The spectral theorem for euclidean spaces is very well known. Let me establish the notation.
Let $V \neq \{0\}$ be a finite dimensional vector space over $\mathbb{R}$ endowed with a symmetric, positive definite, bilinear form $p:V \times V \to \mathbb{R}$.
Suppose $f:V \to V$ is an endomorphism such that for all $v,w \in V$, it is true that $p(f(v),w)=p(v,f(w))$, so $f$ is symmetric (self adjoint).
So, what the spectral theorem says is that there exists an orthonormal basis of $V$ made of eigenvectors of $f$.
Consider this claim:
"If $V$ and $f$ are as above, then $f$ is diagonalizable (equivalently, $V$ admits a basis of eigenvectors of $f$)"
This is much weaker since it does not say anything about $f$ being orthogonally diagonalizable.
Obviously, this claim is pointless if compared with the spectral theorem, because it is just an uninteresting corollary.
But the question I have is:
can the claim be proved without using the spectral theorem? In other words, is there a nice and elementary proof for this fact which does not use the same techniques for proving the spectral theorem?
The proofs I have seen for the spectral theorem do not use anything too fancy, but I just wanted to know because I read this question from someone taking a first course on linear algebra and I could not give an answer.