By defining linear independence and the span $\langle S\rangle$, explain what it means to say that $S$ is a basis of $V$. (3 marks)
I'm not entirely sure if I've got this correct so I'm going to give you my answer and please tell me if/where I'm going wrong. If you think I can shorten/simplify please say so as well:
Let $V$ be a finite-dimensional vector space over a field $F$.
A nonempty set of vectors $S = \{a_1,...,a_n\}$ is linearly independent if the equation $\alpha_1a_1 + \alpha_2a_2 + ... + \alpha_na_n = \underline 0$ has only one solution that is the trivial solution $(\alpha_1 = ... = \alpha_n = 0)$ where $\alpha_1,\alpha_2,...\alpha_n \in F$.
The subspace of $V$ that is formed from all possible linear combinations of the vectors in a nonempty set $S = \{a_1, a_2, ... , a_n\}$ is called the span of $S$ (denoted by $\langle S\rangle$ or $\langle a_1, a_2, ... ,a_n\rangle$) and we say that the vectors in $S$ span that subspace.
$S$ is a basis if it is linearly independent and spans all space ($S$ spans $V$). Am I right?
I'll leave an answer in order not to leave this question unanswered.
You're right. The only confusion you may have is when we say "$S$ spans the whole space" or "$S$ spans all space". Notice that the linear subspace spanned by $S$ always satisfy $\langle S\rangle \subseteq V$. It may be the case that $\langle S\rangle \subsetneq V$ (for example, if we let $V=\mathbb{R}^2$ and $S=\{(1,0)\}$), i.e. the spanned linear subspace of $S$ may not be equal to the whole space, it's natural to say in such a case that "$S$ doesn't span the whole space". We simply say $S$ does when the contrary occurs, i.e. $\langle S\rangle = V$.