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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?

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Rodrigo de Azevedo
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soap
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  • Have you tried to do Gram-Schmidt? – Keshav Oct 22 '19 at 13:13
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    @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
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    @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
  • @Soap The crucial thing is orthogonal vs orthonornal. For any symmetric (even degenerate) bilinear form over any field of $\operatorname{char}\neq 2$ there is an orthogonal basis. However, only positive-definite forms possess an orthonormal basis (in general, you can't really say anything specific about the norms of basis vectors). I'm still in doubt that Gram-Schmidt is important here; probably just other properties of orthonormal bases are used. – lisyarus Oct 22 '19 at 15:17
  • Related: https://math.stackexchange.com/questions/1900620/gram-schmidt-process-in-minkowski-space-bbb-ln – mr_e_man May 16 '23 at 20:56

2 Answers2

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This is a modification of the Gram-Schmidt process. As usual, the projection of $x$ onto $a$ is defined as $\frac{\langle x,a\rangle}{\langle a,a\rangle}a$, and the projection of $x$ away from $a$ is $x-\frac{\langle x,a\rangle}{\langle a,a\rangle}a$.

Take any basis vector $a\neq0$. If $\langle a,a\rangle\neq0$, then project the other $n-1$ basis vectors away from $a$, and continue the process in that subspace. If $\langle a,a\rangle=0$ and there's another basis vector $b$ not orthogonal to $a$, then project the other $n-2$ basis vectors away from the span of $a$ and $b$ (which is hyperbolic), and continue the process in that subspace. If $\langle a,a\rangle=0$ and all other basis vectors are orthogonal to $a$, then nothing needs to be done in this step; continue the process in the span of the other basis vectors.

(And any hyperbolic plane produced in the process can be given an orthonormal basis. Given $\langle a,a\rangle=0\neq\langle b,a\rangle$, define $b'=\frac{b}{\langle b,a\rangle}-\frac{\langle b,b\rangle a}{2\langle b,a\rangle^2}$, so that $\langle b',b'\rangle=0$ and $\langle b',a\rangle=1$. Then take $c=a+\tfrac12b'$ and $d=a-\tfrac12b'$ as the new basis, with $\langle c,c\rangle=1=-\langle d,d\rangle$ and $\langle c,d\rangle=0$.)

Note that this works as well with degenerate forms. It always produces an orthogonal basis.

mr_e_man
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  • It also generalizes to Hermitian forms, though the case of the hyperbolic plane needs further modification: $$b'=\frac{1}{\langle b,a\rangle}b-\frac{\langle b,b\rangle}{2\langle b,a\rangle\langle a,b\rangle}a$$ Here I assume the form is left-linear. – mr_e_man May 30 '23 at 19:03
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The proof basically consists of repeated applications of completing the square (the method sometimes referred to as Lagrange reduction), and can be found here.

lisyarus
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