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

• I
• schniefen
schniefen
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?

## Answers and Replies

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.

schniefen