[Math] Let A be an $m\times n$ matrix. Prove that $\operatorname{rank}(AA^T) = \operatorname{rank}(A)$.

linear algebramatrix-ranktranspose

  1. Let $A$ be an $m\times n$ matrix. Prove that $\operatorname{rank}(AA^T) = \operatorname{rank}(A)$.

The problem tells me to prove it with the theorem that $\operatorname{rank}(A^TA) = \operatorname{rank}(A)$.

I'm a bit lost here…$AA^T$ and $(A)$ don't even have the same number of columns. I'm thinking maybe to prove it by showing that $[m – \operatorname{nullity}(AA^T)] = [n – \operatorname{nullity}(A)],$ but then I'm stuck here.

  1. Let A be an $m\times n$ matrix. Prove that the column space and row space of $A^TA$ are the same.

The problem tells me to prove it also with the theorem – $\operatorname{rank}(A^TA) = \operatorname{rank}(A)$. But I'm really running out of ideas.

Help?

Best Answer

$\newcommand{\rank}{\operatorname{rank}}$Considering $A$ as a linear map, $\operatorname{Im}(AA^T)\subset \operatorname{Im}(A)$ implies that $\rank(AA^T)\leq \rank(A)$.

Consider the canonical scalar product associated to the basis used to write the matrix. Then, $AA^T(x)=0$ implies that $\langle AA^T(x),x\rangle=\langle A^T(x),A^T(x)\rangle=0$ implies that $A^T(x)=0$. this implies that $\ker(AA^T)\subset \ker(A^T)$. since $\rank(A^T)=\rank(A)$ and $\rank(A)+\dim\ker(A)=n$, we deduce that $\rank(AA^T)=\rank(A)$.

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