Something that's been frustrating me

  • Context: Graduate 
  • Thread starter Thread starter Manchot
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the relationship between the nullspace of a matrix A and the nullspace of the product transpose(A)*A, specifically in the context of Singular Value Decomposition in Linear Algebra. Participants explore the implications of this relationship and seek to understand the proof behind it, examining both conceptual and computational aspects.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant notes that while it is clear that any member of N(A) is in N(transpose(A)*A), they struggle to prove the converse.
  • Another participant suggests that the assertion may hold true in real vector spaces but expresses uncertainty regarding more general spaces.
  • A participant proposes a hypothetical scenario involving a non-zero vector in the nullspace of transpose(A)*A but not in the nullspace of A, leading to a contradiction when analyzed.
  • Another participant provides a mathematical manipulation of the equations to further explore the relationship between the nullspaces.
  • A later reply delves into abstract concepts of linear maps and inner products, discussing how these relate to the nullspaces and providing an alternative computational perspective.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement regarding the proof and implications of the relationship between the nullspaces. There is no consensus on the generality of the assertion, and the discussion remains unresolved regarding the broader applicability beyond real vector spaces.

Contextual Notes

Some participants highlight the complexity of the concepts involved, particularly when abstractly composing linear maps and the role of inner products, indicating potential limitations in understanding without further clarification.

Manchot
Messages
470
Reaction score
5
Ok, I was reading the proof for Singular Value Decomposition in my Linear Algebra textbook, when the author made an assertion (without proof). Basically, he said that if A is an m x n matrix, then the nullspace of A is equal to the nullspace of transpose(A)*A.

Now, it's obvious to me that any member of N(A) is in N(transpose(A)*A), since Ax=0 implies that transpose(A)*Ax=0. Nevertheless, I can't prove the converse of this statement to myself. Any tips?
 
Physics news on Phys.org
It seems to be true if the underlying vector spaces are real. I don't know about more general spaces. But here's what I'm thinking. What would happen if there were a column vector [itex]x \neq 0[/itex] that is in the null space of [itex]A^TA[/itex] but not that of A?

[tex]A^TAx = 0[/tex] ,

but [tex]Ax = y \neq 0[/tex]

Well, premultiply the first equation by [itex]x^T[/itex] to get

[tex]x^T A^T A x = y^T y = 0[/tex].

But if y is real vector, this equation implies that y = 0, which contradicts our assumption that it is nonzero.
 
Multiply both sides by [itex]x^T[/itex] so you get:

[tex]x^TA^TAx=0[/tex]

Take it from there.
 
Alright, I thank both of you for the help!
 
This is kind of confusing conceptually.




I.e. abstractly, transpose means "precede by". I.e. a linear map A:V-->W induces a linear map AT:W*-->V*, where W* is linear functioins on W, and if L is such a thing then AT(F) = FoA, a linear function on V.

So if AT(F) = FoA = 0, it means that F vanishes on the image of A, since preceding F by A, gives zero.

But now how do we precede AT by A? i.e. it makes no sense abstractly to compose a map into W with a map out of W*. But this is where an inner product comes in, giving us an isomorphism of W with W* and also of V with V*.


So we compose A:V-->W-->W*-->V*, where the map in the middle takes a vector in W to a functional on W by dotting with that vector.

so if this composition kills v, then it means that "preceding by A", kills "dotting with Av".

I.e. that for every x in V, we have Av.A(x) = 0.

applying this to v gives Av.Av= 0, so Av = 0.

so the point is: the only way that dotting with Av, can kill everything of form Ax, is if Av=0.



it is much easier computationally as follows: AT is the unique map such that for all x,y, we have Ax.y =x.ATy.

hence if ATAv = 0, then Av.A( ) is zero no matter what goes in the blank. putting in v gives Av = 0.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K