It is clear to me that a self-adjoint linear map on a real/complex inner product space has real eigenvalues, and its eigenspaces for distinct eigenvalues are orthogonal.
In order to prove it is diagonalisable, we also need that the geometric multiplicity of the eigenspaces equals the algebraic multiplicity. How is this done?
In particular, if the arguments differ, how is this demonstrated over $\mathbb{R}$? I have seen proofs over $\mathbb {C}$, and they use algebraic closure, so I imagine the real case is more involved, and would like to see a proof that works here.
By induction. Let $T$ be a real self-adjoint endomorphism of a real IPS, $V$. Let $\lambda$ be an eigenvalue of $T$ since $T$ is self-adjoint, $T$ has a real eigenvalue. Then by definition, there is a $\lambda$-eigenvector $v$ for $T$. Let $V'=\{v\}^\perp$. Then since $T$ is self-adjoint, if $w\in V'$, then $\langle Tw,v\rangle = \langle w,Tv\rangle=\langle w,\lambda v\rangle= 0$. Thus $V'$ is $T$-invariant. Hence $V$ splits as a direct sum $\Bbb{R}v\oplus V'$ and $T$ splits with it as $T=\lambda \oplus T|_{V'}$. Then by induction $T|_{V'}$ has an orthonormal basis of eigenvectors, and together with $v$, these give an orthonormal basis for $V$ consisting of eigenvectors of $T$.
This argument doesn't depend on the base field being $\Bbb{R}$ or $\Bbb{C}$, but I assumed it was $\Bbb{R}$ to make the writing simpler.