Linear Algebra: Subspace Sum: What is U + U?

[tylerc1991]

Homework Statement

Suppose U is a subspace of V. What is U + U?

Homework Equations

There are two definitions of a subspace sum that I know of (the first is the definition given in my book):
(1) U_1 + U_2 = \{ u_1 + u_2 : u_1 \in U_1, u_2 \in U_2 \}
(2) U_1 + U_2 = \text{ span} ( U_1 \, \bigcup \, U_2)

The Attempt at a Solution

Before I tried to solve the general problem, I thought about a specific example. Suppose that U = \{ (x,y) \in \mathbb{R}^2 : y = x \} \subseteq \mathbb{R}^2. Now U + U = U. So I should expect the same answer in general.

Using the first definition of a subspace sum:
U + U = \{ u + u : u \in U \} = \{ 2u : u \in U \}

Using the second definition of a subspace sum:
U + U = \text{ span} (U \, \bigcup \, U) = \text{ span}(U) = \{ au : a \in \mathbb{F}, u \in U \}.

While these are very similar(one is a special case of the other), I am leaning towards the second answer. That being said, how could I have come up with the second answer using the first definition of a subspace sum?

 

[micromass]

Hi tylerc!

Homework Statement

Suppose U is a subspace of V. What is U + U?

Homework Equations

There are two definitions of a subspace sum that I know of (the first is the definition given in my book):
(1) U_1 + U_2 = \{ u_1 + u_2 : u_1 \in U_1, u_2 \in U_2 \}
(2) U_1 + U_2 = \text{ span} ( U_1 \, \bigcup \, U_2)

The Attempt at a Solution

Before I tried to solve the general problem, I thought about a specific example. Suppose that U = \{ (x,y) \in \mathbb{R}^2 : y = x \} \subseteq \mathbb{R}^2. Now U + U = U. So I should expect the same answer in general.

Yes, U+U=U holds in general!

Using the first definition of a subspace sum:
U + U = \{ u + u : u \in U \} = \{ 2u : u \in U \}

You have used the first definition incorrectly here. It should be

U+U=\{u+v~\vert~u,v\in U\}

Now, you must prove that this is contained in U. Hint: U is a subspace and hence closed under addition.

Using the second definition of a subspace sum:
U + U = \text{ span} (U \, \bigcup \, U) = \text{ span}(U) = \{ au : a \in \mathbb{F}, u \in U \}.

Isn't it easy to see that span(U)=U because U is a subspace? Try to prove it otherwise.
By the way, your definition of span is also not correct (well, it is in this case, but whatever). In general, the span of X is

span(X)=\{\alpha_1x_1+...+\alpha_nx_n~\vert~i\in \mathbb{N}_0, \alpha_i\in \mathbb{F}, x_i\in X\}

If you don't know the thing I mentioned before (if X is a subspace, then span(X)=X), try to prove it. Hint: a subspace is closed under linear combinations...

 

[tylerc1991]

U+U=\{u+v~\vert~u,v\in U\}

I see. So, since u + v \in U for any u,v \in U, \{ u + v : u,v \in U \} = \{ u : u \in U \}? And hence U + U = U?

span(X)=\{\alpha_1x_1+...+\alpha_nx_n~\vert~i\in \mathbb{N}_0, \alpha_i\in \mathbb{F}, x_i\in X\}

I am guessing this would be shown similar to above, only this way is more general. Formal proofs aside, I can definitely see that subspaces are closed under linear combinations. So \{ a_1u_1 + a_2u_2 + \dots + a_nu_n : n \in \mathbb{N}, a_n \in \mathbb{F}, u_n \in U \} = \{ u : u \in U \}?

 

[micromass]

I see. So, since u + v \in U for any u,v \in U, \{ u + v : u,v \in U \} = \{ u : u \in U \}? And hence U + U = U?



I am guessing this would be shown similar to above, only this way is more general. Formal proofs aside, I can definitely see that subspaces are closed under linear combinations. So \{ a_1u_1 + a_2u_2 + \dots + a_nu_n : n \in \mathbb{N}, a_n \in \mathbb{F}, u_n \in U \} = \{ u : u \in U \}?

Yes, that is correct. I may complain however that you've only proved

\{ u + v : u,v \in U \} \subseteq \{ u : u \in U \}

You still may want to show that every u can be written in the form u+v. But this is quite easy, and you probably knew it already. But I just wanted to make it clear...

 

[tylerc1991]

Yes, that is correct. I may complain however that you've only proved

\{ u + v : u,v \in U \} \subseteq \{ u : u \in U \}

You still may want to show that every u can be written in the form u+v. But this is quite easy, and you probably knew it already. But I just wanted to make it clear...

Got it! I will definitely formalize it when I am writing the final answer. Thank you very much for your help!