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