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For a finite dimensional vector space $V$, is it true that $\bigwedge^{n - 1}V \otimes V = \bigwedge^{n}V \oplus \ker(\bigwedge^{n - 1}V \otimes V \overset{\psi}{\rightarrow}\bigwedge^{n}V)$ where $\psi$ is just the natural map?

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If you have a surjective map of vector spaces $f: W \rightarrow W'$, then you have an exact sequence $0 \rightarrow \text{ker}(f) \rightarrow W \rightarrow W' \rightarrow 0$. Since every short exact sequence of vector spaces splits, you get an isomorphism $W \cong W' \oplus \text{ker}(f)$.

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This is just the Rank-Nullity theorem :) – Alex Youcis May 8 '12 at 3:41

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