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I have a linear transformation $T : V \to W$, and $Z$ is a subspace of $W$. We also have $U = \{v \in V\}$ such that $ T(v) \in Z$, and we want to show that U is a subspace of V.

So basically they're telling me that T maps V into W, and U is a subset of V that maps into Z, which is a subset of W.

For $U$ to be a subspace of $V$, it must be closed under addition and scalar multiplication. For any two vectors $u,v \epsilon U$, it's clear that $T(u + v)$ maps into Z because we know T is a linear transformation, so it's also closed under addition. The same argument holds for scalar multiplication.

Is that sufficient to show that U is a subspace of V? Is there a simpler way to do it? I also thought of maybe using a proof by contradiction, but I don't think that's possible due to insufficient information. Are there other ways to show that U is a subspace of V?

Thanks,

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up vote 1 down vote accepted

Your method works just fine (the only missing argument is that $Z$ itself is closed under addition and scalar multiplication).

It's hard to imagine a shorter proof than that. Also, proof by contradiction is better avoided when there is a direct proof.

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Got it, thanks! –  John Doe Dec 14 '11 at 17:38
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