Linear Transformation and linear independence

Question

I have the following question: - We proved that if $T$ is 1-1 and $\{v_1...v_n\}$ is linearly independent then $\{T(v_1)...T(v_n)\}$ is linearly independent! I understood the proof! But can’t $\{T(v_1)...T(v_n)\}$ be linearly independent without having $T$ 1-1? The image I uploaded shows my work on proving that the set of the images is linearly independent without having $T$ 1-1. Can someone help me?

Answer 1

The problem in your own writing is, that you assume that $c_1 v_1 + \ldots + c_n v_n=0$. To show that $T(v_1),\ldots ,T(v_n)$ are linear independent, you have to assume that there are $c_1,\ldots,c_n \in \mathbb{K}$ such that $c_1 T(v_1)+\ldots + c_n T(v_n)=0$ and then show that $c_1=\ldots=c_n=0$. You are somehow using that $c_1v_1+\ldots+c_nv_n=0$, but since the map is not 1-1, there is no reason for that. If you assume that the map is 1-1, then $T(c_1v_1+\ldots+c_nv_n)=0$, so by injectiveness, $c_1v_1+\ldots+c_nv_n=0$. Then you can use the linear independence of $v_1,\ldots,v_n$ to conclude that $c_1=\ldots=c_n=0$.

So the problem is really that you start with the wrong precondition.

A simple example would be $$T\colon \mathbb{R}^2 \to \mathbb{R}^2,~~ x\mapsto \begin{pmatrix}1&1\\0&0\end{pmatrix}x$$ The base $\left\{\,e_1=\begin{pmatrix}1\\0\end{pmatrix}, e_2=\begin{pmatrix}0\\1\end{pmatrix}\,\right\}$ gets mapped to $\{\,e_1,e_1\,\}$, which is no longer linearly independent.

Answer 2

If T is not 1-1 you can’t guarantee that all of the T(v_i) are different. If any two are the same then {T(v_i)} won’t be a linearly independent set.