Orthogonal eigendecomposition of self-adjoint operator with indefinite scalar product Let $V$ be a real vector space, of finite dimension $d$, equipped with a nondegenerate symmetric bilinear form $q$, and let $A$ be a linear map $V\to V$ that is self-adjoint with respect to $q$, i.e., such that $q(A(x),y)=q(x,A(y))$ for all $x,y \in V$.
I know that, if $q$ is positive definite, then the classic spectral theorem applies; thus $A$ is diagonalizable and the eigenvectors can be chosen to be orthonormal. I also know that, without the assumption of positive definiteness, diagonalizability of $A$ is not guaranteed, as shown here.
I would like to ask the following questions about the general case.
Question 1. Suppose that $A$ is diagonalizable. Is it then possible to choose an orthonormal basis of eigenvectors? Here by orthonormal I mean a basis $(v_{1},\dotsc,v_{d})$ such that $\lvert q(v_{i},v_{j}) \rvert = \delta_{ij}$.
Question 2. Suppose that $F,G$ are two commuting linear maps on $V$, i.e., $F(G(x))=G(F(x))$ for every $x\in V$. If each of $F$ and $G$ has an orthonormal basis of eigenvectors, do they then share a common orthonormal basis of eigenvectors?
Clearly, the answer to both questions is yes in the positive definite case.
 A: To question $1$: if $A$ is $q$-self-adjoint, then eigenvectors of $A$ associated with distinct eigenvalues will necessarily be $q$-orthogonal. Indeed, it suffices to note that if $v,w \in V$ and $\mu,\lambda \in \Bbb F$ are such that $Av = \lambda v$ and $Aw = \mu w$ with $\lambda \neq \mu$, then
$$
\lambda q(v,w) = q(Av,w) = q(v,Aw) = \mu q(v,w) \implies (\lambda - \mu)q(v,w) = 0 \implies q(v,w) = 0.
$$
It follows that $A$ will have an "orthonormal" eigenbasis if and only if all of its eigenspaces have orthonormal bases. Notably, some subspaces do not have orthonormal bases. For example, if $q$ over $\Bbb R^2$ is given by $q(x,y) = x_1y_1 - x_2y_2$, then the span of $(1,1)$ has no orthonormal basis because all elements of this subspace have "norm" $q(x,x) = 0$.
To question 2: if $x \in V, \lambda \in \Bbb F$ are such that $F(x) = \lambda x$, then
$$
F(G(x)) = G(F(x)) = G(\lambda x) = \lambda G(x).
$$
That is, $G(x)$ is also an eigenvector of $F$ associated with $\lambda$. One consequence of this fact is that if $F$ has no repeating eigenvalues and has an orthonormal eigenbasis, then the same basis will diagonalize $G$.
I am not sure about the more general case.
A: The answer to both questions is yes.
For the first question, assume that $A$ is diagonalizable and write $V = V_{\lambda_1} \oplus \dots \oplus V_{\lambda_k}$ where $V_{\lambda_i}$ are the eigenspaces of $A$. The fact that $A$ is $q$-self-adjoint implies that the eigenspaces are mutually $q$-orthogonal. Choose a $q$-orthogonal basis $\mathcal{B}_i = (v_1^i, \dots, v_{d_i}^i)$ for each $V_{\lambda_i}$ such that $q(v_j^i, v_k^i) = 0$ if $j \neq k$ and $q(v_j^i,v_j^i) \in \{ -1, 0, 1 \}$. Then $(\mathcal{B}_1, \dots, \mathcal{B}_k)$ will be a $q$-orthogonal basis of eigenvectors for $V$. Now since $q$ is non-degenerate, it cannot be the case that $q(v_j^i, v_j^i) = 0$ for some $i,j$ and so you must have $\left| q(v_j^i, v_j^i) \right| = 1$. Note that this shows in particular that the bilinear forms $q|_{V_{\lambda_i} \times V_{\lambda_i}}$ must be non-degenerate.
For the second question, assume $F$ and $G$ commute and are both diagonalizable and $q$-self-adjoint. Write $V = V_{\lambda_1}(F) \oplus \dots \oplus V_{\lambda_k}(F)$. Again, this is a $q$-orthogonal decomposition and each $q|_{V_{\lambda_i} \times V_{\lambda_i}}$ is non-degenerate. Since $F$ and $G$ commute, each $V_{\lambda_i}(F)$ is $G$-invariant and $G|_{V_{\lambda_i}(F)}$ is diagonalizable. By the previous item, we can choose a "$q$-orthonormal" basis of eigenvectors of $G|_{V_{\lambda_i}}$ for each $V_{\lambda_i}$ and concatenating those bases will give you a $q$-orthonormal basis which consists of eigenvectors of both $F$ and $G$.
