# Finding the dimension of the span ##u_1-u_2, ...., u_n-u_1##?

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If ##\textbf{u}_1,...,\textbf{u}_n## form a basis in a linear space, how does one determine the dimension of the span ##\textbf{u}_1-\textbf{u}_2, \textbf{u}_2-\textbf{u}_3,...,\textbf{u}_n-\textbf{u}_1##? Since ##\textbf{u}_1,...,\textbf{u}_n## form a basis, they're linearly independent. If one finds the number of vectors that make up the basis of the span ##\textbf{u}_1-\textbf{u}_2,...,\textbf{u}_n-\textbf{u}_1##, then that number is also its dimension. To find the basis of the span, I check the linear independence of the vectors:

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?

Hiero
How does one proceed from here?
Eliminate one vector from the dependent set and check again.

Loosely speaking, eliminating ##(u_k-u_{k+1})## from the set and checking for independence is like setting ##x_k=0## in all of your steps which then makes all coefficients zero, but you should repeat the process and see.

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