> [!NOTE] > Let $\tau: V \times W \rightarrow K$ be a [[Bilinearity|bilinear map]]. Then the $n \times m$ matrix $A=\left(\tau(\mathbf{e}_{i}, \mathbf{f}_{j})\right)$ is defined to be the matrix of $\tau$ with respect to the bases $\mathbf{e}_1, \ldots, \mathbf{e}_n$ and $\mathbf{f}_1, \ldots, \mathbf{f}_m$ of $V$ and $W$. **Compatibility with matrix multiplication**: For $\mathbf{v} \in V, \mathbf{w} \in W$, let $\mathbf{v}=x_1 \mathbf{e}_1+\cdots+x_n \mathbf{e}_n$ and $\mathbf{w}=y_1 \mathbf{f}_1+\cdots+y_m \mathbf{f}_m$, so the coordinates of $\mathbf{v}$ and $\mathbf{w}$ with respect to our bases are $ \underline{\mathbf{v}}=\left(\begin{array}{r} x_1 \\ x_2 \\ \cdot \\ \cdot \\ x_n \end{array}\right) \in K^{n, 1}, \quad \text { and } \quad \underline{\mathbf{w}}=\left(\begin{array}{r} y_1 \\ y_2 \\ \cdot \\ \cdot \\ y_m \end{array}\right) \in K^{m, 1} $ Then it follows from the defiitomn of $\tau$ that $ \tau(\mathbf{v}, \mathbf{w})=\sum_{i=1}^n \sum_{j=1}^m x_i \tau\left(\mathbf{e}_i, \mathbf{f}_j\right) y_j=\underline{\mathbf{v}}^{\mathrm{T}} A \underline{\mathbf{w}} . $ So once we've fixed bases of $V$ and $W$, every bilinear map on $V$ and $W$ corresponds to an $n \times m$ matrix, and conversely every matrix determines a bilinear map.