I am pondering over the question:
Whether $\mathrm{rank}\mathbf{AA^T}\stackrel{?}{=}\mathrm{rank}\mathbf{A^TA}\stackrel{?}{=}\mathrm{rank}\mathbf{A}$ for any matrix $\mathbf{A}\in M_{m\times n}$
I looked through these two answers: Gram Matrix Rank and Prove rank$(A^TA)$=rank$(A)$ for any $A\in M_{m\times n}$.
In the second proof of Gram Matrix Rank, and this proof if we replace $A$ with the $A^T$, we get $$\mathrm{rank}\mathbf{A^T}=\mathrm{rank}AA^T$$ But $\mathrm{rank }A^T=\mathrm{rank}A$. So, we have the following:$$\mathrm{rank}A=\mathrm{rank}A^T=\mathrm{rank}AA^T=\mathrm{rank}A^TA\enspace\enspace\cdots (1)$$
Since the proofs were done in the above are independent of the entries of the matrix. So, consider$$B^T=\begin{bmatrix}1& i\end{bmatrix}$$ where $i^2=-1$.
For the matrix $B$ the result in $(1)$ fails to hold.
I am unable to understand is there a problem with those proofs or I am missing something? Thank you in advance.
Well, in my eyes, the most useful tool to analyze this type of questions, in the non-square matrix case, is the SVD decomposition.
Let us decompose our matrix: $A=U\Sigma V^\star$, then the rank of $A$ is equal to the amount of non zero entries in $\Sigma$. However, each non zero entry is the square of an eigenvalue of $A A^\star$. Thus yielding a perfect correspondence between non zero singular values of $A$ to non zero eigenvalues of $AA^\star$ and $A^\star A$ (Which are actually the same). Hence, the ranks are indeed equal.