Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{\bf{0}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{{}}{\bf{4}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{{}}{\bf{4}}\\{\bf{0}}\end{aligned}} \right)\), and\(p = \left( {\begin{aligned}{{}}{\bf{2}}\\{\bf{1}}\end{aligned}} \right)\). Confirm that

\({\bf{p}} = \frac{{\bf{1}}}{{\bf{3}}}{{\bf{v}}_{\bf{1}}} + \frac{{\bf{1}}}{{\bf{3}}}{{\bf{v}}_{\bf{2}}} + \frac{{\bf{1}}}{{\bf{6}}}{{\bf{v}}_{\bf{3}}} + \frac{{\bf{1}}}{{\bf{6}}}{{\bf{v}}_{\bf{4}}}\)and \({{\bf{v}}_{\bf{1}}} - {{\bf{v}}_{\bf{2}}} + {{\bf{v}}_{\bf{3}}} - {{\bf{v}}_{\bf{4}}} = {\bf{0}}\).

Use the procedure in the proof of Caratheodory’s Theorem to express p as a convex combination of three of the \({{\bf{v}}_i}\)’s. Do this in two ways.

\(\begin{aligned}{}\frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4} &= \frac{1}{3}\left( {\begin{aligned}{{}}1\\0\end{aligned}} \right) + \frac{1}{3}\left( {\begin{aligned}{{}}1\\2\end{aligned}} \right) + \frac{1}{6}\left( {\begin{aligned}{{}}4\\2\end{aligned}} \right) + \frac{1}{6}\left( {\begin{aligned}{{}}4\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{3} + \frac{1}{3} + \frac{4}{6} + \frac{4}{6}}\\{0 + \frac{2}{3} + \frac{2}{6} + 0}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}2\\1\end{aligned}} \right)\\ &= {\bf{p}}\end{aligned}\)

Substitute value in the equation \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\).

\(\begin{aligned}{}{{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} &= \left( {\begin{aligned}{{}}1\\0\end{aligned}} \right) - \left( {\begin{aligned}{{}}1\\2\end{aligned}} \right) + \left( {\begin{aligned}{{}}4\\2\end{aligned}} \right) - \left( {\begin{aligned}{{}}4\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\ &= 0\end{aligned}\)

From two steps, the sum of the coefficients is 0, so the combination is an affine dependence relationship. Both the given equations are valid.

From the equations \({\bf{p}} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\) and \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\), the coefficients \({{\bf{v}}_1}\) and \({{\bf{v}}_3}\) are \(\frac{1}{3}\) and \(\frac{1}{6}\).

Subtract \(\frac{1}{6}\) times \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\) from \({\bf{p}} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\).

\(\begin{aligned}{}{\bf{p}} - 0 = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4} - \frac{1}{6}\left( {{{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4}} \right)\\{\bf{p}} = \left( {\frac{1}{3} - \frac{1}{6}} \right){{\bf{v}}_1} + \left( {\frac{1}{3} + \frac{1}{6}} \right){{\bf{v}}_2} + \left( {\frac{1}{6} - \frac{1}{6}} \right){{\bf{v}}_3} + \left( {\frac{1}{6} + \frac{1}{6}} \right){{\bf{v}}_4}\\ = \frac{1}{6}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_4}\end{aligned}\)

Similarly, add \(\frac{1}{6}\) times \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\) to \({\bf{p}} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\).

\(\begin{aligned}{}{\bf{p}} - 0 = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4} + \frac{1}{6}\left( {{{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4}} \right)\\{\bf{p}} = \left( {\frac{1}{3} + \frac{1}{6}} \right){{\bf{v}}_1} + \left( {\frac{1}{3} - \frac{1}{6}} \right){{\bf{v}}_2} + \left( {\frac{1}{6} + \frac{1}{6}} \right){{\bf{v}}_3} + \left( {\frac{1}{6} - \frac{1}{6}} \right){{\bf{v}}_4}\\ = \frac{1}{2}{{\bf{v}}_1} + \frac{1}{6}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_3}\end{aligned}\)

So, the expressions for p are\({\bf{p}} = \frac{1}{6}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_4}\) and \({\bf{p}} = \frac{1}{2}{{\bf{v}}_1} + \frac{1}{6}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_3}\).