How to show $S=\{u_1,u_2,\cdots,u_n\}$ is linearly independent if $S$ spans $V$? Let $S=\{u_1,u_2,\cdots,u_n\}$ be a spanning set for an n-dimensional vector space $V$. Show $S$ is linearly independent.
My try:
Since $S$ is spanning set for $V$, any vector $x$ in $V$ can be written as the linear combination of vectors in $S$ as
$$
x = \alpha_1 u_1 + \cdots + \alpha_n u_n
$$ 
Since $V$ is a vector space, $0 \in V$. Let $x=0$.
$$
0= \alpha_1 u_1 + \cdots + \alpha_n u_n
$$
We need to show $\alpha_i$'s are zero.
 A: Basically, if $\alpha_j\neq 0,$ then $u_j$ is in the span of the other $n-1$ vectors, which means $Span(S)$ could not have been $n$ dimensional in the first place. You can write this in more detail and get yourself a proof by contradiction. 
A: Proving the contrapositive is not easier in my opinion, but it does give you another way to think about the problem.
$$ \text{$S$ spans $V$} \implies \text{$S$ is linearly independent} \tag{Original} $$
is equivalent to
$$ \text{$S$ is linearly dependent} \implies \text{$S$ does not span $V$} \tag{Contrapositive}$$
Let's march through $u_1, u_2, \dots u_n$ and keep $u_i$ if it is linearly independent of the preceding vectors ($u_1 \dots u_{i-1}$). Let's call the new collection of linearly independent vectors $w_1, \dots, w_m$ where $m < n$. We know $m$ is less than $n$ because at least one $u_i$ must be excluded. We know that $w_1, \dots, w_m$ and $u_1, \dots, u_n$ span the same subspace because we only excluded $u_i$s that were in the "span so far".
$w_1, \dots, w_m$ spans a subspace of exactly dimension $m$, but $m < n$ by our hypothesis.
