The title pretty much says it all, I need to prove that $\text{rank}(A) = \text{rank}(A^T)$ using SVD. It seems quite trivial, but I'd like to hear a second opinion. My thoughts are exposed below.
Based on the SVD theorem, for a matrrix $A \in \mathbb{R}^{m \times n}$ there exist othogonal matrices $U \in \mathbb{R}^{m \times m}$ and $V \in \mathbb{R}^{n \times n}$ such that $$ U^TAV = \text{diag}(\sigma_1, \sigma_2, ..., \sigma_p), \quad p = \min\left\{m, n\right\} $$
and $\sigma_i$ are the singular values of $A$. If it holds that $\sigma_1 \geq \sigma_2 \geq ~ ... ~ \geq \sigma_r > \sigma_{r+1} = ~ ... = \sigma_p = 0$, then it is true that $\text{rank}(A) = r$, meaning that the rank of $A$ is the count of non-zero singular values.
Trivially, if we consider that that $$ U^TAV = \text{diag}(\sigma_1, \sigma_2, \dots, \sigma_p) = \text{diag}(\sigma_1, \sigma_2, ..., \sigma_p)^T = V^TA^TU $$
it is obvious that the singular values of $A$ are also singular values of $A^T$, hence the count of non-zero singular values for both coincide. Bottom line is that $\text{rank}(A) = \text{rank}(A^T)$.
I think it would be OK, but I might be missing something, so please correct me if so. Thank you in advance for your time! :)