Unique Linear Transformation Between Modules

Theorem
Let $G$ and $H$ be unitary $R$-modules.

Let $\left \langle {a_n} \right \rangle$ be an ordered basis of $G$.

Let $\left \langle {b_n} \right \rangle$ be a sequence of elements of $H$.

Then there is a unique linear transformation $\phi: G \to H$ satisfying $\forall k \in \left[{1 \,. \, . \, n}\right]: \phi \left({a_k}\right) = b_k$

Corollary
Let $G$ be a finite-dimensional $K$-vector space.

Let $H$ be a $K$-vector space (not necessarily finite-dimensional).

Let $\left \langle {a_n} \right \rangle$ be a linearly independent sequence of vectors of $G$.

Let $\left \langle {b_n} \right \rangle$ be a sequence of vectors of $H$.

Then there is a unique linear transformation $\phi: G \to H$ satisfying $\forall k \in \left[{1 \,. \, . \, n}\right]: \phi \left({a_k}\right) = b_k$

Proof
By Isomorphism from R^n via n-Term Sequence, the mapping $\phi: G \to H$ defined as:
 * $\displaystyle \phi \left({\sum_{k=1}^n \lambda_k a_k}\right) = \sum_{k=1}^n \lambda_k b_k$

is well-defined.

Thus:
 * $\forall k \in \left[{1 \, . \, . \, n}\right]: \phi \left({a_k}\right) = b_k$

By Linear Transformation of Generated Module, $\phi$ is the only linear transformation whose value at $a_k$ is $b_k$ for all $k \in \left[{1 \,. \, . \, n}\right]$.

Proof of Corollary
From Results concerning Generators and Bases of Vector Spaces, $\left\{{a_1, \ldots, a_m}\right\}$ is contained in a basis.

The result then follows from the above result.