A'A and A have the same null space

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Discussion Overview

The discussion revolves around proving that the null space of the matrix product A'A is equivalent to the null space of the matrix A. Participants are exploring the validity of a proof and examining the implications of certain mathematical steps related to linear algebra concepts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that the null space of A'A can be shown to be the same as the null space of A through a series of steps, questioning the validity of the fourth step in their argument.
  • Another participant seeks clarification on the notation used, specifically whether A' denotes the transpose of A.
  • A later reply provides an alternative perspective on the argument, suggesting that if Ax = 0, then it leads to the conclusion that ||Ax||^2 = 0, which implies Ax = 0.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the proof presented. There are differing views on the reasoning behind certain steps, particularly the fourth step, and the discussion remains unresolved.

Contextual Notes

Some assumptions about the properties of the matrices and the definitions of the null space may not be explicitly stated, and the implications of the scalar product and induced norm are introduced but not fully explored.

aanabtawi
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I'm trying to prove that the null space of A'A is the null space of A, this is what I have so far,

1) A'Ax=0, non trivial solutions are a basis for the null space of A'A

2) x'A'Ax=0

3) (Ax)'Ax=0

4) Since (Ax)'A is a linear combination of the col's of A, we see that the null space of A can be written as a linear combination of the basis for the null space of A'A.

Therefore, they have the same null space.

--> Is this proof valid? I am unsure if argument 4 holds ground, but it seems to make sense to me =P
 
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Is A' your notation for the transpose of A?
 
It is, sorry for not making that clear!
 
didnt think throguh your 4), but this is what you can do once you get to 3):

Ax= 0 => (Ax)'Ax = (Ax|Ax) = ||Ax||^2 = 0 <=> Ax = 0

where (,|,) denotes the scalar product and ||.|| is the norm induced by the scalar product ( ||u|| = sqrt(u|u) ).
 

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