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not sure how to word my answer for this question:

Let V be a vector space and let H and K be two subspaces of V. Show that the following set W is a subspace of V:

W={u+v: u ∈ H, v ∈ K}

I'm pretty sure the answer is because H and K are two subspaces of V, meaning they are closed under addition. So when you add u and v together, they are also a subspace of V, but I'm not sure how to phrase this! Any help would be appreciated.

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  • $\begingroup$ Write down a general element of $W$. Write down another general element of $W$. Write down the sum. Check that it is in $W$. $\endgroup$ – Peter Franek Jun 14 '16 at 15:31
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Suppose that $u\in W$ and $v\in W$. We wish to prove that $\alpha u - v$ is also in $W$.

Since $u\in W$, we know by definition of $W$ that $u$ can be written as $u=h_1+k_1$ where $h_1\in H$ and $k_1\in K$

Similarly, $v\in W$ implies that $v$ can be written as $v=h_2+k_2$ with $h_2\in H$ and $k_2\in K$.

Then $\alpha u - v = \alpha (h_1+k_1)-h_2-k_2 = (\alpha h_1 - h_2) + (\alpha k_1 - k_2)$

Since $H$ is a subspace, $(\alpha h_1 - h_2)\in H$ and similarly since $K$ is a subpsace $(\alpha k_1 - k_2)\in K$

Thus $\alpha u - v$ can be written as the sum of vectors with the first vector from $H$ and the second from $K$ and is therefore an element of $W$, proving that it is indeed a subspace.


As an aside: checking that $\alpha u - v \in W$ is equivalent to checking the following conditions simultaneously:

  • $0\in W$
  • $u\in W, v\in W\Rightarrow u+v\in W$
  • $u\in W, \alpha\in \Bbb F\Rightarrow \alpha u \in W$

Also worth mentioning is that here, we made no assumptions about $H\cap K$. In a future problem, you might use the additional assumption that $H\cap K=\{0\}$ in which case you will conclude that $\dim(W) = \dim(H)+\dim(K)$ and that there is a unique way to write $u=h+k$.

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Check the necessary and sufficient conditions: is $W$ non-empty and is it closed under addition and scalar multiplication?

Take $w_1,w_2 \in W$, this means $w_1=u_1+v_1$ with $u_1 \in H$ and $v_1 \in K$ and $w_2=u_2+v_2$ with $u_2 \in H$ and $v_2 \in K$. Now look at $w_1+w_2$; is it of the required form to be in $W$?

Do the same for $w \in W$ (this means $w = \ldots$) and consider $\lambda w$.

You can also combine these two conditions by working with a more general linear combination $\lambda w_1 + \mu w_2$.

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