1. Injective linear transformation

I got the following question wrong in the final exam : Tell whether the affirmation is true or false by justificating.
$T:\mathbb{R}^n \to \mathbb{R}^n$ is an injective linear transformation if and only if for all subspaces $W$ of $\mathbb{R}^n$ with dimension 2, $T:W \to \mathbb{R}^n$ is injective.
I said it was true but my explanation was wrong. I had a small doubt : I thought it might be false because $n$ can be uneven but then this doubt disappeared.
Can you help me with this exercise?

2. If $T$ is not injective, take $0\neq u \in \mbox{Ker T}$, $v \notin \mbox{Ker T}$; then $u, u+v$ are linearly independent and $T(u)=T(u+v)$. This implies that $T(k_1u+k_2v)=k_1T(u)+k_2T(v) = k_2T(v)$; in other words $T(\mbox{span }\{u,u+v\})=\mbox{span }\{T(v)\}$, so that if $T$ is not injective, there exists a two-dimensional subspace mapping to a one-dimensional subspace.

3. Originally Posted by Bruno J.
If $T$ is not injective, take $0\neq u \in \mbox{Ker T}$, $v \notin \mbox{Ker T}$; then $u, u+v$ are linearly independent and $T(u)=T(u+v)$. This implies that $T(k_1u+k_2v)=k_1T(u)+k_2T(v) = k_2T(v)$; in other words $T(\mbox{span }\{u,u+v\})=\mbox{span }\{T(v)\}$, so that if $T$ is not injective, there exists a two-dimensional subspace mapping to a one-dimensional subspace.
Ok so if I understand well, $T$ has to be injective in order to go from $W \to \mathbb{R}^n$.
So you've proved $\Leftarrow$ ?

4. Not sure what you mean by "T has to be injective in order to...", but yeah I proved $\Leftarrow$.

If $T$ is injective, then clearly the restriction of $T$ to any $W$ is also injective.

If $T$ is not injective, then there exists some 2-dimensional $W$ whose image is 1-dimensional; hence surely if all 2-dimensional $W$ have 2-dimensional images then $T$ must be injective.

This is a proof by the "contrapositive" - in order to prove that A implies B, we prove that (not B) implies (not A).