More than basis vectors in a space are dependent, less can't span the space proof Let $V$ be a vector space, $ B = \{ \mathbf{b}_{1}, \ldots, \mathbf{b}_{n} \}$ a basis for $V$ (independent and spans $V$).
Prove that fewer vectors than $n$ can't span $V$, while more vectors are necessarily dependent.
I am looking for a clearer proof than the one on this page: https://math.kennesaw.edu/~plaval/math3260/basis.pdf (middle of third page, Theorem 306), and/or an explanation as to why that system of equations helps prove linear dependence (says to consider said linear combination, but doesn't really explain anything).
Thanks!
 A: Firstly, assume for contradiction that $B \setminus \{ \mathbf{b}_{n} \} = \{ \mathbf{b}_{1}, \ldots, \mathbf{b}_{n - 1} \}$ spans $V$. Then there exists $c_{1}, \ldots, c_{n - 1} \in \mathbb{R}$ such that $\mathbf{b}_{n} = c_{1} \mathbf{b}_{1} + \cdots + c_{n - 1} \mathbf{b}_{n - 1}$. But this would mean that $c_{1} \mathbf{b}_{1} + \cdots + c_{n - 1} \mathbf{b}_{n - 1} + (-1) \mathbf{b}_{n} = \mathbf{0}$, so $B$ is linearly dependent, a contradiction.
Similarly, let $B' = B \cup \{ \mathbf{b}_{n + 1} \} = \{ \mathbf{b}_{1}, \ldots, \mathbf{b}_{n + 1} \}$, where $\mathbf{b}_{n + 1} \in V$. Then since $V = \operatorname{span} (B)$, there exist constants $c_{1}, \ldots, c_{n} \in \mathbb{R}$ such that $c_{1} \mathbf{b}_{1} + \cdots + c_{n} \mathbf{b}_{n} = \mathbf{b}_{n + 1}$. But this would mean that $c_{1} \mathbf{b}_{1} + \cdots + c_{n} \mathbf{b}_{n} + (- 1) \mathbf{b}_{n + 1} = \mathbf{0}$, so $B'$ is dependent.
A: Let me give a (slightly) different answer.

Let $V$ be a vector space, $ B = \{ b_1, \ldots, b_n \}$ a basis for $V$ (independent and spans $V$).
  Prove that (1) fewer vectors than $n$ can't span $V$, while (2) more vectors are necessarily dependent.

Proof: (1) $B$ is a linearly independent set, so if we take one element out, say $b_i \in B \subset V$,
$$b_i = \sum_{j = 1\\ j \neq i}^n c_j b_j$$
would imply a contradiction.
So $b_i \in V$ is surely not in $\text{Span}(\{ b_1, \ldots, b_{i-1}, b_{i+1}, \ldots, b_n \})$.
(2) Choose a non zero $w \in V$, $w \neq b_i$ for all $i$.
Since $w \in \text{Span}(B) = V$,
$$w = \sum_{i = 1}^n a_i b_i $$
implies that $C = B \cup \{w\}$ is a linearly dependent set. $\square$
