# Prove the sum of two subspaces is also a subspace.

1. Sep 6, 2009

### PhillipKP

1. The problem statement, all variables and given/known data

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

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

3. 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?

Last edited: Sep 7, 2009
2. Sep 7, 2009

### 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

3. Sep 7, 2009

### 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: $$v \in U + W \Rightarrow \exists u \in U, w \in W : v = u + w[/itex] (2). Ok, this is good, but you really need to show why it contains the additive identity, instead of just reasoning like that. You can go like this: Since U, and W are subspaces, that means: [tex]\left\{ \begin{array}{c} 0 \in U \\ 0 \in W \end{array} \right. \Rightarrow 0 = 0 + 0 \in U + V$$ (due to (1)) (the underline part is just for you to understand why)

As I said above, you should make this part clearer.

Just start out as normal. So, what you want is to prove that:

$$\forall v_1 \in U + W, v_2 \in U + W$$, we must have $$v_1 + v_2 \in U + W$$

Hint: Use (2), and the fact that both U, and W are already subspaces. :)

The same proof goes for "closed under scalar multiplication".

4. Sep 7, 2009

### PhillipKP

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

5. Sep 7, 2009

### PhillipKP

OK I think I got it. Thanks guys!!