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I have a simple question.

Actually I just tried to solve the question 'Is range of $A$ equal to range of $AA^TA$'.

But it looks like much general question to ask 'Is range of $A$ equal to range of $AB$'.

In my first impression, I think range of $A$ is same as range of $AB$ because no matter which vectors come after, matrix $A$ would linearly transform it to the column space of $A$.

But, it looks wrong.

Can you help me to understand it?

And how can I prove range of $A$ is equal to $AA^TA$?

I am studying power iteration in randomized SVD and it said they are same but I cannot get it.

https://dl.acm.org/doi/pdf/10.1145/3004053?download=true (in page 4)

I guess something like '$Null(A)$ is equal to $Null(A^TA)$'would be helpful, but hard to apply it.

Thank you very much.

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    $\begingroup$ Consider the range of $AB$ if $B=0$. $\endgroup$ – Arthur May 20 at 8:09
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In general, you can say that $$Range(AB) \subseteq Range(A)$$ because if $v\in Range(AB)$, then $$ABx=v\implies A(Bx) = v\implies v\in Range(A).$$ The opposite is generally false, since, as Arthur suggested, if $B=0$, then $AB=0$ that has usually a very different range from the one of $A$.


For your case, here's a simple explanation using SVD. If $A=U\Sigma V$ is the SVD, then (assuming $A$ real) $AA^TA = U\Sigma^3 V$. From the outer expansion $$ A =\sum_i \sigma_i u_iv_i^T $$ you can see that the range of $A$ is the span of $u_1,\dots,u_r$, where $r$ is the rank of $A$, or equivalently the number of non-zero singular values. Since $$ AA^TA = U\Sigma^3 V = \sum_i \sigma_i^3 u_iv_i^T $$ you see that $AA^TA$ has the same rank of $A$, and the range of $AA^TA$ is still the span of $u_1,\dots,u_r$.

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As to the general question, $Range(AB)\subseteq Range(A)$, but $Range(AB)=Range(A)$ if $B$ is square and full rank: $$Range(A)=Range(ABB^{-1})\subseteq Range(AB)\subseteq Range(A)$$

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  • $\begingroup$ @twosigma: Right! Edited. Thanks. $\endgroup$ – Sergio May 20 at 20:36

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