If $A$ is a compact operator, then the rank of $A^*A$ is equal to the rank of $A$.

Question

Let $A$ be a compact operator on a Hilbert space and $r(A)$ be the rank of $A$, that is, the dimension of the image of $A$.

The inequality $r(A^*A)\le r(A^*)=r(A)$ is rather straightforward, but how do I get the other direction?

Answer 1

Note that $$ \langle x,A^\ast Ax\rangle=\langle Ax,Ax\rangle $$ Thus, if $Ax\ne0$, then $A^\ast Ax\ne0$. This implies that $r(A^\ast A)\ge r(A)$.