# 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.

• It is not difficult to prove that if $\{v_1, \cdots, v_k\}$ is a basis of $\text{span}(A)$, then $\{\overline{v_1},\cdots, \overline{v_k}\}$ is a basis of $\text{span}(\overline{A})$. Commented Oct 19, 2014 at 6:10
• Thanks. Obviously the dimension of the span of $\{ \bar{v_1}, \ldots, \bar{v_k} \}$ can't be greater than $k$, but I guess I'm wondering if it can be less thank $k$, which is what I should try to prove.
– ted
Commented Oct 19, 2014 at 6:24
• As @DongRyulKim wrote, if $v_k$ form a basis of $\text{span}(A)$, then $\overline{v_k}$ form a basis of $\text{span}(\overline{A})$, so the dimensions of both spaces must be $k$. Commented Oct 19, 2014 at 6:38

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)