Linear Algebra showing a subspace

Click For Summary

Homework Help Overview

The problem involves demonstrating that a set U, formed by the sum of vectors from two subspaces U1 and U2 of a vector space V, is itself a subspace of V. The context is linear algebra, specifically focusing on the properties of vector spaces and subspaces.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to show closure under vector addition and scalar multiplication for the set U. There are attempts to clarify what it means for U to be a subspace of V, with some participants suggesting specific properties to demonstrate.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, emphasizing the importance of defining the terms and conditions necessary for U to be a subspace. There is an acknowledgment of the need for the original poster to engage with the problem further.

Contextual Notes

There are indications that some posts may have provided excessive guidance, which has been noted and addressed by participants. The original poster is encouraged to explore the problem independently.

ibkev
Messages
131
Reaction score
61
I'm having trouble getting started on this one and I'd really appreciate some hints. This question comes from Macdonald's Linear and Geometric Algebra book that I'm using for self study, problem 2.2.4.

Homework Statement


Let U1 and U2 be subspaces of a vector space V.
Let U be the set of all vectors of the form u1 + u2, where u1 is a member of U1 and u2 is a member of U2.
Show that U is a subspace of V.

Homework Equations


U inherits it's operations from V (through U1 and U2) and so we have to show that scalar.mult and vec.add are closed for U.
a(bv)=(ab)v
av + bv = (a+b)v

The Attempt at a Solution


Since U1 and U2 are a subset of V, then adding any u1 to u2 may not be closed wrt either U1 or U2 but I think they should be closed wrt V. Intuitively that seems true but I get stuck trying to be more concrete than that.
 
Physics news on Phys.org
It often helps to write down explicitly what you're trying to show. For example, one of the things you want to show is if ##\vec{x},\vec{y} \in U##, then ##\vec{x}+\vec{y}\in U##.

When you say ##\vec{x} \in U##, it means there are vectors ##\vec{x}_1 \in U_1## and ##\vec{x}_2 \in U_2##, such that ##\vec{x} = \vec{x}_1 + \vec{x}_2##. And so on…
 
  • Like
Likes   Reactions: ibkev
ibkev said:
I'm having trouble getting started on this one and I'd really appreciate some hints.

The Attempt at a Solution


Since U1 and U2 are a subset of V, then adding any u1 to u2 may not be closed wrt either U1 or U2 but I think they should be closed wrt V. Intuitively that seems true but I get stuck trying to be more concrete than that.

Can you say what it means for ##U## to be a subspace of ##V##?

Hint: Let ##u, v \in U## and ##a## be a scalar ...

And @vela has given you even more of a helping hand!
 
  • Like
Likes   Reactions: ibkev
We define:

a and b vectors, W1 and W2 subsets of a vectorspace V.

Definition: W1 + W2 = {a+b|a ∈ W1 and b ∈ W2}

Choose 2 vectors a,b element of W1. Choose 2 vectors c,d elements of W2. W1 and W2 are subsets of a vectorspace V.
Then $$a + c ∈ (W1 + W2)$$ and $$b + d ∈ (W1 + W2)$$. Try to continue.

You need to proof that (a+b) + (c+d) ∈ (W1 + W2), this means: a + b ∈ W1 and c + d ∈ W2
 
  • Like
Likes   Reactions: ibkev
Math_QED said:
We define:

a and b vectors, W1 and W2 subsets of a vectorspace V.

Definition: W1 + W2 = {a+b|a ∈ W1 and b ∈ W2}

Choose 2 vectors a,b element of W1. Choose 2 vectors c,d elements of W2. W1 and W2 are subsets of a vectorspace V.
Then $$a + c ∈ (W1 + W2)$$ and $$b + d ∈ (W1 + W2)$$. Try to continue.

You need to proof that (a+b) + (c+d) ∈ (W1 + W2), this means: a + b ∈ W1 and c + d ∈ W2

That's giving too much away. You need to let the OP do the problem himself.
 
What an amazing community.
Thanks everyone for the replies! I'll take it from here
 
PeroK said:
That's giving too much away. You need to let the OP do the problem himself.
So noted, and addressed. I would have deleted the offending post, but the OP has already seen it.
 

Similar threads

Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K