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

## 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?

Last edited:

## Answers and Replies

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

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: $$v \in U + W \Rightarrow \exists u \in U, w \in W : v = u + w[/itex] (2). ## The Attempt at a Solution 1. Since [tex]U$$ and $$W$$ both contain the additive identity, $$U+W$$ contains the additive identity

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:

$$\left\{ \begin{array}{c} 0 \in U \\ 0 \in W \end{array} \right. \Rightarrow 0 = 0 + 0 \in U + V$$ (due to (1)) (the underline part is just for you to understand why)

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).

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

I'm having a hard time thinking about how to justify that U+W is closed under addition.

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".

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

OK I think I got it. Thanks guys!!