Rank of $A$ and $A^{\theta}A$ are same

Question

The original question was If $A$ is a real $m \times n$ matrix then show that rank of $A$ and $A^TA$ are same. The proof follows by showing that nullspace of both $A$ and $A^TA$ are same which is given as below: $x\in N(A)\Rightarrow x\in N(A^TA)$ \begin{align} x\in N(A^TA) &\Rightarrow A^TA(x)=0 \\ &\Rightarrow \langle x,A^TAx\rangle =0 \\ &\Rightarrow \langle Ax,Ax\rangle =0 \\ &\Rightarrow Ax=0 \\ &\Rightarrow x\in N(A). \end{align} Now with the same argument can we show that rank of $A$ and $A^{\theta}A$ are same where $A$ in a complex matrix? Here $\theta$ means conjugate transpose

Answer 1

Hint: show that in the complex case $\langle Au,v \rangle _m = \langle u,A^\theta v \rangle _n$ and repeat the argument from the real case using Rank Nullity.