Column space of the conjugate of a matrix Let $\mathbf{A} \in \mathbb{C}^{m\times n}$ and let $\mathbf{A}^* \in \mathbb{C}^{m\times n}$ be its complex conjugate. Is the column space of $\mathbf{A}$ equal to the column space of $\mathbf{A}^*$?
 A: Assuming that an element in $\mathbb C^m$ is an ordered m-uple $(z_1,z_2,\ldots,z_m)$ with $z_i \in \mathbb C$, clearly in general $A$ and $A^*$ don't share the same column space.
Indeed, let consider for example $(i,1)$ and $(-i,1)$ and let $z\in \mathbb C$ such that
$$z(i,1)=(iz,z)=(-i,1)$$
that is


*

*$iz=-i$

*$z=1$
which is impossible.
A: Consider the case $m = 2, n = 1$. We have $\begin{pmatrix}i \\ 1 \end{pmatrix}, \begin{pmatrix}-i \\ 1 \end{pmatrix} \in \mathbb{C}^{2 \times 1}$ are complex conjugates of each other, but they are linearly independent (with respect to both $\mathbb{R}$ and $\mathbb{C}$), so their column spaces are different.
I.e. $\text{Span} \left \{ \begin{pmatrix}i \\ 1\end{pmatrix} \right \} \neq \text{Span}\left\{\begin{pmatrix}-i \\ 1\end{pmatrix} \right\} $.

Edit: To address the comments by OP, firstly, $\mathbb{C}^2$ is a $2$ dimensional vector space over $\mathbb{C}$, so the span of any particular vector in $\mathbb{C}^2$ cannot be all of $\mathbb{C}^2$.
Next, more detail on why $\text{Span} \{v_1\} \neq \text{Span} \{v_2\}$, where $v_1 = \begin{pmatrix} i \\ 1 \end{pmatrix},$ and $v_2 = \begin{pmatrix} -i \\ 1 \end{pmatrix}$. Working over $\mathbb{C}$, it can be shown that $v_1$ and $v_2$ are linearly independent. This is enough to imply that $\text{Span} \{v_1\} \neq \text{Span} \{v_2\}$.
For if it were the case that $\text{Span} \{v_1\} = \text{Span} \{v_2\}$, since $v_1 \in \text{Span}\{v_1\}$, we would have $v_1 \in \text{Span} \{v_2\}$. So for some $\lambda \in \mathbb{C}$, we would have $v_1 = \lambda v_2 \iff v_1 - \lambda v_2 = 0$. But this contradicts the fact that $v_1$ and $v_2$ are linearly independent, so indeed $\text{Span} \{v_1\} \neq \text{Span} \{v_2\}$.
