**Proof plan.** Given a linear map $\alpha: U \to V$ of rank $r$, we construct bases for $U$ and $V$ in which the matrix of $\alpha$ takes the block form $\begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$. The strategy is to choose a basis for $\ker\alpha$, extend it to a basis for $U$, and then show that the images of the extension vectors form a basis for $\mathrm{im}\,\alpha$, which we extend to a basis for $V$. **Step 1: Choose a basis for $\ker\alpha$ and extend.** Let $r = r(\alpha) = \dim\mathrm{im}\,\alpha$ and let $s = n(\alpha) = \dim\ker\alpha$, so that $r + s = m = \dim U$ by the [Rank-Nullity Theorem](/theorems/385). Choose a basis $\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_m\}$ for $\ker\alpha$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v), extend this to a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_r, \mathbf{e}_{r+1}, \dots, \mathbf{e}_m\}$ for $U$. **Step 2: The images $\alpha(\mathbf{e}_1), \dots, \alpha(\mathbf{e}_r)$ form a basis for $\mathrm{im}\,\alpha$.** [claim:Images Form Basis] The set $\{\alpha(\mathbf{e}_1), \dots, \alpha(\mathbf{e}_r)\}$ is a basis for $\mathrm{im}\,\alpha$. [/claim] [proof] **Spanning:** Every $\mathbf{v} \in \mathrm{im}\,\alpha$ has the form $\mathbf{v} = \alpha(\mathbf{u})$ for some $\mathbf{u} = \sum_{i=1}^m \lambda_i \mathbf{e}_i$. Then $\mathbf{v} = \sum_{i=1}^r \lambda_i \alpha(\mathbf{e}_i) + \sum_{i=r+1}^m \lambda_i \alpha(\mathbf{e}_i) = \sum_{i=1}^r \lambda_i \alpha(\mathbf{e}_i)$, since $\alpha(\mathbf{e}_i) = \mathbf{0}$ for $i > r$. **Independence:** Suppose $\sum_{i=1}^r \lambda_i \alpha(\mathbf{e}_i) = \mathbf{0}$. Then $\alpha(\sum_{i=1}^r \lambda_i \mathbf{e}_i) = \mathbf{0}$, so $\sum_{i=1}^r \lambda_i \mathbf{e}_i \in \ker\alpha = \langle \mathbf{e}_{r+1}, \dots, \mathbf{e}_m \rangle$. Write $\sum_{i=1}^r \lambda_i \mathbf{e}_i = \sum_{i=r+1}^m \mu_i \mathbf{e}_i$, which gives $\sum_{i=1}^r \lambda_i \mathbf{e}_i - \sum_{i=r+1}^m \mu_i \mathbf{e}_i = \mathbf{0}$. Since $\{\mathbf{e}_1, \dots, \mathbf{e}_m\}$ is a basis, all coefficients vanish, so $\lambda_i = 0$ for $i = 1, \dots, r$. [/proof] **Step 3: Extend to a basis for $V$.** Set $\mathbf{f}_i = \alpha(\mathbf{e}_i)$ for $i = 1, \dots, r$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v), extend $\{\mathbf{f}_1, \dots, \mathbf{f}_r\}$ to a basis $\{\mathbf{f}_1, \dots, \mathbf{f}_n\}$ for $V$. **Step 4: Compute the matrix.** With respect to $(\mathbf{e}_1, \dots, \mathbf{e}_m)$ and $(\mathbf{f}_1, \dots, \mathbf{f}_n)$: \begin{align*} \alpha(\mathbf{e}_i) = \mathbf{f}_i \quad &\text{for } i = 1, \dots, r, \\ \alpha(\mathbf{e}_i) = \mathbf{0} \quad &\text{for } i = r+1, \dots, m. \end{align*} The matrix of $\alpha$ in these bases is therefore $\begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$.

**Step 5: Express via invertible matrices.**

Let $P$ be the change-of-basis matrix from $(\mathbf{e}_i)$ to the original basis for $U$, and $Q$ the change-of-basis matrix from $(\mathbf{f}_j)$ to the original basis for $V$. By the [Change of Basis Formula](/theorems/387), $Q^{-1}AP = \begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$, where $A$ is the matrix of $\alpha$ in the original bases. $\blacksquare$