Help with linear algebra

[Durandal]

I have a proof to do for my linear algebra course, and it involves the use of a concept which has never been discussed in class, the concept of two subspaces intersecting each other. This is the proof.

Prove that if V and W are 3-dimensional subspaces of R^5, then V (intersecting with) W =/= {0} (the zero subspace).

The (intersecting with) is represented with an upside-down U without a tail.

The hint that my professor gave us on this one was to consider the bases of each subspace, so I've defined A and B as the bases for V and W, respectively, such that

A = {u1, u2, u3}
and
B = {v1, v2, v3}.

In order for these subspaces to intersect, they must share a common vector (that much is obvious), but I can't figure out how to represent their intersection as a single set so that I can prove it isn't the zero subspace. Should I represent that common vector as a spanning set? How can I prove that the common vector isn't the zero vector?

Any help is much appreciated.

[Nova Andromeda]

Maybe this might be of help...

R^5 = {u1, u2, u3, u4, u5}
A = {u1, u2, u3} or {u1, u2, u4} or {u1, u2, u5} or ...
B = (same as A)

-When you do this you will see that no matter which base set you choose V and W will have a common vector.
-You might also subtract V from R^5 and show that W is larger than the resulting R^2 subspace (meaning you have subtracted away part of subspace W when you subtracted V from R^5).

[Durandal]

Maybe this might be of help...

R^5 = {u1, u2, u3, u4, u5}
A = {u1, u2, u3} or {u1, u2, u4} or {u1, u2, u5} or ...
B = (same as A)

-When you do this you will see that no matter which base set you choose V and W will have a common vector.

That helps out a lot. Thank you.

-You might also subtract V from R^5 and show that W is larger than the resulting R^2 subspace (meaning you have subtracted away part of subspace W when you subtracted V from R^5).

I don't know how that'd help. What I really need to do is represent V (intersecting with) W as one subspace.

[Glass Pearl Player]

Yes, linear algebra, the warmup round for anyone who wishes to study mathematics at an university! Say, did you prove the following theorem somewhen, somewhere?

Let V, W be subspaces of a vector space U. Let X denote the intersection of V and W, and Y the sum (read: all vectors x that can be written as a+b, a in V, b in W) of them. Then I claim that:
dim Y = dim V + dim W - dim X

Proof:
Let n=dim X, k=dim V, l=dim W.
Let x={x1,x2,...,xn} be a base set of X. We complete it to a base set of V respectively W, denoted by v={x1...xn,v1...vk} and w={x1...xn, w1...wl}. The rest is basically the exercise you are asking help for, and should be found in any decent textbook on that topic.

In your case, we know that dim Y <= 5, dim V = dim W = 3, so we get:

3 + 3 - dim X <=5, or
dim X >= 1

But this obviously implies that X cannot be the Nullspace.

Remark 1: Proving the theorem above needs the existence theorem for base sets and the theorem that a base set of a subspace can be completed to a base of the whole space. You did see them in your course, didn't you?
Remark 2: Yes, I am studying mathematics at an university.
Remark 3: As the proof shows, the trick is to start with a base set of X. Well, your teacher said "consider the set of each subspace".
Remark 3.14159.... : Finite dimensional spaces are dull and boring. Go study Hibert spaces

@Nova Andromeda: You had the Correct Idea(tm), but you simply can't assume that you can pick such nice base sets for V and W. It might be that u1...u5 are not even contained in V and W..

[Durandal]

Yes, linear algebra, the warmup round for anyone who wishes to study mathematics at an university! Say, did you prove the following theorem somewhen, somewhere?

Physics, actually.

Let V, W be subspaces of a vector space U. Let X denote the intersection of V and W, and Y the sum (read: all vectors x that can be written as a+b, a in V, b in W) of them. Then I claim that:
dim Y = dim V + dim W - dim X

Proof:
Let n=dim X, k=dim V, l=dim W.
Let x={x1,x2,...,xn} be a base set of X. We complete it to a base set of V respectively W, denoted by v={x1...xn,v1...vk} and w={x1...xn, w1...wl}. The rest is basically the exercise you are asking help for, and should be found in any decent textbook on that topic.

In your case, we know that dim Y <= 5, dim V = dim W = 3, so we get:

3 + 3 - dim X <=5, or
dim X >= 1

But this obviously implies that X cannot be the Nullspace.

I've never seen that before.

Remark 1: Proving the theorem above needs the existence theorem for base sets and the theorem that a base set of a subspace can be completed to a base of the whole space. You did see them in your course, didn't you?

Yup. That would be the extension principle.

Remark 2: Yes, I am studying mathematics at an university.

I don't believe you.

Remark 3: As the proof shows, the trick is to start with a base set of X. Well, your teacher said "consider the set of each subspace".

That's what I inferred from it.

Remark 3.14159.... : Finite dimensional spaces are dull and boring. Go study Hibert spaces

I've heard that most physics majors end up doing horribly in linear algebra. Personally, I just can't wait to get to differential equations ... back to stuff that makes sense to me.

