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I have a linear algebra question /sci/

Anonymous Mon 14 Jun 16:37:37 2021 No.13277618 View Reply Original Report

Quoted By: >>13277648 >>13277760 >>13277762

If the null space is directly dependent upon all the other vectors which are in the rank of the matrix, then how is this vector {alpha_1,.........,alpha_n} linearly independent? Like, if you take AX=0 and A is an mxn matrix with r<m<n, then shouldn't the n-r matrices be linearly dependent upon the r matrices for them to be in the null space?

Anonymous

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Anonymous Mon 14 Jun 2021 16:46:32 No.13277648 Report

Quoted By: >>13277757

>>13277618
Losing my mind over this please help

Anonymous

Anonymous Mon 14 Jun 2021 17:32:28 No.13277757 Report

Quoted By: >>13277814

>>13277648
Stop crying and state your question properly. What exactly about the proof don't you understand? The stuff you wrote above makes little sense.

Anonymous

Anonymous Mon 14 Jun 2021 17:34:40 No.13277760 Report

Quoted By:

>>13277618
>If the null space is directly dependent upon all the other vectors which are in the rank of the matrix,
>other vectors which are in the rank of the matrix,
Do you mean range? What you typed doesn't really make sense.

Anonymous

Anonymous Mon 14 Jun 2021 17:35:41 No.13277762 Report

Quoted By:

>>13277618
>then how is this vector {alpha_1,.........,alpha_n} linearly independent?
That's not a vector. It is a list of vectors. Those vectors were picked to be a basis for the VS, so they are LI.

Anonymous

Anonymous Mon 14 Jun 2021 17:47:28 No.13277781 Report

Quoted By: >>13277820

Also, I think the main thing you need to grok in order to understand this proof is that in a finite dimensional VS any LI set can be extened to a basis by (possibly) adding vectors. I'm sure this book has a proof of that, so go back and learn that proof. A sister theorem to that is that every spanning set can be reduced to a basis by (possibly) deleting vectors. That is another theorem you should understand.

Another thing, maybe practice showing sets are/are not LI with some concrete examples. Might help with your intuition.

Anonymous

Anonymous Mon 14 Jun 2021 17:59:42 No.13277814 Report

Quoted By: >>13278474

>>13277757
There is an n-tuple: {alpha_1,......,alpha_n} which forms the basis of V here. The author says that {alpha_1,.........,alpha_k} is the basis of the null space. The proof asserts that {alpha_k+1,.......,alpha_n} is the basis of the range of T. I remember studying that the dimension of the null space is n-r, because that's exactly the number of free variables in A when we solve AX=0. This implies that those n-r rows are linear combinations of the r rows. So if take this n-tuple as a matrix, then shouldn't {alpha_k+1,.....,alpha_n} be dependent upon {alpha_1,......,alpha_k}?

Anonymous

Anonymous Mon 14 Jun 2021 18:00:56 No.13277820 Report

Quoted By:

>>13277781
Yes, I do know about that theorem, but my problem still doesn't make sense in the context of that theorem

Anonymous

Anonymous Mon 14 Jun 2021 22:02:27 No.13278474 Report

Quoted By: >>13280215

>>13277814
Anon, what you are saying about the matrix is true but relies on the theorem you're trying to understand, and which is more important to understand.

First, it is not {alpha_k+1,.......,alpha_n} which is claimed as the basis of range(T), but {Talpha_k+1,.......,Talpha_n}. It is easy to show that Talpha_i span range(T), the first k are 0, and you fill in the rest of the proof.

Next, you need to show {Talpha_k+1,.......,Talpha_n} is a basis for range(T). Follow the argument for that, and you get \sum c_ialpha_i is in null(T). Since you picked alpha_i to be LI (coz it is a basis), this is only possible if all c_i=0, or that the set is LI.

From there, you should be able to get to the conclusion. Hoffman and Kunze is based, keep at it and you'll understand eventually.

Anonymous

Anonymous Tue 15 Jun 2021 10:14:42 No.13280215 Report

Quoted By: >>13280257

>>13278474
I finally do, I just misunderstood the difference between range and the rank of T

Anonymous

Anonymous Tue 15 Jun 2021 10:39:54 No.13280257 Report

Quoted By:

>>13280215
The range is the image of the matrix when viewed as a linear map.
The rank is the dimension of the image.

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