Theorem 3.2.1. Let $A$ be the matrix of the bilinear map $\tau: V \times W \rightarrow K$ with respect to the bases $\mathbf{e}_1, \ldots, \mathbf{e}_n$ and $\mathbf{f}_1, \ldots, \mathbf{f}_m$ of $V$ and $W$, and let $B$ be its matrix with respect to the bases $\mathbf{e}_1^{\prime}, \ldots, \mathbf{e}_n^{\prime}$ and $\mathbf{f}_1^{\prime}, \ldots, \mathbf{f}_m^{\prime}$ of $V$ and $W$. Let $P$ and $Q$ be the basis change matrices, as defined above. Then $B=P^{\mathrm{T}} A Q$. Proof: Let $B$ be the matrix of $\tau$ with respect to these new bases. Let $\mathbf{v}$ be any vector in $V$. Then we know that if $\underline{\mathbf{v}} \in K^{n, 1}$ is the column vector of coordinates of $\mathbf{v}$ with respect to the old basis $\mathbf{e}_1, \ldots, \mathbf{e}_n$, and $\underline{\mathbf{v}}^{\prime}$ the coordinates of $\mathbf{v}$ in the new basis $\mathbf{e}_1^{\prime}, \ldots, \mathbf{e}_n^{\prime}$, then we have $P \underline{\mathbf{v}}^{\prime}=\underline{\mathbf{v}}$. Similarly, for any $\mathbf{w} \in W$, the coordinates $\underline{\mathbf{w}}$ and $\underline{\mathbf{w}}^{\prime}$ of $\mathbf{w}$ with respect to the old and new bases of $W$ are related by $Q \underline{\mathbf{w}}^{\prime}=\underline{\mathbf{w}}$. We know that we have $ \underline{\mathbf{v}}^{\mathrm{T}} A \underline{\mathbf{w}}=\tau(\mathbf{v}, \mathbf{w})=\left(\underline{\mathbf{v}}^{\prime}\right)^{\mathrm{T}} B \underline{\mathbf{w}}^{\prime} $ By substituting $P \underline{\mathbf{v}}^{\prime}=\underline{\mathbf{v}}$, we have $ \begin{aligned} \left(\underline{\mathbf{v}}^{\prime}\right)^{\mathrm{T}} B \underline{\mathbf{w}}^{\prime} & =\left(P \underline{\mathbf{v}}^{\prime}\right)^{\mathrm{T}} A\left(Q \mathbf{w}^{\prime}\right) \\ & =\left(\underline{\mathbf{v}}^{\prime}\right)^{\mathrm{T}} P^{\mathrm{T}} A Q \underline{\mathbf{w}}^{\prime} \end{aligned} $ Since this relation must hold for all $\underline{\mathbf{v}}^{\prime} \in K^{n, 1}$ and $\underline{\mathbf{w}}^{\prime} \in K^{m, 1}$, the two matrices in the middle must be equal (exercise!): that is, we have $B=P^{\mathrm{T}} A Q$. Remark: matrices that represent the same bilinear form are called congruent.