The adjoint and inner product space

Question

Do these real spectral theorems hold for a general inner product?

Version 1: Let $V$ be a finite dimensional inner product space. Suppose that $T\in L(V, V)$ is self-adjoint. Then, there exists an orthonormal basis $B\subset V$ consisting of eigenvectors of $T$.

Version 2 (Matrix version): Let $A\in M_n(\mathbb R)$ be a symmetric matrix. Then, there exists an orthogonal matrix $Q\in M_n(\mathbb R)$ such that $Q^{-1}AQ$ is a diagonal matrix.

I am struggling to understand whether these theorems hold for a general inner product space or only the typical Euclidian inner product space (dot product in the real numbers).

After looking at proofs for version 1, my understanding is that these proofs do not rely on any specific definition of the inner product. I have concluded that version one holds for ANY inner product space, regardless of how the inner product is defined (so long as it satisfies the axioms). Is this conclusion correct?

After looking at proofs of version 2, it appears to me that they all rely on the inner product being defined as the Euclidian inner product. Should version 2 be revised to say: "... there exists an orthogonal(with respect to the Euclidian inner product) matrix $Q\in M_n(\mathbb R)$ such that... " In other words, I believe that version 2 does not hold for any inner product on $\mathbb R^n$, so we need to specify that the inner product is the usual dot product. Is this correct?

Answer 1

You are right (in the finite dimensional case). The key point here is that in the second version, symmetry is equivalent to being self-adjoint only when the basis we pick to present the operator as a matrix is orthonormal, hence the standard inner product is imposed on the underlying space represented as $\mathbb R^n$. And also orthogonal matrices only keep the standard inner product untouched.

Answer 2

For 1, A nice example:

Real vector space $L^2[0,1]$
inner product $\langle f,g\rangle = \int_0^1 f(x)g(x)\;dx$
operator "multiply by $x$" given by $T(f)(x) = xf(x)$.
Then $T$ is a bounded self-adjoint operator, but has no eigenvalues at all.

So, the next step is to decide in what sense $T$ is already a "diagonal" operator. For explanation, and a version of 1, look for "Spectral Theorem" https://en.wikipedia.org/wiki/Spectral_theorem