Basis for Vector Space iff can be Expressed Uniquely as Linear Combo of Basis Let $V$ be a vector space and $\beta= \{ u_1,\dots ,u_n \}$ be a subset of $V$. 
$\Rightarrow$ $\beta$ is a basis for $V$ iff each vector $v\in V$ can be unquiley expressed as a linear combination of vectors in $\beta$ , that is , can be expressed in the form
$$v=a_1*u_1+\dots +a_n u_n  $$

Def of basis $\beta$ for a vector space $V$ is linearly independent subset of $V$ that generates $V$

Assume $V$ is a vector space and  $\beta= \{ u_1,\dots ,u_n \}$ be a subset of $V$. 
$\Rightarrow$ ] let $\beta$ be basis. Suppose representation is not unique so 
$$\begin{aligned} v&=a_1 \hat{u_1}+\dots +a_n  \hat{u_n}
                 \\ &=b_1\hat{u_1}+\dots +b_n \hat{u_n} 
   \end{aligned} $$
Now 
   $$ \begin{aligned}
           0 &=v-v=(a_1-b_1) \hat{u_1}+\dots +(a_n -b_n)  \hat{u_n}
           \\ &= c_1 u_1 + \dots + c_n u_n  
      \end{aligned}$$
ofcourse $c_i = a_i-b_i $ for $i=1,\dots,n $ 
Since $\beta$ is a basis $\Rightarrow$ Linearly Indep. the only solution is the trivial one so $c_i=a_i-b_i=0$ that is $a_i=b_i$ so our vectors unique
$\Leftarrow$] each vector can be expresed  uniquely as a linear combo of vectors $\beta$ 
$\vdots$
$\therefore \beta$ is a basis

Can use guidance for $\Leftarrow$] part.

Span is clear. For linear independence, use $0$ can be expressed uniquely.
$$0 = 0v_1 + \cdots + 0v_n $$
so if $c_1 v_1 + \cdots + c_n v_n = 0$, then $c_i = 0$ for all $i$. 
