- #1

schniefen

- 178

- 4

##x_1(\textbf{u}_1-\textbf{u}_2)+x_2(\textbf{u}_2-\textbf{u}_3) +...+x_n(\textbf{u}_n-\textbf{u}_1)=0\qquad(1)##

Rearranging terms:##(x_1-x_n)\textbf{u}_1+(x_2-x_1)\textbf{u}_2+...+(x_n-x_{n-1})\textbf{u}_n=0##

From the linear independence of the vectors ##\textbf{u}_1,...,\textbf{u}_n##, ##(x_1-x_n)=(x_2-x_1)=...=(x_n-x_{n-1})=0##, which means that ##x_1=x_2=...=x_n=k## for some ##k\in\mathbf{R}##. (1) can thus be written ##k((\textbf{u}_1-\textbf{u}_2)+(\textbf{u}_2-\textbf{u}_3) +...+(\textbf{u}_n-\textbf{u}_1))=k(\textbf{0})=0##. Thus they're all linear dependent. How does one proceed from here?