 at 19:45
  • Perhaps your version treats the matrix as a sesquilinear form and diagonalizes it in the sense of finding an orthogonal basis, whereas the answers below treat is as an operator and diagonalize it in the sense of finding an eigenbasis? – WillG Feb 09 '22 at 19:48

3 Answers3

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No, consider $\Bbb{Q}(\sqrt{2})$ and a matrix $\begin{pmatrix} 0 & \sqrt{2} \\ -\sqrt{2} &0\end{pmatrix}.$ This matrix has characteristic equation $\lambda^2+2=0$ so it has eigenvalues $\pm\sqrt{2}i.$

The diagonalization of any conjugate-symmetric $A\in M_{n\times n}(\Bbb{F})$ would have the roots of its characteristic polynomial as the diagonal entries, and we'd need to make sure those are elements of the field $\Bbb{F}.$

Chickenmancer
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Let $(K, *) = (\mathbb C, \text{id})$ and $M=\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}$. Then $M$ is not diagonalizable.

(See Why a complex symmetric matrix is not diagonalizible?)

It's pretty harsh for the spectral theorem to hold. For example, you need the field to be almost algebraically closed, as argued by @Chickenmancer. In general, it's hard to generalize results from linear algebra over $\mathbb R$ or $\mathbb C$ to other fields if the result depends on positivity, like the dot product or conjugation with the property $x^*x\ge 0$.

Just a user
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  • I'm confused because this counterexample is over an algebraically closed field, hence it seems to contradict this post. Am I missing something, or is there a mistake in one of these? – WillG Feb 09 '22 at 20:21
  • My guess is that the answer in the linked post misinterpreted the question, and I posted some comments there to that effect. But I'm not sure. – WillG Feb 09 '22 at 20:36
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    Yeah, I think you are right. In the linked post, the selected answer tried to answer whether $X^tAX$ is diagonal, not $X^{-1}AX$. – Just a user Feb 09 '22 at 21:40
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For arbitrary fields of positive characteristic the answer is no.

with $\text{char }\mathbb K = p$, let $J$ be the $p\times p$ matrix of all ones. Then $\text{rank}\big(J\big) =1$ and $\text{trace}\big(J\big) =0$ so $J$ is non-zero nilpotent-- i.e. it is not diagonalizable. But $J=J^T$ and the any field homomorphism sends $1$ to $1$ so $J$ is conjugate-symmetric.

user8675309
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  • Care to take a look at this problem? https://math.stackexchange.com/q/4389947/64809 – Hans Feb 24 '22 at 23:16