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  "body": "\n## Proof\n\nWe compute the entries of $M'$ directly.\n\n1. Write $\\mathcal{B} = (b_1, \\dots, b_n)$ and $\\mathcal{B}' = (b'_1, \\dots, b'_n)$, and let $P = (p_{ij})$.\n2. By the definition of the [[change of basis matrix|0KWXYZPJKN]], $b'_k = \\sum_{i=1}^n p_{ik} b_i$ for each $k$.\n3. By the definition of the [[matrix of a bilinear form|0P24YW565F]], $M_{ij} = B(b_i, b_j)$ and $M'_{kl} = B(b'_k, b'_l)$.\n4. By step 2, $M'_{kl} = B\\bigl(\\sum_i p_{ik} b_i, \\sum_j p_{jl} b_j\\bigr)$.\n5. A [[bilinear form|0P799D6AMN]] is linear in each argument. Expanding both arguments gives $M'_{kl} = \\sum_{i=1}^n \\sum_{j=1}^n p_{ik}\\, B(b_i, b_j)\\, p_{jl}$.\n6. By step 3 and the definition of the [[transpose|0W809K6AW0]], $p_{ik} = (P^T)_{ki}$. So $M'_{kl} = \\sum_{i,j} (P^T)_{ki} M_{ij} P_{jl}$.\n7. The right side is the $(k, l)$ entry of the product $P^T M P$, by the definition of matrix multiplication.\n8. This holds for all $k$ and $l$, so $M' = P^T M P$.\n\n## Idea\n\nEach new basis vector is a combination of old ones, with coefficients in a column of $P$. Bilinearity expands $B(b'_k, b'_l)$ into a double sum of old entries. That double sum is exactly a matrix product with $P$ on the right and $P^T$ on the left.\n",
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