Thread: Kernal and Image of a Linear Transformation

1. Kernal and Image of a Linear Transformation

Given $\{v_1, ... ,v_n\}$ in a vector space $V$, define $T: \mathbb{R}^n \rightarrow V$ by $T(r_1, ..., r_n) = r_1v_1+...+r_nv_n$

Show that T is one-to-one if and only if $\{v_1,...,v_n\}$ is independent.

2. Originally Posted by Scopur
Given $\{v_1, ... ,v_n\}$ in a vector space $V$, define $T: \mathbb{R}^n \rightarrow V$ by $T(r_1, ..., r_n) = r_1v_1+...+r_nv_n$

Show that T is one-to-one if and only if $\{v_1,...,v_n\}$ is independent.
$T$ is a linear transformation. Therefore, $T$ is one-to-one if and only if $\ker (T) = \{ \bold{0} \}$ if and only if $r_1\bold{v}_1+...+r_n\bold{v_n} = \bold{0} \implies r_1=r_2=...=r_n=0$

3. uhh.. 2nd part i get one and i have the answer to this but.. I don't quite understand onto as well as the kernel.

2. T is onto iff $V = span\{v_1,...,v_n\}$

4. Originally Posted by Scopur
uhh.. 2nd part i get one and i have the answer to this but.. I don't quite understand onto as well as the kernel.

2. T is onto iff $V = span\{v_1,...,v_n\}$
A linear transformation $T: \mathbb{R}^n \to V$ is onto iff for any $\bold{v}\in V$ we can find $(r_1,...,r_n)\in \mathbb{R}^n$ so that $T(r_1,...,r_n) = \bold{v}$. It should be clear now how to finish the proof.