This was a question brought up by a classmate of mine;
In the general case (i.e. when not neccessarily considering $\mathbb{R}^n$ but just a smooth manifold $\overline{M}$ of dimension $m \in \mathbb{N}_{> 0}$), we consider a semi-riemannian manifold $(\overline{M},\overline{g})$ and a semi-riemannian submanifold $(M,g)$ of dimension $n$, that is, $M \subset \overline{M}$, as well as a normal field $\xi \in \mathfrak{X}(M)^{\perp}$, we also consider the shape operator $$S^{\xi}:\mathfrak{X}(M) \to \mathfrak{X}(M).$$
Now, it is said (in the notes I have) that since for each $p \in M$, $S^{\xi}$ is a self-adjoint linear transformation between finite-dimensional vector spaces $T_pM$ of dimension $n \in \mathbb{N}$, by the spectral theorem it admits $n$ distinct eigenvalues $\lambda_1 \leq \ldots \leq \lambda_n$ and that $T_pM$ admits a pseudo-orthonormal basis of eigenvectors of $S^{\xi}_p$.
My friends question was this; Since we only assume that $M \subset \overline{M}$ is semi-riemannian, we are only guaranteed a scalar-product $g(p)$ on $M$, for each $p \in M$ ($g$ being the metric on $M$). But the spectral theorem requires an inner-product. Yes, $T_pM$ is isomorphic to $\mathbb{R}^n$, which gives us an inner product, $g(\varphi(p))_{\text{eucl}}(T_p\varphi(v),T_p\varphi(w))$ for $p \in M$ $v,w \in T_pM$, and a chart $(U,\varphi)$ of $M$, but is that sufficient?
To clarify, since I believe there to be different conventions; By scalar-product, I mean a non-degenerate, symmetric bilinear form $g$, where the index of $g$ is not necessarily $0$, while for an inner product, we require that $g$ is positive-definite ($g(v,v) > 0, \quad \forall v \in V, v \neq 0$, of a vector space V), i.e. that the index of $g$ must be $0$.