The spectral theorem (well, some version thereof) says that if $A$ is a self-adjoint linear operator on a (finite-dimensional) Hilbert space $H$, there exists a basis of $H$ consisting of eigenvectors of $A$. My question is this: say I have an operator $A$ which is self-adjoint with respect to some inner product $\langle \cdot | \cdot \rangle$ on a (finite-dimensional) vector space $V$, except that this inner product is not positive definite (which I mean in the strict sense that there exist vectors $|v\rangle$, $|u\rangle$ such that $\langle v | v \rangle > 0$ and $\langle u | u \rangle < 0$). Is it known whether the spectral theorem generalizes to this case? I.e. can I conclude that there exists a basis of $V$ consisting of eigenvectors of $A$?

I thought perhaps there might be a way of defining a new positive-definite inner product $\langle \cdot | \cdot \rangle_+$ from the old one in such a way that $A$ is still self-adjoint with respect to $\langle \cdot | \cdot \rangle_+$, and then the usual spectral theorem can be applied, but I'm hoping that someone already knows the answer.

EDIT: The answer to my question as stated is no, as Robert's counterexample shows. Comparing with @Joppy's comment, it seems the issue is that with an indefinite inner product, the orthogonal complement to a null vector (i.e. a nonzero vector with vanishing norm) contains that same vector. Thus if $A$ has a null eigenvector, the usual proof of the spectral theorem doesn't go through. With this observation, let me modify my question: is this the only obstacle to there existing a basis of eigenvectors of $A$? In other words, if $A$ is self-adjoint and none of its eigenvectors are null, does there exist an basis of eigenvectors of $A$?

  • $\begingroup$ Your Hilbert space must be finite-dimensional for the Spectral Theorem to imply a basis of eigenvectors. Otherwise, you can have continuous spectrum with no eigenvectors. $\endgroup$ Nov 5, 2018 at 1:58
  • $\begingroup$ I'm happy to have an answer even in just the finite-dimensional case; I'll edit the question to clarify. $\endgroup$
    – Sebastian
    Nov 5, 2018 at 1:59
  • $\begingroup$ Following the usual argument, $A$ has an eigenvector since $H$ is a complex vector space. Then the orthogonal complement of this eigenvector is still stable under $A$, and so by induction we get a full basis of eigenvectors. For this to work, I think the only constraint on the inner product is that it is non degenerate (so that the eigenvector plus it’s orthogonal complement is the full space). Symmetry also helps so that you don’t have to worry about left vs right complements. $\endgroup$
    – Joppy
    Nov 5, 2018 at 5:27
  • $\begingroup$ Do you allow non-zero vectors $v$ such that $\langle v,v\rangle = 0$? $\endgroup$ Nov 5, 2018 at 18:02
  • $\begingroup$ @DisintegratingByParts: yes, that follows automatically from linearity and the statement that there are vectors with $\langle v , v \rangle > 0$ and $\langle u , u \rangle < 0$. An appropriate linear combination of these two will have zero norm. $\endgroup$
    – Sebastian
    Nov 6, 2018 at 2:09

1 Answer 1


EDIT: Oops, it's not true. In dimension $2$, consider the indefinite inner product $$ \langle u, v \rangle = u_1 v_1 - u_2 v_2$$ The matrix $$A = \pmatrix{1 & -1\cr 1 & -1\cr}$$ is "self-adjoint" with respect to this, i.e. $$ \langle u, A v \rangle = \langle A u, v \rangle = (u_1 - u_2)(v_1 - v_2)$$ but it is not diagonalizable: its eigenvalue $0$ has algebraic multiplicity $2$ but geometric multiplicity $1$, its only eigenvectors being scalar multiples of $\pmatrix{1\cr 1\cr}$.

  • $\begingroup$ Great, thanks. Your counterexample brings up an interesting observation which I've used to modify my question: namely, say that $A$ is self-adjoint and also has no null eigenvectors (which excludes your counterexample); is this now sufficient to guarantee the existence of a basis of eigenvectors of $A$? $\endgroup$
    – Sebastian
    Nov 6, 2018 at 5:22
  • $\begingroup$ @Sebastian A late response, but for you and future visitors the answer is no; this updated condition is still insufficient. As a counterexample, it suffices to consider the matrix $$ A + I = \pmatrix{2 & -1\\1 & 0}. $$ $\endgroup$ Nov 8, 2021 at 11:06

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