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quackensack

Member

Registered: 2007-02-27

Posts: 47

Linear algebra problem

This proof was confusing me and I have a test next Thursday on this type of thing, so if anyone could explain how this is done, I would really appreciate it!  Thanks in advance!

Let

be a vector space and let

and 

be subspaces of 

.  Prove that

is a subspace of 

.  (Do not forget to show that

is nonempty.)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra problem

Just go through the properties one by one.  Many you won't have to do.  For example, being associative in V automatically means being associative in any subset of V, and the intersection of two subspaces is at least a subset.

So for example, to be a vector space, it must contain the 0 (I say "the" because you can prove that 0 is unique.  i.e. there aren't two or more 0's).  So we need to show that 0 is in the intersection.  But part of being a subspace means that you have to contain the 0.  So each subspace must contain the 0, and the intersection must therefore contain the 0.

Now lets try it for a + b.  Let a and b be vectors in the intersection.  This means that a is in W1 and a is in W2, and b is in W1 and b is in W2.  So this means that a + b has to be in W1, as W1 must be closed.  Same for W2.  So a + b is in W1 and a + b is in W2.  So a + b must be in the intersection.

All the rest are excruciatingly similar.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

quackensack

Member

Registered: 2007-02-27

Posts: 47

Re: Linear algebra problem

Just go through the properties one by one.  Many you won't have to do.  For example, being associative in V automatically means being associative in any subset of V, and the intersection of two subspaces is at least a subset.

So for example, to be a vector space, it must contain the 0 (I say "the" because you can prove that 0 is unique.  i.e. there aren't two or more 0's).  So we need to show that 0 is in the intersection.  But part of being a subspace means that you have to contain the 0.  So each subspace must contain the 0, and the intersection must therefore contain the 0.

Now lets try it for a + b.  Let a and b be vectors in the intersection.  This means that a is in W1 and a is in W2, and b is in W1 and b is in W2.  So this means that a + b has to be in W1, as W1 must be closed.  Same for W2.  So a + b is in W1 and a + b is in W2.  So a + b must be in the intersection.

All the rest are excruciatingly similar.

Thank you!

So from what you said, I understand that I also have to prove it for scalar multiplication.  So would it be right if I were to say:

Let a be a vector in the intersection.  This means that a is in W1 and a is in W2.  Therefore, a times a scalar x will be in W1 as well as W2 by the closure property of scalar multiplication.  So a*x must be in the intersection.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra problem

Yep.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

quackensack

Member

Registered: 2007-02-27

Posts: 47

Re: Linear algebra problem

Thanks!  I was wondering...part of the problem says to show that the intersection of W1 and W2 is nonempty.  Did I already prove this by saying that ax and a+b are in the intersection, or is there more to it than that?

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Linear algebra problem

It's nonempty because as subspaces, 0 must be in both W[sub]1[/sub] and W[sub]2[/sub], and hence 0 ∈ W[sub]1[/sub] ∩ W[sub]2[/sub].

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra problem

Did I already prove this by saying that ax and a+b are in the intersection, or is there more to it than that?

No, because for a + b to be in the intersection, there has to be stuff in the intersection.  That is, both a and b must be in there.  What you're saying is "If a and b are in the intersection".  This first assumes the intersection is non-empty.  You must show that is true.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."