This is my question. How to show that the column space of matrix A is just equal to the column space of AA'?.. A' represents the transpose of A. I know that the column space of AA' is a subset of the column space of A which is just trivial. But the other way around I still used inclusion but it seemed that it is going nowhere. Maybe it can be done by some manipulation or the barbaric way of doing this. I just want to obtain a simple proof. Anyone?
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By the way the size of matrix A here is nxn real matrix. – Aintegral Oct 07 '13 at 16:34
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Hint. If you can show that $A'x=0$ whenever $AA'x=0$, then $\operatorname{nullity}(AA')\le\operatorname{nullity}(A')$ and hence $\operatorname{rank}(AA')\ge\operatorname{rank}(A')=\operatorname{rank}(A)$, i.e. the dimension of the column space of $A$ does not exceed the dimension of the column space of $AA'$.
If you really want a super-duper simple proof, use singular value decomposition.
user1551
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I think this is a more straight forward proof: Since rank(AA')f = rank(A) and trivially column space of AA' is a subspace of A. Therefore they have the same rank implies that the dimension of their column space is equal. Then CL(AA')=CL(A). – Aintegral Oct 08 '13 at 03:14
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@JCAintegral Yes, the proof would be easy if you know that rank(AA')=rank(A), provided that this equality has been proved independently (so that you won't end up with a cyclic proof). – user1551 Oct 09 '13 at 15:04
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