Let $V$ be a complex vector space. A complex conjugation on $V$ is an antilinear map $\sigma : V \to V$ such that $\sigma\circ\sigma = \operatorname{id}_V$. Not every complex vector space comes with a complex conjugation, in general it is an extra piece of data.
Complex Euclidean space $\mathbb{C}^n$ has a complex conjugation given by $\sigma(z_1, \dots, z_n) = (\overline{z_1}, \dots, \overline{z_n})$. As any finite dimensional complex vector space is non-canonically isomorphic to $\mathbb{C}^n$, we can always obtain a complex conjugation $\sigma_V$ on $V$ by choosing an isomorphism $\phi : V \to \mathbb{C}^n$ and setting $\sigma_V = \phi^{-1}\circ\sigma\circ\phi$.
Note that $\mathbb{C}^n$ is not the only complex vector space which carries a natural complex conjugation. If $W$ is a real vector space, the complex vector space $V = W\otimes_{\mathbb{R}}\mathbb{C}$ is called the complexification of $W$ and it carries a natural complex conjugation given by $\sigma(w\otimes z) = w\otimes\overline{z}$. This generalises the previous example as $\mathbb{C}^n = \mathbb{R}^n\otimes_{\mathbb{R}}\mathbb{C}$; you can check that the complex conjugation given by the complexification is the same as the one given above. Once you choose a complex conjugation on $V$, which as we showed above is always possible, you can write $V$ as the complexification of a real vector space $W$ (the set of vectors fixed by $\sigma$).
If $V$ has a complex conjugation $\sigma$, then given any subspace $W$, the conjugate subspace, often denoted $\overline{W}$, is $\sigma(W)$; you should check that this is indeed a subspace.
