"Every self adjoint operator $T$ of a hermitian space $V$ is diagonalisable and its eigenvalues are real"
The proof consists of 3 parts:
Proof that all the eigenvalues of $T$ are real.
Induction proof that there exists a basis of eigenvectors of $V$ for any dimension of $V$.
Proof that all the eigen-subspaces are orthogonal.
Why is that enough to say that T is diagonalisable
I understand that this means that we created a diagonal matrix who's coefficients are $T$'s eigenvalues and that we created an orthogonal matrix who's component are T's eigenvectors but how is that enough to show that T is indeed diagonalisable?