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The full problem is:

Given $T: W\to V$, a linear transformation of $F$-vector spaces, such that $\text{ker} T = 0$, show that the kernel of $\bigwedge^r(T): \bigwedge^r(W)\to \bigwedge^r(V)$ is also zero.

I'm confused on how to deal with this problem when $T$ is not an endomorphism. It would also be helpful to know what $\bigwedge^r(T)$ is supposed to look like.

Thanks

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    $\begingroup$ HINT: If $\ker T = \{0\}$, then $T$ maps sets of $r$ linearly independent vectors to linearly independent vectors. $\endgroup$ – Ted Shifrin Mar 19 at 23:32
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Hint: Use the fact that a linear transformation $T \colon W \rightarrow V$ is injective if and only if there exists a linear transformation $S \colon V \rightarrow W$ such that $S \circ T = \operatorname{id}_{W}$. Then use the functoriality of $\Lambda$ (that is, $\Lambda^r(S \circ T) = \Lambda^r(S) \circ \Lambda^r(T)$ and $\Lambda^r(\operatorname{id_{U}}) = \operatorname{id}|_{\Lambda^r(U)}$).

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