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.
Best Answer
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 $G$.