I'm finding some trouble with diagonalization of quadratic forms using the "eigenvectors way".
Let $\Phi:V\longrightarrow\mathbb{R}$ be a quadratic form and $A$ an associated matrix (symmetrical) of $\phi$ in a certain basis. I understand that a way to diagonalize $\Phi$ is to consider the symmetrical endomorphism $f:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ raprensented by $A$ in the standard base and with the standard inner product. $f$ is diagonalisable, so let $\Delta=Diag(\lambda_1, ... , \lambda_n)$ with $\lambda_1, ..., \lambda_n$ eigenvalues of $f$ and $B={b_1, ..., b_n}$ a basis of eigenvectors of $f$. $A$ and $\Delta$ are congruent (and similar), so $\Delta$ is a diagonal assocciated matrix of $\Phi\ $. Now, if I ortogonalise $B$ obtaining $B'$, the associated matrix in base $B'$ of $\Phi$ is diagonal, but I don't understand, why? And what is this new matrix? Is $\Delta$ or anything different?
Thanks for responses