There are two versions of the spectral theorem that I see in sources (for the real case):
For an inner product space $(V,g)$ with a s.a operator $T$, there exists an o.n.b s.t. each basis vector is an eigen-vector of $T$.
Given any symmetric matrix $T$, there exists an orthogonal matrix $P$ s.t. $P^{-1}TP$ is diagonal with the diagonal elements of the matrix corresponding to the eigenvalues of $T$.
The second condition is essentially the diagonalisability condition. It is fairly obvious that if one expresses $T$ in an arbitrary basis, then the existence of an o.n.b implies by change of basis the second definition except for one point: $P$ is orthogonal. What am I missing?