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.
Best Answer
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)