Let $F$ be a field whose characteristic is not $2$. Let $X$ be an $n(<\infty)$-dimensional vector space over $F$, equipped with a symmetric bilinear form $\langle\cdot,\cdot\rangle$ on $X$. Let $u_1,\ldots,u_{n-1}$ be an orthogonal (i.e., $\langle u_i,u_j\rangle=0$ if $i\neq j$) linearly independent vectors of $X$. Does there exist $u_n\in X$ such that $u_1,\ldots,u_{n-1},u_n$ are orthogonal linearly independent?
I know how to prove that $X$ has an orthogonal basis. But I do not know how to extend an already given set $\{u_1,\ldots,u_{n-1}\}$ to an orthogonal basis.
This need not be possible! A single nonzero vector is an orthogonal set, but not every nonzero vector is part of an orthogonal basis: this depends on the type of symmetric bilinear form you are using. See Remark 4.5 and exercise 5 if Section 4 here.
Something nice about the dot product in Euclidean space is that (i) it is nondegenerate and (ii) its restriction to each subspace is nondegenerate. This is not true of general symmetric bilinear forms, and that is why things like Gram-Schmidt are not as generally valid for all symmetric bilinear forms (including nondegenerate ones).