Let $ u_1,u_2 $ be linearly independent elements of $U$. There is a linear map $T:U \to V$. Given this $T(u_1), T(u_2)$ are a linearly independent elements of $V$. Here, $U$ and $V$ are vector spaces over a field $K$.
My attempt:
$a_1u_1+a_2u_2=0_U$
(*) $T(a_1u_1+a_2u_2)=T(0_U)$
(*) is equivalent to,
(**): $a_1T(u_1)+a_2T(u_2)=T(0_U)=0_V$
We know $a_1=a_2=0$ as $u_1, u_2$ is a linearly independent set
Suppose $a_3,a_4$ are also solutions to (**)
Then $ T^{-1}( a_3T(u_1)+a_4T(u_2))=T^{-1}(0_V) $
This means $ a_3(u_1)+a_4(u_2)=(0_U) $, but this is not possible as $u_1,u_2$ are linearly independent
So $a_1,a_2$ are the only solutions to (**) $\Rightarrow$ linear independence.
a) Have I made any oversights/mistakes in my attempted proof? b)Can someone direct me towards proving this statement is true/false?
Best Answer
This is not true. Let $V =\mathbb R$, and let $T:U\to \mathbb R$ be any linear map. It doesn't matter what $u_1,u_2\in U$ you choose, $T(u_1),T(u_2)$ will always be linearly dependent because $\mathbb R$ has dimension $1$: It's impossible to find two or more linearly independent vectors of $\mathbb R$.
On the other hand, if $T:U\to V$ is injective, and if $T(u_1),T(u_2)$ are linearly dependent, then there exists $\alpha,\beta$ such that:
$$\alpha T(u_1)+\beta T(u_2)=0_V$$ $$\iff T(\alpha u_1+\beta u_2)=0_V$$ $$\iff \alpha u_1+\beta u_2\in \ker (T)$$
But, because $T$ is injective, $\ker(T)=\{0_U\}$. This means that $$\alpha u_1+\beta u_2=0_U$$
So, in other words $T(u_1),T(u_2)$ are linearly dependent $\implies$ $u_1,u_2$ are linearly dependent
This is the same as saying: $u_1,u_2$ are linearly independent $\implies $ $T(u_1),T(u_2)$ are linearly independent (Contraposition)
So in the case where $T$ is injective, that holds.