I saw in the spectral theorem that any real symmetric matrix $A$ is diagonalizable by orthogonal matrices; i.e., given a real symmetric matrix $A$, $Q^{T} AQ$ is diagonal for some orthogonal matrix $Q$.
I was wondering if we could prove that any real symmetric matrix $A$ is diagonalizable only by using tools and properties of diagonalization (i.e. $P_{car}(X)=\det(A-X\cdot In)$, $A^T=A$, etc...). We do not care about the fact that proving that it is diagonalizable by orthogonal matrices, we just want to show that it is diagonalizable.
I started by thinking about the fact that any real symmetric matrix $A$ has a real eigenvalue (by showing that the characteristic polynomial has a root (complex or real) and this eigenvalue must be real with a small computation proof). But what about multiplicity stuff ? I think I can answer my own question just by saying that if $x_0$ is an eigenvalue then the characteristic polynomial $P(X)$ can be written as $P(X)=(X-x_0)Q(X)$. Then Q(X) has also a complex root that also must be real. But we are only working on the multiplicity of the roots here and nothing tells us that the sum of the geometric multiplicity (= dimension of an eigenvector space associated $x_0$ of another root) is equal to $\dim(A)$ $\iff A$ is diagonalizable.