Linear Transformation from Ordered Basis less Kernel

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

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

Let $$G$$ be $n$-dimensional.

Let $$\left \langle {a_n} \right \rangle$$ be any ordered basis of $$G$$ such that $$\left\{{a_k: r + 1 \le k \le n}\right\}$$ is the basis of the kernel of $$\phi$$.

Then $$\left \langle {\phi \left({a_r}\right)} \right \rangle$$ is an ordered basis of the image of $$\phi$$.

Proof

 * Suppose:
 * $$\sum_{k=1}^r \lambda_k \phi \left({a_k}\right) = 0$$

Then:
 * $$\phi \left({\sum_{k=1}^r \lambda_k a_k}\right) = 0$$

So $$\sum_{k=1}^r \lambda_k \phi \left({a_k}\right)$$ belongs to the kernel of $$\phi$$ and hence is also a linear combination of $$\left\{{a_k: r + 1 \le k \le n}\right\}$$.

Thus $$\forall k \in \left[{1 \,. \, . \, r}\right]: \lambda_k = 0$$ since $$\left \langle {a_n} \right \rangle$$ is linearly independent.

Thus the sequence $$\left \langle {\phi \left({a_r}\right)} \right \rangle$$ is linearly independent.


 * We have $$\forall k \in \left[{r + 1 \, . \, . \, n}\right]: \phi \left({a_k}\right) = 0$$.

So let $$x \in G$$. Let $$x = \sum_{k=1}^n \mu_k a_k$$.

Then $$\phi \left({x}\right) = \sum_{k=1}^n \mu_k \phi \left({a_k}\right) = \sum_{k=1}^r \mu_k \phi \left({a_k}\right)$$.

Therefore $$\left \langle {\phi \left({a_r}\right)} \right \rangle$$ is an ordered basis of the image of $$\phi \left({G}\right)$$.