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This seems right, but I cannot think of a proof.

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Are the spaces finite dimensional? Try using rank-nullity? –  Ben West Feb 9 '13 at 4:01
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@BenW. There is no need for finite dimensionality. If $T:V\to W$ is an $F$-linear map ($F$ is some field) then $0\to \ker V\to V\to T(V)\to 0$ is an exact sequence, and since $T(V)$ is a free $F$-module the sequence splits so that $V\cong \ker V\oplus T(V)$. –  Alex Youcis Feb 9 '13 at 4:05
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@AlexYoucis Thanks! I just wanted to ask if they were finite dimensional to give an answer at the appropriate level. –  Ben West Feb 9 '13 at 4:08
    
@AlexYoucis this was a buge sledgehammer –  Norbert Feb 9 '13 at 4:49
    
@Norbert Yes, yes it was. I didn't have much space though--I hope you'll forgive me. –  Alex Youcis Feb 9 '13 at 4:53
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3 Answers

up vote 4 down vote accepted

By rank-nullity, $$ \dim(V)=\dim(\ker(L))+\dim(\operatorname{Im}(L)). $$ So $\dim(\ker(L))\neq 0$, otherwise $$ \dim(\operatorname{Im}(L))=\dim(V)>\dim(W) $$ which is impossible since $\operatorname{Im}(L)\leq W$.

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By assumption, the image in $W$ of a basis for $V$ cannot be an independent set of vectors, so some nontrivial combination of images of basis vectors must be zero, so the image of some nontrivial combination of basis vectors is zero.

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Hint:

1) $L: V \to W$ is $1-1$ iff $$\ker L =\{0\}$$ now $$\dim V > \dim W$$ then $L$ cannot be $1-1$ therefore $\ker L\neq 0$.

2) let $A$ be matrix of this linear transformation respect to standard base (The Matrix of a Linear Transformation), so $$[A]_{\dim w×\dim v}$$ according to this theorem you can conclude $\ker L\neq 0$, if $m\lt n$ then has $$A_{m×n}X=0$$ nontrivial solution .

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