[proofplan] We construct $\alpha$ explicitly by extending linearly from the prescribed values on the basis: every vector in $U$ has a unique representation as a linear combination of basis elements (by [Unique Representation by a Basis](/theorems/372)), so we define $\alpha$ on an arbitrary vector by applying the same linear combination to the target vectors $v_1, \dots, v_m$. Uniqueness of the basis representation makes this well-defined, and linearity follows by direct computation. For uniqueness, any two linear maps agreeing on a basis must agree on all linear combinations, hence everywhere. [/proofplan] [step:Construct $\alpha$ by extending linearly from the basis] By [Unique Representation by a Basis](/theorems/372), every $u \in U$ can be written uniquely as $u = \sum_{i=1}^{m} \lambda_i e_i$ for scalars $\lambda_1, \dots, \lambda_m \in \mathbb{F}$. Define \begin{align*} \alpha: U &\to V \\ u = \sum_{i=1}^{m} \lambda_i e_i &\mapsto \sum_{i=1}^{m} \lambda_i v_i. \end{align*} This map is well-defined because the representation of $u$ in the basis $\{e_1, \dots, e_m\}$ is unique: if $u = \sum_{i=1}^{m} \lambda_i e_i = \sum_{i=1}^{m} \mu_i e_i$, then $\lambda_i = \mu_i$ for all $i$, so the output is the same. By construction, $\alpha(e_i) = v_i$ for each $i$ (take $\lambda_i = 1$ and $\lambda_j = 0$ for $j \neq i$). [guided] We need to produce a linear map $\alpha: U \to V$ satisfying $\alpha(e_i) = v_i$ for each $i$. Since the basis $\{e_1, \dots, e_m\}$ spans $U$, every vector $u \in U$ is a linear combination of basis elements: $u = \sum_{i=1}^{m} \lambda_i e_i$. If $\alpha$ is to be linear and send $e_i$ to $v_i$, then we are forced to define $\alpha(u) = \sum_{i=1}^{m} \lambda_i v_i$. This is the only possible definition -- the question is whether it is well-defined. By [Unique Representation by a Basis](/theorems/372), the scalars $\lambda_1, \dots, \lambda_m$ are uniquely determined by $u$. Define \begin{align*} \alpha: U &\to V \\ u = \sum_{i=1}^{m} \lambda_i e_i &\mapsto \sum_{i=1}^{m} \lambda_i v_i. \end{align*} Well-definedness follows from the uniqueness of the basis representation: two different expressions for the same $u$ must have identical coefficients, so the output $\sum_{i=1}^m \lambda_i v_i$ does not depend on the representation chosen. Setting $\lambda_i = 1$ and $\lambda_j = 0$ for $j \neq i$ gives $\alpha(e_i) = 1 \cdot v_i = v_i$, confirming the prescribed values. [/guided] [/step] [step:Verify that $\alpha$ is linear] [claim:Alpha Is Linear] The map $\alpha$ is $\mathbb{F}$-linear. [/claim] [proof] Let $u = \sum_{i=1}^{m} \lambda_i e_i$ and $w = \sum_{i=1}^{m} \mu_i e_i$ be elements of $U$, and let $a, b \in \mathbb{F}$. Then \begin{align*} au + bw = \sum_{i=1}^{m} (a\lambda_i + b\mu_i) e_i, \end{align*} so by definition of $\alpha$: \begin{align*} \alpha(au + bw) = \sum_{i=1}^{m} (a\lambda_i + b\mu_i) v_i = a \sum_{i=1}^{m} \lambda_i v_i + b \sum_{i=1}^{m} \mu_i v_i = a\,\alpha(u) + b\,\alpha(w). \end{align*} [/proof] [/step] [step:Prove uniqueness by evaluating any linear extension on basis representations] Suppose $\beta: U \to V$ is another [linear map](/page/Linear%20Map) with $\beta(e_i) = v_i$ for all $i$. For any $u = \sum_{i=1}^{m} \lambda_i e_i \in U$, linearity of $\beta$ gives \begin{align*} \beta(u) = \beta\!\left(\sum_{i=1}^{m} \lambda_i e_i\right) = \sum_{i=1}^{m} \lambda_i \beta(e_i) = \sum_{i=1}^{m} \lambda_i v_i = \alpha(u). \end{align*} Since $\beta(u) = \alpha(u)$ for every $u \in U$, we conclude $\beta = \alpha$. [guided] To establish uniqueness, suppose $\beta: U \to V$ is any linear map satisfying $\beta(e_i) = v_i$ for each $i$. We must show $\beta = \alpha$, i.e., $\beta(u) = \alpha(u)$ for every $u \in U$. Take an arbitrary $u \in U$. By [Unique Representation by a Basis](/theorems/372), write $u = \sum_{i=1}^{m} \lambda_i e_i$. The key point is that linearity of $\beta$ completely determines its value on $u$ from its values on the basis: \begin{align*} \beta(u) = \beta\!\left(\sum_{i=1}^{m} \lambda_i e_i\right) = \sum_{i=1}^{m} \lambda_i \beta(e_i) = \sum_{i=1}^{m} \lambda_i v_i = \alpha(u). \end{align*} Since this holds for every $u \in U$, we have $\beta = \alpha$. The essential insight is that a linear map has no freedom once its values on a basis are fixed: linearity forces the value on every other vector. [/guided] [/step]