# Orthogonal basis for bilinear forms

I found the following claim all over the physics literature and online:

Let $$B$$ be a non-degenerate symmetric bilinear form on a finite-dimensional $$\mathbb K$$-vector space with $$\mathbb K$$ a field of characteristic $$\ne 2$$. Then $$B$$ has an orthogonal basis.

(The specific context in the physics books is when choosing an orthogonal basis for the Killing form of a semisimple Lie algebra (and sometimes they also assume the Lie algebra to be compact, I do not know why)).

Where can I find a proof of this claim?

Related: This post

• Have you tried to do Gram-Schmidt? – Keshav Oct 22 '19 at 13:13
• @Keshav Gram-Schmidt fails for forms that are not positive-definite, since it requires dividing by $B(x,x)$ which may be $0$. – lisyarus Oct 22 '19 at 13:19
• @lisyarus That would actually explain why physicists choose compact Lie algebras, for which the killing form is negative-definite, and thus the Gram-Schmidt procedure gives the orthogonal basis. – Soap Oct 22 '19 at 13:53
• @Soap This is hardly the case. As I've said in the answer, existence of an orthonormal basis doesn't depend on Gram-Schmidt. Physicists choose compact Lie algebras because they correspond to compact Lie groups, which are usually a lot easier to work with than non-compact ones. Furthemore, the groups that show up in physics are pretty much always either compact or can be reduced to compact ones with the unitarian trick. – lisyarus Oct 22 '19 at 14:03
• @lisyarus That is true, but there are also physics papers where they say "since the Lie algebra is compact, we can choose an orthonormal basis". It seems that there is the more general case that you referred to in your answer, but in the compact case it is even easier because we can use the Gram-Schmidt process. – Soap Oct 22 '19 at 14:22