Does the dimension of the row space equal dimension of the column space for complex matrices? In the case of real matrices, the dimension of the row space equals the dimension of the column space:  $\text{dim } \text{span}(A) = \text{dim }\text{span}(A^T)$. But for complex matrices the transpose gets replaced by the conjugate transpose so $$\text{dim } \text{span}(A) = \text{dim }\text{span}(A^H).$$
Nonetheless, I'm wondering if for complex matrices it is also true that $\text{dim } \text{span}(A) = \text{dim }\text{span}(A^T)$. I.e., if we use the regular transpose instead of the conjugate transpose do we still have equality?
For this to be true, it would also have to be that $\text{dim }\text{span}(A) = \text{dim }\text{span}(\bar{A})$. In other words, that the elementwise complex conjugate operation preserves rank. Is this true?
Thanks.
 A: For the proof of $\dim \text{span}(A)=\dim \text{span} (\overline{A})$, we first consider of basis $\{{v_1}, \cdots, {v_k}\}$ of $\text{span}(A)$. Since $\text{span}(A) = \text{span}(v_1, \cdots, v_k)$, we have $\text{span}(\overline{A}) = \text{span}(\overline{v_1}, \cdots, \overline{v_k})$. If there exist complex numbers $c_1, \cdots, c_k$ such that $c_1 \overline{v_1} + \cdots + c_k \overline{v_k} = 0$, then we can consider the conjugate of each side to obtain $\overline{c_1} v_1 + \cdots + \overline{c_k} v_k = 0$. Becuase $v_1, \cdots, v_k$ is linearly independent in $\mathbb{C}$, we have $c_1 =\cdots = c_k = 0$. Hence $\overline{v_1}, \cdots , \overline{v_k}$ is also linearly independent. Thus $\{\overline{v_1}, \cdots, \overline{v_k} \}$ is a basis and the dimension of $\text{span}(A)$ is also $k$. 
Actually the proof of $\dim \text{span}(A) =\dim \text{span}(A^T)$ for real numbers will also work in the same way for complex numbers. (because it only uses the fact that $\mathbb{R}$ is a field)
