Span, linear independence problem

[pyroknife]

The problem is attached.

I don't know why he called all 4 vectors V1, I guess it was a typo.

Anyways, part I) This is not linearly independent as the determinant of the matrix containing those 4 vectors is 0

I am having trouble with part II)
I think I know the answer, but I don't think my explanation is right.

I said you can find a subset of vectors that are linearly independent, BUT not have the same span. The span of the original four vectors is c1v1+c2v2+c3v3+c4v4, where c1,c2,c3, c4 are real #s. The subset is independent, thus, the vectors cannot be written as a linear combination, thus, it doesn't have a span?

 

[LCKurtz]

The problem is attached.

I don't know why he called all 4 vectors V1, I guess it was a typo.

Anyways, part I) This is not linearly independent as the determinant of the matrix containing those 4 vectors is 0

I am having trouble with part II)
I think I know the answer, but I don't think my explanation is right.

I said you can find a subset of vectors that are linearly independent, BUT not have the same span.

Of course, you can almost always do that. Just take one of the vectors. Its span will be 1 dimensional and won't span the same space unless it happens to be 1 dimensional in the first place. But you are missing the point.

The span of the original four vectors is c1v1+c2v2+c3v3+c4v4, where c1,c2,c3, c4 are real #s. The subset is independent, thus, the vectors cannot be written as a linear combination, thus, it doesn't have a span?

That last sentence makes no sense. What subset? What vectors can't be written as a linear combination of what? What doesn't have a span?

Look carefully at the first two vectors and see if you notice anything.

 

[pyroknife]

I guess I'ma bit confused about spans and subsets.

The first two vectors are multiples of each other, but I'm not sure what this means.

They're multiples of each other, thus, they're linear combos of each other.

 

[LCKurtz]

I guess I'ma bit confused about spans and subsets.

The first two vectors are multiples of each other, but I'm not sure what this means.

They're multiples of each other, thus, they're linear combos of each other.

So could you drop one of them out and have the same span as the original four vectors?

 

[pyroknife]

So could you drop one of them out and have the same span as the original four vectors?

Original span is c1v1+c2v2+c3v3+c4v4
Dropping either of the 4 vectors will not have the same spam.

 

[LCKurtz]

Original span is c1v1+c2v2+c3v3+c4v4
Dropping either of the 4 vectors will not have the same spam.

Why not?

 

[pyroknife]

If you dropped, say v4
that'd give you a span c1v1+c2v2+c3v3 which isn't equal to c1v1+c2v2+c3v3+c4v4 or is that wrong?

 

[LCKurtz]

I guess I'ma bit confused about spans and subsets.

The first two vectors are multiples of each other, but I'm not sure what this means.

They're multiples of each other, thus, they're linear combos of each other.

So could you drop one of them out and have the same span as the original four vectors?

Original span is c1v1+c2v2+c3v3+c4v4
Dropping either of the 4 vectors will not have the same spam.

If you dropped, say v4
that'd give you a span c1v1+c2v2+c3v3 which isn't equal to c1v1+c2v2+c3v3+c4v4 or is that wrong?

But what if you dropped v2, which you know is a multiple of v1. Couldn't you get everything you could get before by using that multiple of v1 everywhere where you had v2 before?

 

[pyroknife]

Oh I see what you're saying.
So I would pretty much have two options since the 1st 2 vectors are linear combinations of each other.

One of my options is to pick the last 3 vectors.
Another option is I can pick the 1st vector and the last two vectors.
Since v1 is a linear combination of v2 and vise versa, I get drop either one of those, but not both and still have the span of the remaining 3 vectors include v1, v2, v3, v4. So the answer would be yes.

 

[LCKurtz]

That's the idea. Now I haven't worked through the numbers in your problem. If you drop, for example, v1 that doesn't reduce the span. If the remaining 3 vectors are linearly independent you can't drop any more. But if the remaining 3 turn out to still be linearly dependent, you would be able to drop another one, etc. If they were all multiples of each other you could drop all but one of them and keep the same span.

 

[pyroknife]

I think the answer is still no.

If I drop the 1st vector, and put the other 3 vectors into RREF form, a free variable exists thus those 3 aren't linearly independent.

If I drop the 2nd vector, i still get a free variable.

Since the question asks for the subset to still have linearly independent and have the same span, I guess my answer would be no?

 

[LCKurtz]

I think the answer is still no.

If I drop the 1st vector, and put the other 3 vectors into RREF form, a free variable exists thus those 3 aren't linearly independent.

If I drop the 2nd vector, i still get a free variable.

Since the question asks for the subset to still have linearly independent and have the same span, I guess my answer would be no?

I think you should check your arithmetic. If a finite set of vectors is linearly dependent you can always remove one like we did above without changing the span. If what is left is still linearly dependent, you can remove another one. Eventually what is left will be linearly independent at which point you can't remove any without changing the span. The results of your first row reduction should determine the dimension of the space and hence the number of vectors you can remove.

 

[pyroknife]

Ugh I think I got it. I put the original 4 vectors in matrix form and found the RREF. V1 and V3 are the only pivot columns in the RREF form. V2 and V4 are the free variable columns. Thus if I take the vectors V1 and V2, the span of these 2 vectors is the span of the original 4 vectors since V1 and V2 can be written as linearly combinations of V2 and/or V4.