If $\{ u_1,...,u_n\}$ is a linearly independent of $U$, show that $\{ T(u_1), ..., T(u_n)\}$ is linearly independent My Professor gives us a homework that contains this question:

Let $U$ and $V$ be a vector spaces on a field $F$, and let $T:U \to V$ be a linear transformation on $F$.
If $\{ u_1,...,u_n\}$ is a linearly independent of $U$, show that $\{ T(u_1), ..., T(u_n)\}$ is linearly independent.

Is that true? I took $T:\mathbb{R}^2 \to \mathbb{R}$, such that $T((x,y))=0$ for all $(x,y) \in \mathbb{R}^2$
Then I found that $\{ (1,0),(0,1)\}$ is linearly independent in $ \mathbb{R}^2$, but $\{T(1,0)=0,T(0,1)=0\}$ is not linearly independent in $\mathbb{R}$ ?
What do you think?
 A: Your example is a valid counterexample to the claim.
So, this is not true without additional hypotheses on the linear transformation $T$. If you assume, for instance, that $T$ is injective, then the claim is indeed true. Or, as @ThomasAndrews says in the comments under the question, perhaps you meant to claim that if $T(u_1),\dots,T(u_n)$ are linearly independent, then $u_1,\dots,u_n$ are linearly independent.

Additionally, a small note on terminology: it is preferable to say that $\{ u_1,\dots,u_n \}$ is a linearly independent subset of $U$, and that $\{ T(1,0), T(0,1) \}$ is not linearly independent over $\mathbb{R}$.
A: You are correct. This also needs to assume that $T$ is invertible (or as Brahadeesh points out, injectivity suffices). In that case, if
$$\alpha_1 T(u_1) + \cdots + \alpha_n T(u_n) = 0$$
linearity tells us
$$T(\alpha_1 u_1 + \cdots + \alpha_n u_n) = 0,$$
and hence if we apply $T^{-1}$ to both sides
$$\alpha_1 u_1 + \cdots + \alpha_n u_n = 0.$$ Since $\{u_1,\ldots,u_n\}$ are linearly independent this implies that $\alpha_1=\cdots=\alpha_n=0$, and then by definition $\{T(u_1),\ldots,T(u_n)\}$ are linearly independent as well.
A: In general, this isn't true as your counterexample shows. 
If $T$ is invertible then the statement holds. Start from
$$c_1T(u_1)+\cdots+c_nT(u_n)=0$$
Apply $T^{-1}$ on both sides and get
$$c_1u_1+\cdots+c_nu_n=0$$
and $c_1=\cdots=c_n=0$ by assumption. Actually it is enough for $T$ to have a left inverse i.e. $T$ is injective.
