Prove A' is the Generalized Inverse of A When A'A is Idempotent

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Discussion Overview

The discussion revolves around the conditions under which the transpose of a matrix \( A \) (denoted \( A' \)) serves as its generalized inverse, specifically when the product \( A'A \) is idempotent. Participants explore the implications of idempotency, properties of projections, and the relationships between the ranks and column spaces of the matrices involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants question whether \( A'A \) being idempotent implies \( A'AA'A = A'A \) and discuss the implications of this property.
  • There is uncertainty about the invertibility of \( A \) and \( A' \), with one participant noting that if they are invertible, one could derive \( AA' = I \).
  • One participant clarifies that \( A' \) refers to the transpose of \( A \) and seeks to understand the meaning of "generalized" inverse in this context.
  • Another participant suggests that the relationship \( A'AA'A = A'A \) leads to the equation \( A'(AA'A - A) = 0 \), indicating a need to show that \( AA'A - A = 0 \).
  • It is noted that showing \( A' \) is the generalized inverse hinges on proving the relationships \( AA'A = A \) and \( A'AA' = A' \), with emphasis on the projection properties of \( AA' \).
  • Participants discuss the rank of \( A'A \) being equal to the rank of \( A \) and the implications for the column space of \( AA' \).
  • There is a suggestion that \( AA' \) acts as a projection onto a subspace related to \( A \), prompting further exploration of how this can be shown equationally.
  • One participant emphasizes the need to clarify that \( A \) is a matrix and not a subspace, and encourages a stronger characterization of the column space.

Areas of Agreement / Disagreement

Participants express various viewpoints regarding the implications of idempotency and the properties of projections, with no consensus reached on the conditions under which \( A' \) is the generalized inverse of \( A \).

Contextual Notes

Participants note the importance of ranks and column spaces in understanding the relationships between the matrices, but do not resolve the mathematical steps or assumptions necessary for a complete proof.

siucw
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generalized inverse of A is equal to A' if A'A is idempotent?
 
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A'A idempotent means A'AA'A=A'A, don't know what you mean by ' though - if A and A' are invertible then you can cancel and get AA'=I but you don't know A and A' are invertible do you?
 
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A' is transpose of A,
A'A idempotent should mean A'AA'A=A'A, right?
 
Yes, realized and corrected myself.

A'A is thus symmetric and idempotent... exactly what do you mean by "generalized" inverse?
 
AA-A=A,A- is generalized inverse, and no assumption about full rank or not for A
 
All I've figured is that:

A'AA'A=A'A => A'(AA'A - A) =0

So it boils down to showing this implies AA'A - A = 0

which must use the fact that A' is the transpose, some how... can't quite explain it, sorry.
 
If you're trying to show A' is the generalized inverse, the symmetric properties are essentially freebies. The hard part is showing AA'A=A and A'AA'=A', but if you've got one you've got the other, so you can concentrate on the first one.

You know AA' is idempotent, so it's a projection, the question is onto what? What can you say about the column space of AA'? What can you say about it's rank? Do you have any guess as to what it should be projecting onto?
 
rank A'A is equal to rank A
 
Good, now what about the column space of AA' (or it's image if you prefer to think of it that way)?
 
  • #10
it is a subspace of A
 
  • #11
so, you can prove it conceptually: since AA' is a projection, AA'A is projecting A to a space which include AA', then...?
is there any way to show it equationally?
 
  • #12
A is a matrix not a subspace. I think you left out the words 'column space'? You should also be able to say something stronger than 'subspace' at this point-remember the bit about the ranks.

A general thing about projections- if T is a projection onto a subspace U, and v is any vector in U then T(v)=v. You should be able to prove this. You should then be able to prove (AA')v=v where v is any column of A.
 
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