A'A and A have the same null space

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The discussion focuses on proving that the null space of A'A is equivalent to the null space of A. The initial steps involve showing that if A'Ax = 0, then Ax must also equal 0, indicating that non-trivial solutions form a basis for the null space of A'A. The argument is supported by demonstrating that (Ax)'Ax = 0 leads to the conclusion that Ax = 0. There is some uncertainty about the validity of the fourth argument, but it ultimately reinforces the relationship between the null spaces. The proof hinges on understanding the properties of the transpose and the scalar product.
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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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