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The linear algebra course that I took was fairly consistent about assuming that the scalar field is either the reals or the complex numbers. The theory about linear maps, basis, their matrices, eigenvalues and eigenvectors, trace and determinant clearly generalize to a general field without any changes. Similarly, the Jordan canonical form only seems to require algebraic closedness of a field.

The definition of an inner-product seems to explicitly require either the reals or the complex numbers, but even then one should be able to replace it by a bilinear pairing $V\times V\to \mathbb{F}$, where $\mathbb{F}$ is our field. However, why do we then want conjugate symmetry in the complex case? Do we need something similar for fields which have a "similar" automorphism? Is there a precise way to formalize this?

My questions is essentially the following: How can we generalize spectral theorems to general fields? What would the results look like and what do we need to assume? What's the right way to generalize inner-product spaces and what can we translate unchanged from the setting of an undergraduate linear algebra class?

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    The conjugate symmetry is so that $\langle x,x\rangle$ is a positive *real* even if $x$ is a nonzero vector in a vector space over the *complex* numbers, hence we can obtain a translation invariant *metric* compatible with scalar multiplication up to absolute value.2012-06-04
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    OK, so that requirement can then pretty much be scrapped for a general pairing of a space with a scalar field that does not embed in the complex numbers, since there's no hope of getting a metric.2012-06-04
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    I think this is a duplicate but I can't find it right now.2012-06-04
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    Though if $[L:K] = 2$, one has conjugation (the nontrivial element of the Galois group) and could still be interested in sesquilinear forms over $L$, and we would have $\langle x,x \rangle \in K$. Alas, we can't have $L$ algebraically closed unless $K$ is real closed, and thus characteristic 0.2012-06-04
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    As an example of how irritating the positive characteristic can be, there are symmetric nilpotent matrices, such as $$\left( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} \right)$$ in characteristic 2. I had not realized this before today. :(2012-06-06

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