# Confused about the proof of U = U + U, where U is a subspace

1. Aug 29, 2009

### PhillipKP

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

Hi I need help understanding a proof. This is my first time in a pure math class, so proofs of this type are a little weird to me.

If U is a subspace of the vectorspace V, what is U+U?

2. Relevant equations

The proof:

$$(v_{1}+v_{2})\in U+U$$

As $$v_{1},v_{2}\in U$$and $$U$$ is a subspace of $$V$$,

$$(v_{1}+v_{2})\in U$$

Thus $$U+U\subseteq U$$

Now let
$$v\in U$$.

Then as $$0\in U$$,

$$v=(v+0)\in U+U.$$

Thus $$U\subseteq U+U$$

$$\therefore U=U+U$$

3. The attempt at a solution

I don't understand why the first part proves $$U+U\subseteq U$$ rather than $$U\subseteq U+U$$.

Similarly, I don't understand why the second part proves $$U\subseteq U+U$$ rather than proving $$U+U\subseteq U$$.

I guess I just don't understand why each part proves 1 direction of the equality by not the other direction.

2. Aug 29, 2009

### Dick

I don't understand what's confusing you. In the first part you picked a general element of U+U and showed that it's in U, right? If I pick any element of A and show that it is in B, that means A is a subset of B, also right?

3. Aug 29, 2009

### PhillipKP

Ok I agree with you. But why is the second part of the proof so much different from the first part?

Couldn't I just exchange the 1st and 3rd line of the proof to prove the other direction?

4. Aug 29, 2009

### Dick

The second part is different, uh, because it's different. Now you want to take an element of U and show it's in U+U. The first part uses the fact that vector spaces are closed under addition. The second part uses that they have an additive identity. They really are different.

5. Aug 29, 2009

### VietDao29

To prove that 2 set A and B are equal, one has to prove that A is a subset of B, and B is also a subset of A. Or written formally:

$$A = B \Leftrightarrow \left\{ \begin{array}{ccc} A & \subset & B \\ B & \subset & A \end{array} \right.$$

Or in words, A are B are equal, iff every element of A is contained in B, and vice-versa.

Vector-space (or sub-space) can be understood as a set of vectors with 2 predefined operators: vector addition, and scalar multiplication.

So to prove 2 subspace A, and B are equal, one just needs to prove:

$$A = B \Leftrightarrow \left\{ \begin{array}{ccc} A & \subset & B \\ B & \subset & A \end{array} \right.$$

We'll now be proving that every element of U + U is in U. First, take out a random vector in U + U, namely v1 + v2.

We have proved that this vector is also contained in U. And hence, we have:

Our second step is to prove that every element of U is contained in U + U.

Consider a random vector v in U.

Our second goal is here.. And hence:

Are there any more steps that confuse you?