[proofplan] Given a [linear map](/page/Linear%20Map) $\alpha: U \to V$ of rank $r$, we construct adapted bases for $U$ and $V$ in which the matrix of $\alpha$ is $\begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$. We start with a basis for $\ker\alpha$, extend it to a basis for $U$, show that the images of the extension vectors form a basis for $\mathrm{im}\,\alpha$, and extend that to a basis for $V$. The change-of-basis formula then yields the desired matrix equation. [/proofplan] [step:Choose a basis for $\ker\alpha$ and extend to a basis for $U$] 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 $\{e_{r+1}, \dots, e_m\}$ for $\ker\alpha$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v), extend this to a basis $\{e_1, \dots, e_r, e_{r+1}, \dots, e_m\}$ for $U$. [guided] The strategy is to split $U$ into two complementary pieces: the kernel of $\alpha$ (which maps to zero and contributes the zero block) and a complement (whose images will form the identity block). By the [Rank-Nullity Theorem](/theorems/385), $\dim\ker\alpha = s = m - r$. Choose any basis $\{e_{r+1}, \dots, e_m\}$ for $\ker\alpha$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v) (every linearly independent set in a finite-dimensional space can be extended to a basis), extend this to a basis $\{e_1, \dots, e_r, e_{r+1}, \dots, e_m\}$ for $U$. The ordering is deliberate: we place the kernel vectors last so that the first $r$ basis vectors have non-zero images under $\alpha$, producing the identity block in the top-left corner. [/guided] [/step] [step:Show $\alpha(e_1), \dots, \alpha(e_r)$ form a basis for $\mathrm{im}\,\alpha$] [claim:Images Form Basis] The set $\{\alpha(e_1), \dots, \alpha(e_r)\}$ is a basis for $\mathrm{im}\,\alpha$. [/claim] [proof] **Spanning:** Every $v \in \mathrm{im}\,\alpha$ has the form $v = \alpha(u)$ for some $u = \sum_{i=1}^m \lambda_i e_i \in U$. Then \begin{align*} v = \sum_{i=1}^r \lambda_i \alpha(e_i) + \sum_{i=r+1}^m \lambda_i \alpha(e_i) = \sum_{i=1}^r \lambda_i \alpha(e_i), \end{align*} since $\alpha(e_i) = \mathbf{0}$ for $i > r$ (these vectors lie in $\ker\alpha$). **Independence:** Suppose $\sum_{i=1}^r \lambda_i \alpha(e_i) = \mathbf{0}$. Then $\alpha\!\left(\sum_{i=1}^r \lambda_i e_i\right) = \mathbf{0}$, so $\sum_{i=1}^r \lambda_i e_i \in \ker\alpha = \langle e_{r+1}, \dots, e_m \rangle$. Write $\sum_{i=1}^r \lambda_i e_i = \sum_{i=r+1}^m \mu_i e_i$, giving $\sum_{i=1}^r \lambda_i e_i - \sum_{i=r+1}^m \mu_i e_i = \mathbf{0}$. Since $\{e_1, \dots, e_m\}$ is a basis, all coefficients vanish: $\lambda_i = 0$ for $i = 1, \dots, r$. [/proof] [/step] [step:Extend the image basis to a basis for $V$ and compute the matrix] Set $f_i = \alpha(e_i)$ for $i = 1, \dots, r$. By [Properties of Finite Dimensional Bases](/theorems/374) part (v), extend $\{f_1, \dots, f_r\}$ to a basis $\{f_1, \dots, f_n\}$ for $V$. With respect to the bases $(e_1, \dots, e_m)$ for $U$ and $(f_1, \dots, f_n)$ for $V$: \begin{align*} \alpha(e_i) = f_i \quad &\text{for } i = 1, \dots, r, \\ \alpha(e_i) = \mathbf{0} \quad &\text{for } i = r+1, \dots, m. \end{align*} The matrix of $\alpha$ in these bases is $\begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}$. [/step] [step:Express the canonical form via invertible change-of-basis matrices] Let $P$ be the change-of-basis matrix from $(e_i)$ to the original basis for $U$, and $Q$ the change-of-basis matrix from $(f_j)$ to the original basis for $V$. By the [Change of Basis Formula](/theorems/387): \begin{align*} Q^{-1}AP = \begin{pmatrix} I_r & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{pmatrix}, \end{align*} where $A$ is the matrix of $\alpha$ in the original bases. The canonical form is determined entirely by $r$, so it is unique. [/step]