Proof of the change of basis rule for a bilinear form

Proof

We compute the entries of $M'$ directly.

1. Write $\mathcal{B} = (b_1, \dots, b_n)$ and $\mathcal{B}' = (b'_1, \dots, b'_n)$, and let $P = (p_{ij})$.
2. By the definition of the change of basis matrix, $b'_k = \sum_{i=1}^n p_{ik} b_i$ for each $k$.
3. By the definition of the matrix of a bilinear form, $M_{ij} = B(b_i, b_j)$ and $M'_{kl} = B(b'_k, b'_l)$.
4. By step 2, $M'_{kl} = B\bigl(\sum_i p_{ik} b_i, \sum_j p_{jl} b_j\bigr)$.
5. A bilinear form is linear in each argument. Expanding both arguments gives $M'_{kl} = \sum_{i=1}^n \sum_{j=1}^n p_{ik}\, B(b_i, b_j)\, p_{jl}$.
6. By step 3 and the definition of the transpose, $p_{ik} = (P^T)_{ki}$. So $M'_{kl} = \sum_{i,j} (P^T)_{ki} M_{ij} P_{jl}$.
7. The right side is the $(k, l)$ entry of the product $P^T M P$, by the definition of matrix multiplication.
8. This holds for all $k$ and $l$, so $M' = P^T M P$.