[Colonel Olrik]

...

Blergh..

I prefer real math. Palpable math. Physical math. First years algebra/math anaylsis is just.. blergh..

I'm glad I won't need to look at this stuff, never again. I have a degree that says so.

[Kuroneko]

@Nova Andromeda: You had the Correct Idea(tm), but you simply can't assume that you can pick such nice base sets for V and W. It might be that u1...u5 are not even contained in V and W..

I believe his method can be made justified in the following way. Let the bases be V, W be A = {v1, v2, v3} and B = {w1, w2, w3}, respectively. Then A U B spans a subspace of R^5, and since it has six members, at least one of them must be linearly dependent on the others. Remove it, at call the resulting set C = {u1, u2, u3, u4, u5}. It must the the case that V and W are subspaces of span(C), and hence their "new" bases can be chosen from C.

Nova Andromeda's combinatorial argument then applies, though we are not guaranteed that span(C) = R^5 as Nova assumes, that itself is not necessary.

[Queeb Salaron]

Am I alone when I say that I have NO FUCKING CLUE what's going on right now?

[Durandal]

@Nova Andromeda: You had the Correct Idea(tm), but you simply can't assume that you can pick such nice base sets for V and W. It might be that u1...u5 are not even contained in V and W..

I believe his method can be made justified in the following way. Let the bases be V, W be A = {v1, v2, v3} and B = {w1, w2, w3}, respectively. Then A U B spans a subspace of R^5, and since it has six members, at least one of them must be linearly dependent on the others. Remove it, at call the resulting set C = {u1, u2, u3, u4, u5}. It must the the case that V and W are subspaces of span(C), and hence their "new" bases can be chosen from C.

Nova Andromeda's combinatorial argument then applies, though we are not guaranteed that span(C) = R^5 as Nova assumes, that itself is not necessary.

Actually, if we take R^5 to be a subspace of itself and C to be a subspace contained within R^5, then we can infer that C = R^5 because they have the same dimension.

[Kuroneko]

Actually, if we take R^5 to be a subspace of itself and C to be a subspace contained within R^5, then we can infer that C = R^5 because they have the same dimension.

I assume you mean "span(C)" instead of "C" above. But in any case, that does not follow from the construction of C. For example, consider the case V = W. Then span(C) = V = W. The point is, we don't know anything about C except that span(C) is at least three-dimensional and at most five-dimensional. Fortunately, that's all we need.

[Durandal]

Actually, if we take R^5 to be a subspace of itself and C to be a subspace contained within R^5, then we can infer that C = R^5 because they have the same dimension.

I assume you mean "span(C)" instead of "C" above. But in any case, that does not follow from the construction of C. For example, consider the case V = W. Then span(C) = V = W. The point is, we don't know anything about C except it span(C) is at least three-dimensional and at most five-dimensional. Fortunately, that's all we need.

Nope, I meant the set C. It's a theorem regarding subspaces within subspaces.

[Kuroneko]

I prefer real math. Palpable math. Physical math. First years algebra/math anaylsis is just.. blergh..

I still contend the pure mathematics is the "real" mathematics, the line between pure and applied is very blurry. I for one was suprised that algebra (of the general kind) has applications in physics.

[Durandal]

I just thought this over. When you're constructing the set C, at least one of the vectors in that set must be linearly dependent upon the others, but no more than three. Does that sound correct?

[Kuroneko]

Nope, I meant the set C.

C is a set with five vectors. It is not a subspace. Its span is a subspace.

It's a theorem regarding subspaces within subspaces.

The construction of C does not guarantee that all five vectors in it are independent. span(C) = R^5 if, and only if, they are all linearly independent.

[Kuroneko]

I just thought this over. When you're constructing the set C, at least one of the vectors in that set must be linearly dependent upon the others, but no more than three. Does that sound correct?

Yes. By combining the bases for V, W, we can eliminate one of the vectors to form C, since there can be at most five. span(C) is a superspace of both V and W, and hence we can choose a new basis for V, and one for W exclusively from C. Nova's argument then follows, except that we don't know if C is a basis for R^5.

[Durandal]

I just thought this over. When you're constructing the set C, at least one of the vectors in that set must be linearly dependent upon the others, but no more than three. Does that sound correct?

Yes. By combining the bases for V, W, we can eliminate one of the vectors to form C, since there can be at most five. span(C) is a superspace of both V and W, and hence we can choose a new basis for V, and one for W exclusively from C. Nova's argument then follows, except that we don't know if C is a basis for R^5.

Hm ... I think I see your point now. I'm lead back to the original question of how the Hell do I do my assignment.

[Kuroneko]

Here it is in excruciating detail... It should be noted that applying the theorem dim(VW) = dimV + dimW - dim(V+W) as Glass Pearl Player pointed out is a lot simpler, but you implied that you've not encountered it yet.

