Linear Transformation from Ordered Basis less Kernel

From ProofWiki
Jump to navigation Jump to search


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 $\sequence {a_n}$ be any ordered basis of $G$ such that $\set {a_k: r + 1 \le k \le n}$ is the basis of the kernel of $\phi$.

Then $\sequence {\map \phi {a_r} }$ is an ordered basis of the image of $\phi$.



$\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k} = 0$


$\ds \map \phi {\sum_{k \mathop = 1}^r \lambda_k a_k} = 0$

So $\ds \sum_{k \mathop = 1}^r \lambda_k \map \phi {a_k}$ belongs to the kernel of $\phi$ and hence is also a linear combination of $\set {a_k: r + 1 \le k \le n}$.

Thus $\forall k \in \closedint 1 r: \lambda_k = 0$ since $\sequence {a_n}$ is linearly independent.

Thus the sequence $\sequence {\map \phi {a_r} }$ is linearly independent.

We have $\forall k \in \closedint {r + 1} n: \map \phi {a_k} = 0$.

So let $x \in G$.

Let $\ds x = \sum_{k \mathop = 1}^n \mu_k a_k$.


$\ds \map \phi x = \sum_{k \mathop = 1}^n \mu_k \map \phi {a_k} = \sum_{k \mathop = 1}^r \mu_k \map \phi {a_k}$

Therefore $\sequence {\map \phi {a_r} }$ is an ordered basis of the image of $\phi \sqbrk G$.