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 $$\phi \left({\sum_{k=1}^n \lambda_k a_k}\right) = \sum_{k=1}^n \lambda_k b_k$$ 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.