Change of basis for a bilinear form

Statement

Let $F$ be a field and let $V$ be a finite-dimensional vector space over $F$, with two bases $\mathcal{B}$ and $\mathcal{B}'$. Let $P$ be the change of basis matrix whose columns are the coordinate vectors of the vectors of $\mathcal{B}'$ in $\mathcal{B}$. Let $B$ be a bilinear form on $V$, with matrix $M$ in $\mathcal{B}$ and $M'$ in $\mathcal{B}'$. Then $$M' = P^T M P.$$

Remarks

Two square matrices $M$ and $M'$ are congruent when $M' = P^T M P$ for an invertible matrix $P$. So two matrices represent the same bilinear form in two bases if and only if they are congruent. Congruence keeps the rank and the symmetry of a matrix, but in general not its eigenvalues. Compare this with the rule $P^{-1} A P$ for the matrix of a linear map.

Explanations

L1 human

When we change the basis, the matrix of a bilinear form changes by $P^T M P$. Both vectors change coordinates, so $P$ appears twice. For a linear map, the rule is $P^{-1} A P$ instead.