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

[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:

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

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?

 

[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.

 

[mathwonk]

isn't the sum of your vectors zero? so their span has dim ≤ n-1. indeed it seems to equal n-1 since if you throw in un, you seem to get everything. but i did not check it on paper.