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Came across this problem in my textbook and am not sure how it would be done. There are so many different key words that I don't know which should be focused on.

Let V and W be vector spaces over a field F and let U be a nontrivial proper subspace of V. Are the subsets $\{ \alpha \in Hom(V,W) | ker(\alpha) \subseteq U \}$ and $\{ \alpha \in Hom(V,W) | U \subseteq ker(\alpha\}$ subspaces of Hom(V,W)?

I know that proving a subspace requires proving vector space and that a homomorphism is closed under addition and scalar multiplication. However, I have no idea how to connect all the information in this question. Any suggestions?

Edited to add further research:

  1. Let V and W be vector spaces over a field F and let $\alpha \in Hom (V,W)$. Then $ker(\alpha)$ is a subspace of V.
  2. $Hom(V,W)$ is a subspace of $W^V$ (the space of all functions from V to W).
  3. Subspaces: contain the zero vector, are closed under addition and scalar multiplication.
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    $\begingroup$ For the first set: notice $\ker(0) = V$. $\endgroup$ – user251257 Oct 17 '16 at 17:46
  • $\begingroup$ @user251257 Why? (I apologize if this should be obvious but I do not see it) $\endgroup$ – AlyraGamer Oct 17 '16 at 22:04
  • $\begingroup$ for every $x\in V$ we have $0(x) = 0$, that is $V = \ker(0)$. $\endgroup$ – user251257 Oct 17 '16 at 23:52
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    $\begingroup$ I'd suggest you to review the vector space structure on $\hom(V,W)$. $\endgroup$ – Alex Provost Oct 18 '16 at 0:27
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    $\begingroup$ @AlexProvost Such as $(f + g)(v)=f(v) +g(v)$ and $(cf)(v)=c(f(v))$? Im clearly missing something as all of the comments make it seem as though this should be simple.... I understand how to prove a subspace but cant seem to understand it for these subsets. $\endgroup$ – AlyraGamer Oct 18 '16 at 0:56
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To show that a subset is a subspace you should check three things. Closed under addition, closed under scalar multiplication and the zero vector in in the subset. But just looking at one, you should get some ideas to get going.

  • Closed under addition

That is take two functionals, $\alpha$ and $\beta$ and check whether there are more elements from $v\in$ $V$ such that $\alpha(v)$ + $\beta(v)$ is in the subset or not (remember that they are $proper$ subsets of $V$)

Edit: As the other comments pointed out it's much more evident to look at the at the condition of having the zero vector. In that case the zero vector is the functional that for all $v\in$ $V$, $0(v) = 0$. The condition tells you that $\{ 0 \in Hom(V,W) | ker(0) \subseteq U \}$. But remembering the condition of being a $proper subspace$, then $ U \subset V\ $ and $U \neq V$ and $U \neq \varnothing\ $. But what happens to $ker(0)$ ?

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    $\begingroup$ The proper subspace piece is part of my confusion I believe. In order to be proper U cannot equal V and for nontrivial U cannot equal the zero vector. How does this affect the problem? $\endgroup$ – AlyraGamer Oct 17 '16 at 22:38
  • $\begingroup$ I edited the answer, taking into account your comment! $\endgroup$ – Leo Lerena Oct 18 '16 at 0:50
  • $\begingroup$ Ker(0)=0 which contradicts the condition of the subset. Correct? $\endgroup$ – AlyraGamer Oct 18 '16 at 1:10
  • $\begingroup$ No! for all $v\in$ $V$, $0(v) = 0$ . So $ker(0)$ = $V$ $\endgroup$ – Leo Lerena Oct 18 '16 at 1:43
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    $\begingroup$ So if the ker(0)=V, does that mean that the subset cannot be a subspace because U is a proper subspace and the condition will give V contains U, which contradicts the proper subspace? $\endgroup$ – AlyraGamer Oct 18 '16 at 1:50

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