$(a_1,a_2,..,a_n)$ be basis for $V$.($e_1,e_2,..,e_n$) different basis for $V$.$(e_1,..,e_n)=(a_1,a_2,..,a_n)\times P$.Then $[x]_e=P\times[x]_a$.Let $[x]_e=X,[x]_a=Y$.
$X=PY$,$X^T=Y^TP^T$.
$q_f(x)=X^TAX=Y^T(P^TAP)Y$=$Y^TBY$=$\sum_{i,j=1}^nb_{ij}y_iy_j$ and from theorem follows that $P^TAP$ is diagonal.
$A=$$\begin{pmatrix}f(e_1,e_1) & f(e_1,e_2) & .. &f(e_1,e_n\\\ f(e_2,e1) & f(e_2,e_2) & .. &f(e_2,e_n) \\\ .. &.. &..& ..\\\ f(e_n,e_1) &f(e_n,e_2) & .. & f(e_n,e_n) \end{pmatrix}$ where $f$ is bilinear form.