Let $\mathbb P_n$ be the complex projective space on $\mathbb C^{n+1}$, i.e. $$\mathbb P_n = \{[x_0,\ldots, x_n]: (x_0,\ldots, x_n) \in \Bbb C^{n+1} -\{0\} \}$$
where $[x_0,\ldots, x_n]$ is the equivalence class of $(x_0,\ldots, x_n)$ under the equivalence relation $\mathbb C^{n+1} \ni x \equiv y \in \Bbb C^{n+1} \iff \exists$ $\lambda \neq 0$: $x = \lambda y$.
Let $\pi$ be the projection map $\mathbb C^{n+1} -\{0\} \to \mathbb P_n$, $\pi(x_0,\ldots,x_n) = [x_0,\ldots,x_n]$. A hyperplane in $\mathbb P_n$ is $\pi(V-\{0\})$ where $V \subset \Bbb C^{n+1}$ is a subspace of dimension $n$.
Let $p_0,\ldots,p_n \in \mathbb P_n$ and suppose that they do not lie on a hyperplane. Let $u_0,\ldots, u_n \in \mathbb C^{n+1}$ such that $\pi(u_i) = p_i$.
Supposedly $\{u_0, \ldots, u_n\}$ is a basis of $\mathbb C^{n+1}$. Why?
I am stuck at this. It is part of a proof in Kirwan's Complex Algebraic Curves p: 39.
Suppose the contrary, that $u_0, \dots, u_n$ do not form a basis of $\mathbb{C}^{n+1}$. Since they are $n+1$ vectors in $\mathbb{C}^{n+1}$ they must be linearly dependent, hence they belong to a $n$-dimensional subspace and their projection belong to a hyperplane in $\mathbb{P}^n$.
Being linearly dependent, there are complex numbers $a_0, \dots , a_n$, not all zero, such that $$ a_0u_0 + \dots + a_nu_n =0 $$ and writing the $u_j$s in coordinates will give you the equation of the hyperplane.