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Under what conditions there exists an orthogonal basis? Or even better, is there a characterization of the existence of an orthogonal basis in terms of a given bilinear form and/or the base field?

For instance, if the characteristic of the field is not 2 and the bilinear form is symmetric, then there exists an orthogonal basis and we can as well extend to an orthogonal basis. Now, the proof is based on the fact that if $B$ is a nonzero symmetric bilinear form, then there exists $v \neq 0$ such that $B(v,v)\neq 0$.

So, is the above possible if the bilinear form is alternating, i.e., $B(v,v) = 0$ for all $v$? Or skew-symmetric, i.e., $B(v,w) = -B(w,v)$ for all $v,w$?

I am primarily interested on non-degenerate bilinear forms, but I am curious about the degenerate case as well.

user 1987
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  • Bilinear form is a function $g:V\times W\to K$, where $K$ is the sclar field. You can express a bilinear form in the following way $\langle Av,v\rangle$, in which A is an operator $A:V\to V$. The inner product as I see is implicit in the bilinear form. I guess that only if the inner-product is non-degenerate you can talk about orthogonal bases. – Pedro Gomes Aug 06 '17 at 17:13

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Let's take $B$ to be a bilinear form $B : V \times V \rightarrow K$ on a finite-dimensional $K$-vector space $V$; where $K$ is a field of characteristic $p$.

Take an orthogonal basis, i.e. a basis $e_1, \ldots, e_n$ such that $B(e_i, e_j) = 0$ if $i \neq j$.

  • If $B$ is alternating, then $B(e_i, e_i) = 0$, so in fact $B$ is zero.

  • If $B$ is skew-symmetric and we have $p \neq 2$, then $B$ is alternating and thus again $B$ is zero.

  • When $p = 2$, the bilinear form $B$ is skew-symmetric iff it is symmetric.

So the only interesting case is the one where $B$ is symmetric. We can reduce to the case where $B$ is non-degenerate: write $V = W \oplus V^\perp$, here $V^\perp$ is the radical of $B$. You can see that $V$ has a orthogonal basis if and only if $W$ has a orthogonal basis, and that the restriction of $B$ to $W$ is non-degenerate.

So we might as well assume that $B$ is a non-degenerate symmetric bilinear form.

Assume that $p \neq 2$ and that $B$ is nonzero. As mentioned in your question, then $B$ does have an orthogonal basis.

We still have the situation where $p = 2$ and $B$ is a non-degenerate symmetric bilinear form. In this case you can write $V = W \oplus W'$, where $W$ has an orthogonal basis, and where restriction of $B$ to $W'$ is alternating. So $V$ has an orthogonal basis if and only if $W' = 0$.

spin
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  • Do you know of a book where they show that for $p\ne 2$, if $B$ is a non-degenerate symmetric bilinear form then one has an orthogonal basis? – soap Oct 22 '19 at 12:54
  • @Soap: I am not sure, but the proof is straightforward. A non-degenerate symmetric bilinear form is not alternating if $p \neq 2$, so you can find some vector $e_1 \in V$ such that $b(e_1, e_1) \neq 0$. We can write $V = \langle e_1 \rangle \oplus W$, where $W = \langle e_1 \rangle^\perp$. Repeat this argument on $W$. – spin Oct 23 '19 at 12:20