Mathematics 21 Online
OpenStudy (anonymous):

) The Gram-Schmidt process produces from a linearly independent set {X1,....,Xn} an orthogonal set {V1,..., Vn} with the property that for every 1 <= k <= n, the vec- tors v1,...,vk span the same subspace as x1,.....,xk. True | False? REASON:

OpenStudy (anonymous):

I feel like they're being tricky here. Certainly \(\{x_1,...x_n\}\) Spans the same subspace as\(\{v_1,...v_n\}\) . But I don't think there is any guarantee that the subsets of those will also span the same subspace.

OpenStudy (anonymous):

Or rather that subsets of those will each span the same subspace as the congruent subset from the other.

OpenStudy (anonymous):

then the answer would be false? because ther can be a situation where they do not span the same subspace?

OpenStudy (anonymous):

I'm trying to think if I can come up with one.

OpenStudy (anonymous):

No, I think they do span the same subspace.

OpenStudy (anonymous):

Although I dislike that they've suggested some sort of ordering in their set notation ;p

OpenStudy (anonymous):

If you think about how gram-shmit works.. the first two vectors will actually be the same so of course they will span the same subspace.

OpenStudy (anonymous):

well let me ask you this, because the gram-schmidt process turned them into orthonormal basises, they have to span the same subspace, because they are basisses

OpenStudy (anonymous):

im pretty sure what u just said was the proper way of saying what i was asking

OpenStudy (anonymous):

Um.. as a whole they will span the same subspace. The question is whether or not you can take smaller subsets and have those smaller subsets span the same subspace.

OpenStudy (anonymous):

So after the first one, you have another new vector that has part in the same span as the first, and part not in that span. So you break it into components and only keep the orthogonal part.

OpenStudy (anonymous):

regardless doesnt it have to be that they are both linear combinations of each other?

OpenStudy (anonymous):

So that new set {x1,x2} will span the same subspace as {v1,v2}, and the process will continue that way.

OpenStudy (anonymous):

It must be that x1 and x2 will both be linear combinations of v1 and v2.

OpenStudy (anonymous):

but if we take x1, x3 and v1, v2? is the dilemma you are presenting? which makes your mind unclear of the exact answer?

OpenStudy (anonymous):

the subspaces then do not match

OpenStudy (anonymous):

I feel as though they precluded that by saying that they take the 'corrisponding' vectors from the set of x's and the set of v's.

OpenStudy (anonymous):

But yes, if they did take x1,x3 and v1,v2 then they would not be the same subspace.

OpenStudy (anonymous):

it's a good question. I am thinking that it's true with the requirement that they be 'corresponding' as I said.

OpenStudy (anonymous):

I think the answer is true then? because they define k to be less than n and both Spaces of V and X go from 1 to k respectively, so they both have dim(k).

OpenStudy (anonymous):

and because of the GS process, they have to span the same subspaces?

OpenStudy (anonymous):

Well the dimensions don't really matter. They can have the same dimension but be different subspaces. But if they are choosing the elements x1, x2 and v1, v2 (which were constructed from x1, x2) then yes. They span the same subspace

OpenStudy (anonymous):

thank you