# Is range of A equal to range of AB?

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.

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

Thank you very much.

• Consider the range of $AB$ if $B=0$. – Arthur May 20 at 8:09

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 $$u_1,\dots,u_r$$.
As to the general question, $$Range(AB)\subseteq Range(A)$$, but $$Range(AB)=Range(A)$$ if $$B$$ is square and full rank: $$Range(A)=Range(ABB^{-1})\subseteq Range(AB)\subseteq Range(A)$$