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For some propositions and proofs it was assumed that there exist a $q$-orthogonal (respect. $q$-orthonormal) basis of $\mathbb{R}^{r+s}\subset Cl_{r,s}$. Here $q$ is a quadratic form. Since we consider Clifford algebras $Cl_{r,s}$ on $\mathbb{R}^{r+s}$, it holds that $q(x)=x_1^2+...+x_r^2-x_{r+1}^2-...-x_{r+s}^2$. Two vectors are $q$-orthogonal if $q(v,w)= 0$, hence if $q(v+w)=q(v)+q(w)$

I was thinking about using Gram-Schmidt to make an arbitrary basis $q$-orthogonal. It holds that $q(v,w)$ is a symmetric bilinear form over $\mathbb{R}$, but it is in general not positive definite, so I cannot define a inner product.

Later it was also said that one can choose a $q$-orthonormal basis s.t. $q(e_i)=1$ for $i\leq r+1$ and $q(e_i)=-1$ for $i> r+1$

How do I show the existence in general and this special case? Thanks for your help

Schrödinger's cat
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    The Clifford algebra is irrelevant here. I don't understand the question though; if $q$ has the form you give then the standard basis is orthonormal. Are you asking how to find an orthonormal basis for generic $q$? In that case I still don't understand the question; if you've already proven that $q$ can be put in the standard form you wrote, then you've already found an orthonormal basis. Or perhaps you're asking how to perform Gram-Schmidt with an indefinite form? – Nicholas Todoroff Jun 08 '23 at 17:25
  • Does this answer your question? Orthogonal basis for bilinear forms – mr_e_man Jun 12 '23 at 17:00

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