Given Information
We have to prove Theorem 6 for an affinely independent set $S = \left\{ \mathbf { v } _ { 1 } , \ldots , \mathbf { v } _ { k } \right\}$ in $\mathbb { R } ^ { n } $

Step-1:
Let there be another set of scalars $d _ { 1 } , d _ { 2 } , \cdots , d _ { k }$ such that $\mathbf { p } = d _ { 1 } \mathbf { v } _ { 1 } + d _ { 2 } \mathbf { v } _ { 2 } + \cdots + d _ { k } \mathbf { v } _ { k } ,$ where$d _ { 1 } + d _ { 2 } + \cdots + d _ { k } = 1$

Write the system of equations:\[\begin{array} { l } { \mathbf { p } = c _ { 1 } \mathbf { v } _ { 1 } + c _ { 2 } \mathbf { v } _ { 2 } + \cdots + c _ { k } \mathbf { v } _ { k } } \\ { \mathbf { p } = d _ { 1 } \mathbf { v } _ { 1 } + d _ { 2 } \mathbf { v } _ { 2 } + \cdots + d _ { k } \mathbf { v } _ { k } } \end{array}\]Subtract both equations:\[\begin{aligned} \mathbf { p } - \mathbf { p } & = \left( c _ { 1 } - d _ { 1 } \right) \mathbf { v } _ { 1 } + \cdots + \left( c _ { k } - d _ { k } \right) \mathbf { v } _ { k } \\ 0 & = \left( c _ { 1 } - d _ { 1 } \right) \mathbf { v } _ { 1 } + \cdots + \left( c _ { k } - d _ { k } \right) \mathbf { v } _ { k } \end{aligned}\]Since S is a linearly independent set, the coefficients should all be 0. \[\begin{array} { c } { \left( c _ { 1 } - d _ { 1 } \right) = 0 } \\ { \left( c _ { 2 } - d _ { 2 } \right) = 0 } \\ { } \\ { } \\ { \left( c _ { k } - d _ { k } \right) = 0 } \end{array}\].This proves that:\[c _ { 1 } = d _ { 1 } , c _ { 2 } = d _ { 2 } , \ldots c _ { k } = d _ { k }\]Therefore,

The set of scalars $c _ { 1 } , c _ { 2 } , \dots , c _ { k }$ s unique for $\mathbf { p } = c _ { 1 } \mathbf { v } _ { 1 } + \cdots + c _ { k } \mathbf { v } _ { k }$