Prove the sum of two subspaces is also a subspace.

[PhillipKP]

Homework Statement

Hi I'm trying to prove that the sum of two subspaces $$U$$ and $$W$$ is also a subspace.

Homework Equations

$$U$$ is a subspace of $$V$$ if $$U$$ is also a vector space and it contains the additive identity, is closed under addition, and closed under scalar multiplication.

The definition of a sum a vector subspace U and W is
$$U+W=\{u+w:\, u\in U,w\in W\}$$

The Attempt at a Solution

1. Since $$U$$ and $$W$$ both contain the additive identity, $$U+W$$ contains the additive identity
3. Since both $$U$$ and $$W$$ are closed under scalar multiplication, any combination of $$u+w$$ is closed under scalar multiplication since multiplication is distributive, associative and commutes (assuming were dealing with the reals here). I'm having a hard time thinking about how to justify that U+W is closed under addition.

Also is my justification for closure under scalar multiplication right?

 

[lanedance]

First let Z = U+W

to show its closed under addition
so take z, z' in Z then show z + z' is an element of Z

your words are close, but I would try a similar thiong for the scalar multiplication to be a little more explicit

 

[VietDao29]

$$U+W=\{u+w:\, u\in U,w\in W\}$$

Writing this means that: U + W is the set of the sum of all possible 2 vectors, of which one is taken out from U, and the other from W, i.e:

• If u is some vector of U, and w is some vector of W, then $u + w \in U + W$. (1)
• And if v is some vector of U + W, then it can be split into the sum of 2 vectors, one in U, and one in W. That is: [tex]v \in U + W \Rightarrow \exists u \in U, w \in W : v = u + w[/itex] <b>(2)</b>.[/tex]

[tex] <br /> <blockquote data-attributes="" data-quote="PhillipKP" data-source="post: 2337030" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> PhillipKP said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>The Attempt at a Solution</h2><br /> <br /> 1. Since $$U$$ and $$W$$ both contain the additive identity, $$U+W$$ contains the additive identity </div> </div> </blockquote><br /> Ok, this is good, but you really need to <b>show</b> why it contains the additive identity, instead of just reasoning like that. You can go like this:<br /> <br /> Since U, and W are subspaces, that means:<br /> <br /> $$\left\{ \begin{array}{c} 0 \in U \\ 0 \in W \end{array} \right. \Rightarrow 0 = 0 + 0 \in U + V$$ <u>(due to <b>(1)</b>)</u> (the underline part is just for you to understand why)<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 3. Since both $$U$$ and $$W$$ are closed under scalar multiplication, any combination of $$u+w$$ is closed under scalar multiplication since multiplication is distributive, associative and commutes (assuming were dealing with the reals here). </div> </div> </blockquote><br /> As I said above, you should make this part clearer.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I'm having a hard time thinking about how to justify that U+W is closed under addition. </div> </div> </blockquote><br /> Just start out as normal. So, what you want is to prove that:<br /> <br /> $$\forall v_1 \in U + W, v_2 \in U + W$$, we must have $$v_1 + v_2 \in U + W$$<br /> <br /> Hint: Use <b>(2)</b>, and the fact that both U, and W are already subspaces. :)<br /> <br /> The same proof goes for "<i>closed under scalar multiplication</i>".[/tex]

 

[PhillipKP]

Ok thank you. Let me think about this some and try to make some progress. I'll be back later in the day.

 

[PhillipKP]

OK I think I got it. Thanks guys!