Is $ \operatorname{rank}A =\operatorname{rank} A^T$?

Question

Assume A is an $m\times n$ matrix with real-valued entries. Is it always true that $\operatorname{rank} A = \dim \operatorname{Col} A = \dim\operatorname{Row} A = \dim \operatorname{Col} A^T = \operatorname{rank} A^T$?

Answer 1

Yes, it's true, you can prove it using the fact that the range of $A^T$ (denoted here by $\text{Im}(A^T)$ is the orthogonal complement of the null space of $A$ (denoted here by $\text{null}(A))$. This tells us that \begin{equation} \text{dim null}(A) + \text{dim Im}(A^T) = n. \end{equation}

From the rank-nullity theorem, we have: \begin{equation} \text{dim null}(A) + \text{dim Im}(A) = n. \end{equation}

Comparing these two equations, we see that $\text{dim Im}(A) = \text{dim Im}(A^T)$.