Linear Algebra: Subspace Sum

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Homework Statement



Suppose [itex]U[/itex] is a subspace of [itex]V[/itex]. What is [itex]U + U[/itex]?

Homework Equations



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

The Attempt at a Solution



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

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

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

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?
 

Answers and Replies

  • #2
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Hi tylerc! :smile:

Homework Statement



Suppose [itex]U[/itex] is a subspace of [itex]V[/itex]. What is [itex]U + U[/itex]?

Homework Equations



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

The Attempt at a Solution



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

Yes, U+U=U holds in general!

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

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

[tex]U+U=\{u+v~\vert~u,v\in U\}[/tex]

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:
[itex]U + U = \text{ span} (U \, \bigcup \, U) = \text{ span}(U) = \{ au : a \in \mathbb{F}, u \in U \}[/itex].

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

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

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...
 
  • #3
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[tex]U+U=\{u+v~\vert~u,v\in U\}[/tex]

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

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

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 [itex]\{ 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 \}[/itex]?
 
  • #4
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3,297
I see. So, since [itex]u + v \in U[/itex] for any [itex]u,v \in U[/itex], [itex]\{ u + v : u,v \in U \} = \{ u : u \in U \}[/itex]? And hence [itex]U + U = U[/itex]?



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 [itex]\{ 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 \}[/itex]?

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

[tex]\{ u + v : u,v \in U \} \subseteq \{ u : u \in U \}[/tex]

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...
 
  • #5
166
0
Yes, that is correct. I may complain however that you've only proved

[tex]\{ u + v : u,v \in U \} \subseteq \{ u : u \in U \}[/tex]

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!
 

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