Proof involving symmetric bilinear forms and matrices

Question

Let $f$ $:$ $V$$\times$$V$ $\rightarrow$ $\Bbb R$ be a symmetric bilinear form.

Let $A, B$ be the matrices of $f$ with respect to possibly different bases.

I want to prove that there exists $P$ such that $B$ $=$ $P^T$$AP$.

Please can anyone lend a hand here?

Answer 1

Suppose $A$ and $A'$ are the matrices of the form with respect to the bases $B$ and $B'$, respectively. Since $B$ and $B'$ are bases for $V$, there exists and invertible square matrix $P$ such that $B'=BP$. We'll show that this means $A'=P^tAP$.

Let $X,X'$ be the coordinate vectors of $v\in V$ with respect to the bases $A,A'$, respectively. Then $v=BX=B'X'$ and with analogous notation $w=BY=B'Y'$ if $w\in V$. Then $$\langle v,w \rangle =X^tAY=(PX')^tA(PY')=X'^t(P^tAP)Y'$$ which identifies $P^tAP$ as the matrix of the form with respect to the basis $B'$.

Note that the symmetry condition of the form was not used in the proof. This is because the result holds for any bilinear form on a real vector space.