Linear Transformation of Vector Space Monomorphism

Theorem
Let $$G$$ and $$H$$ be a $K$-vector space.

Let $$\phi: G \to H$$ be a linear transformation.

Then $$\phi$$ is a monomorphism iff for every linearly independent sequence $$\left \langle {a_n} \right \rangle$$ of vectors of $$G$$, $$\left \langle {\phi \left({a_n}\right)} \right \rangle$$ is a linearly independent sequence of vectors of $$H$$.

Proof

 * Suppose $$\phi$$ is a monomorphism.

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

Let $$\sum_{k=1}^n \lambda_k \phi \left({a_k}\right) = 0$$.

Then $$\phi \left({\sum_{k=1}^n \lambda_k a_k}\right) = 0$$.

So by hypothesis $$\sum_{k=1}^n \lambda_k a_k = 0$$.

Hence $$\forall k \in \left[{1 \,. \, . \, n}\right]: \lambda_k = 0$$.


 * Suppose that for every linearly independent sequence $$\left \langle {a_n} \right \rangle$$ of vectors of $$G$$, $$\left \langle {\phi \left({a_n}\right)} \right \rangle$$ is a linearly independent sequence of vectors of $$H$$.

Let $$\phi \left({a_1}\right) = 0$$. Then $$a_1 = 0$$, otherwise the sequence $$\left \langle {a_1} \right \rangle$$ of one term would be linearly independent but $$\left \langle {\phi \left({a_1}\right)} \right \rangle$$ would not.

Thus $$\mathrm{ker} \left({\phi}\right) = \left\{{0}\right\}$$ and by the Quotient Theorem for Group Epimorphisms $$\phi$$ is an isomorphism and therefore a monomorphism.