# How to proove this

generalized inverse of A is equal to A' if A'A is idempotent?

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matt grime
Homework Helper
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?

matt grime
Homework Helper
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

matt grime
Homework Helper
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.

shmoe
Homework Helper
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

shmoe
Homework Helper
Good, now what about the column space of AA' (or it's image if you prefer to think of it that way)?

it is a subspace of A

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?

shmoe