A'A and A have the same null space

[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

 

[Fredrik]

Is A' your notation for the transpose of A?

 

[aanabtawi]

It is, sorry for not making that clear!

 

[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) ).

 

[blue_raver22]

Yes, your proof is valid. You have correctly shown that the null space of A'A can be written as a linear combination of the basis for the null space of A, and vice versa. This means that any vector in the null space of A'A can also be written as a linear combination of the basis for the null space of A, and vice versa. Therefore, the null space of A'A and the null space of A are equivalent and have the same basis, making them the same null space. Good job!