# 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.

• The answer to Question 1 seems no to me, as $A$ may have null eigenvectors.
– MK7
Nov 8, 2021 at 13:16

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 $$G$$.

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.

• Grossman: This is somewhat surprising but after analyzing the $2 \times 2$ case I convinced myself that it isn't possible that a subspace which does not have a "$q$-orthonormal" basis will be an eigenspace of a $q$-self-adjoint operator and diagonalizable operator. In particular you cannot construct an operator which is both $q$-self adjoint and diagonalizable which has $(1,1)$ as an eigenvector (of geometric multiplicity one). Nov 8, 2021 at 16:59
• @levap Strictly speaking, my answer is consistent with that conclusion. That is interesting, though. In the $2 \times 2$ case it makes sense: if we choose $(1,1)$ as an eigenvector, then there is no available second eigenvector that is both orthogonal and linearly independent to $(1,1)$. Nov 8, 2021 at 20:05