K^n/Linear subspace/Sum is 0/Basis/Example

We consider the $K$-linear subspace $U\subset K^n$ given by

$U=\{v\in K^n\mid \sum _{i=1}^n{v}_i=0\}.$

A basis for $U$ is given by the $n-1$ vectors

$u_1=(1,-1,0,0,\dots ,0),u_2=(0,1,-1,0,\dots ,0),\dots ,u_{n-1}=(0,0,\dots ,0,1,-1),.$

These vectors belong evidently to $U$. The linear independence can be checked in $K^n$. From an equation

$\sum _{i=1}^{n-1}{a}_i{u}_i=0$

we can deduce step by step $a_1=0$, $a_2=0$, etc. That the system is a generating system follows from

$\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\\ ⋮\\ v_{n-1}\\ v_n\end{array}\right)=v_1\left(\begin{array}{c}1\\ -1\\ 0\\ ⋮\\ 0\\ 0\end{array}\right)+(v_1+v_2)\left(\begin{array}{c}0\\ 1\\ -1\\ ⋮\\ 0\\ 0\end{array}\right)+(v_1+v_2+v_3)\left(\begin{array}{c}0\\ 0\\ 1\\ -1\\ ⋮\\ 0\end{array}\right)+\cdots +(v_1+v_2+v_3+\cdots +v_{n-1})\left(\begin{array}{c}0\\ 0\\ 0\\ ⋮\\ 1\\ -1\end{array}\right),$

where the equality in the last row rests on the condition

$\sum _{i=1}^n{v}_i=0.$