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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    $\begingroup$ 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. $\endgroup$ – anon Jun 4 '12 at 22:36
  • $\begingroup$ 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. $\endgroup$ – tkp Jun 4 '12 at 22:39
  • $\begingroup$ I think this is a duplicate but I can't find it right now. $\endgroup$ – lhf Jun 4 '12 at 22:43
  • $\begingroup$ 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. $\endgroup$ – Hurkyl Jun 4 '12 at 22:53
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    $\begingroup$ 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. :( $\endgroup$ – Hurkyl Jun 6 '12 at 9:00

An inner product is a symmetric positive definite bilinear form. In general fields, you'll happily be able to satisfy symmetric bilinear, but you'll struggle with positive definite: over a field of nonzero characteristic, you will not even be able to make sense of $\langle x, x \rangle \ge 0$, much less find a form for which it holds, since there is no ordering compatible with the field. Note that there is also no ordering on the complex numbers that is compatible with the field, but there is the subfield $\mathbb R$ which can be ordered, and so conjugate symmetry rescues us by ensuring $\langle x, x \rangle$ falls inside there, but fields of characteristic $p$ must contain a subfield $\mathbb F_p$ which is already not orderable.

Over fields of characteristic zero, like the rationals, you can find a reasonable inner product, but it's not as useful as you'd expect. For example, you can't get an orthonormal basis from a given basis, because you can't do square roots: you can find the norm squared, but not the norm itself. You could go all the way to the algebraics, or just as far as all the square roots, but at this stage I think you gain very little generality over just using $\mathbb R$ in the first place, and if you find you want completeness at any point, you'll be forced into the reals anyway.

I think I have too found the asymmetry between complex and real inner products to be frustrating. It's possible that there's a unifying theory that I'm unaware of, but I don't think it's unreasonable to view them as basically separate (if highly similar) entities, albeit entities that embed one in the other in a neat way. Essentially, $\mathbb R$ is special: it is, after all, the unique complete totally ordered Archimedean field, so it's not that surprising that we should pay it specific attention.

  • $\begingroup$ Off the top of my head, I think there are only two reasons to desire $\langle x,x \rangle \geq 0$: (1) so that it is square, and (2) because it relates to real geometry. In an algebraically closed field of characteristic $p$, everything is square and we aren't doing real geometry, so I don't think we even care about an analog. $\endgroup$ – Hurkyl Jun 4 '12 at 23:04
  • $\begingroup$ Okay, but surely you want $\langle x, x \rangle\not= 0$ at least? I suspect that, too, is going to be awkward to arrange. I would also wonder, if you're not doing geometry, why are you interested in inner products? $\endgroup$ – Ben Millwood Jun 4 '12 at 23:18
  • $\begingroup$ I think you are confusing the conditions "non-degenerate" and "positive-definite." The former makes sense over any field. $\endgroup$ – Qiaochu Yuan Jun 5 '12 at 17:07
  • $\begingroup$ I said non-degenerate? I meant positive definite. Edited. (I know the former makes sense over any field, but I don't think it's easy to arrange on fields of nonzero characteristic) $\endgroup$ – Ben Millwood Jun 5 '12 at 21:36
  • $\begingroup$ @QiaochuYuan: pinging you just in case it was you who down-voted and it was only because of the confusion I've now fixed. (sorry to bother you if not!) $\endgroup$ – Ben Millwood Jun 6 '12 at 15:57

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