Prove $\operatorname{rank}(A^*A)=\operatorname{rank}(A)$

Question

Let $A$ be a linear operator on a finite-dimensional complex inner product space. Then prove that $$\operatorname{rank}(A^*A)=\operatorname{rank}(A)$$ (the * denotes complex conjugate transpose)

My thoughts:

$$0 = ||Ax||^2 = (Ax)^*Ax = \bar x^T(A^* A) x. $$ Hence

$$Ax = 0 \, \text{iff} \, A^* Ax = 0. \tag 1$$

I am not able to continue further. Are my steps correct till now? Any help would be appreciated. Thanks in advance.

Answer 1

You've just shown that the kernels of $A$ and $A^*A$ coincide, now use Rank-Nullity.