Linear Transformation from Ordered Basis less Kernel

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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 $\phi$.



$\displaystyle \sum_{k \mathop = 1}^r \lambda_k \phi \left({a_k}\right) = 0$


$\displaystyle \phi \left({\sum_{k \mathop = 1}^r \lambda_k a_k}\right) = 0$

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


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

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