Proving matrices of bilinear forms are congruent

Question

Trying to prove that for any two matrices $A,B$ representing the same symmetric bilinear form $f:V \times V \to \mathbb{R}$, there is an invertible matrix $P$, such that $B = P^{T}AP$ ?

I have a method for constructing such $P$ by repeated change of bases (when it is a quadratic form) but can't see a way to prove this generally?

Answer 1

This is just a change of basis. Let $A$ and $B$ be the matrices of the same bilinear form, but with respect to two ordered bases $\{x_1,\ldots,x_n\}$ and $\{y_1,\ldots,y_n\}$ respectively. Let $P$ be the matrix such that $y_j=\sum_ip_{ij}x_i$, i.e. the matrix such that $\sum_jc_jy_j=\sum_i(Pc)_ix_i$ for any $c\in\mathbb R^n$. Then $B=P^TAP$.

Answer 2

You can't prove it because it is false.

$$A = \begin{pmatrix}1 & 0\\ 0 &1\end{pmatrix}$$

$$B = \begin{pmatrix}-1 & 0\\ 0 &-1\end{pmatrix}$$

aren't congruent, since $PAP^T = PP^T$ is definite positive, so can't be $B$.

Another example? let $A$ be the identity matrix, and $B$ the zero matrix. $A\ne PBP^T=B$.