A'A and A have the same null space

1. Oct 28, 2012

aanabtawi

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

2. Oct 29, 2012

Fredrik

Staff Emeritus
Is A' your notation for the transpose of A?

3. Oct 29, 2012

aanabtawi

It is, sorry for not making that clear!

4. Oct 31, 2012

Jomenvisst

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) ).