My question is a finding basis.I used the definition but I am not sure if it did work or not.
Let $ { v_1 ,..., v_n} $ denote a set of linearly independent vectors in V. Let
U={u $ \in V $ : u= $ \sum_{i=1}^n\left (a_iv_i\right )$ where $a_i \in F $ for all i and $ \sum_{i=1}^n\left (a_i\right )$=0
I did the following:
if i=1, u=$a_1v_1$ , $a_1$ =0
if i=2 , u=$a_1v_1+a_2v_2$ , $a_1+a_2$=0
.
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if i=n , u=$a_1v_1+a_2v_2+...+a_nv_n$ , $a_1+a_2+...+a_n=0 $
I use the definition of a basis: $ (v_1, v_2,...,v_n) $ is a basis of U if every $u \in U $ can be written uniquely in the form u= $a_1v_1+a_2v_2+...+a_nv_n$
To show that every $u \in U $ can be written uniquely in the form u= $a_1v_1+a_2v_2+...+a_nv_n$ I used the matrix A has the following form;
\begin{matrix} v_1 & 0 & \ldots & 0 \\ v_1 & v_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ v_1 & v_2 & \ldots & v_n \end{matrix}
since the matrix has a pivot position in every row then its columns are linearly independent. Can we say that every $u \in U $ can be written uniquely in the form u= $a_1v_1+a_2v_2+...+a_nv_n$
any help will be appreciated thanks