Code: Select all

A = {u1,u2,u3}; V = span(A)
B = {u4,u5,u6}; W = span(B)
span(A U B) is at most five-dimensional. Without loss of generality, assume
 u6 is linearly dependent on the rest.
C = {u1,u2,u3,u4,u5};  u6 = c1u1+...+c5u5, c's _fixed_

      V = span{u1,u2,u3}
      W = span{u4,u5,c1u1+...+c5u5}
        = a1 u4 + a2 u5 + a3( c1 u1 + ... + c5 u5 )
        = (a1+a3c4)u4 + (a2+a3c5)u5 + a3( c1u1 + c2u2 + c3u3 )
      W contains z = c1u1+c2u2+c3u3
        if z = 0, then u6 = c4u4 + c5u5, then W is not three-dimensional!
      V contains z != 0, being a combination of u1,u2,u3 !
       --> z != 0 is in the intersection of V, W

[Durandal]

OK, I got it (I saw your post after I finished this, Kuroneko, but feel free to check my work). The assignment is posted here:

http://www.ilstu.edu/~dsorres/Homework10.pdf

[Kuroneko]

The fact that at least one vector in one set is linearly dependent upon a vector from the other set tells us that the two sets have an intersection space V ⋂ W.

If I was the teacher, that just wouldn't fly. It would follow if a vector in one set was linearly dependent on vectors exclusively from the other set, but that is simply false.

[Durandal]

The fact that at least one vector in one set is linearly dependent upon a vector from the other set tells us that the two sets have an intersection space V U W.

If I was the teacher, that just wouldn't fly. It would follow if a vector in one set was linearly dependent on vectors exclusively from the other set, but that is simply false.

But that has to be true, because the sets A and B are bases, and thus composed of all linearly independent vectors. So, if a vector in one set is going to be linearly dependent upon anything, it has to be a vector in the other set.

[Kuroneko]

But that has to be true, because the sets A and B are bases, and thus composed of all linearly independent vectors. So, if a vector in one set is going to be linearly dependent upon anything, it has to be a vector in the other set.

No. Just because, say, v3 is linearly independent of {v1,v2} but is linearly dependent on {u1,u2,u3,v1,v2} does not make it dependent on {u1,u2,u3}.

For example, in three dimensions, let u1 = [1 0 0], u2 = [0 1 0], v1 = [0 0 1], v2 = [0 1 1]. v2 is obviously dependent on {u1,u2,v1}, since v2 = u2 + v1. However, remove v1 from the equation and {u1,u2,v2} become independent. The point is that v2 is does not have to be dependent on anything if v1 is not present. In the case of five dimensions, the same holds true: you can't ignore the other vectors that come from the same set. If you do, you are no longer guaranteed linear dependence.

[Durandal]

But that has to be true, because the sets A and B are bases, and thus composed of all linearly independent vectors. So, if a vector in one set is going to be linearly dependent upon anything, it has to be a vector in the other set.

No. Just because, say, v3 is linearly independent of {v1,v2} but is linearly dependent on {u1,u2,u3,v1,v2} does not make it dependent on {u1,u2,u3}.

For example, in three dimensions, let u1 = [1 0 0], u2 = [0 1 0], v1 = [0 0 1], v2 = [0 1 1]. v2 is obviously dependent on {u1,u2,v1}, since v2 = u2 + v1. However, remove v1 from the equation and {u1,u2,v2} become independent. The point is that v2 is does not have to be dependent on anything if v1 is not present. In the case of five dimensions, the same holds true: you can't ignore the other vectors that come from the same set. If you do, you are no longer guaranteed linear dependence.

Fuck.

[Kuroneko]

Fuck.

I probably wouldn't put it that way, but I understand the sentiment.

Regardless, from the linear dependence of one of the vectors, it is possible to identify a nonzero vector that belongs to both subpaces: if u6 = c1u1+c2u2+c3u3+c4u4+c5u5, then z = c1u1+c2u2+c3u3 is nonzero and belongs to both V and W, and hence is in the intersection.

[Durandal]

Fuck.

I probably wouldn't put it that way, but I understand the sentiment.

Regardless, from the linear dependence of one of the vectors, it is possible to identify a nonzero vector that belongs to both subpaces: if u6 = c1u1+c2u2+c3u3+c4u4+c5u5, then z = c1u1+c2u2+c3u3 is nonzero and belongs to both V and W, and hence is in the intersection.

So we'd be able to infer a minimum dimension of 1 for the intersection space, then, right? That would be enough to show that it isn't the zero subspace.

[Kuroneko]

So we'd be able to infer a minimum dimension of 1 for the intersection space, then, right? That would be enough to show that it isn't the zero subspace.

Correct, though really you don't even have to bring in how many dimensions the intersection has to answer the problem as stated. Just the presence of z in the intersection and the fact that z is not the zero vector proves that V∩W is not the zero subspace.