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

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

${\displaystyle {}U={\left\{v\in K^{n}\mid \sum _{i=1}^{n}v_{i}=0\right\}}\,.}$

A basis for ${\displaystyle {}U}$ is given by the ${\displaystyle {}n-1}$ vectors

${\displaystyle u_{1}=(1,-1,0,0,\ldots ,0),\,u_{2}=(0,1,-1,0,\ldots ,0),\,,\ldots ,u_{n-1}=(0,0,\ldots ,0,1,-1),\,.}$

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

${\displaystyle {}\sum _{i=1}^{n-1}a_{i}u_{i}=0\,}$

we can deduce step by step ${\displaystyle {}a_{1}=0}$, ${\displaystyle {}a_{2}=0}$, etc. That the system is a generating system follows from

${\displaystyle {}{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\\\vdots \\v_{n-1}\\v_{n}\end{pmatrix}}=v_{1}{\begin{pmatrix}1\\-1\\0\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}){\begin{pmatrix}0\\1\\-1\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}+v_{3}){\begin{pmatrix}0\\0\\1\\-1\\\vdots \\0\end{pmatrix}}+\cdots +(v_{1}+v_{2}+v_{3}+\cdots +v_{n-1}){\begin{pmatrix}0\\0\\0\\\vdots \\1\\-1\end{pmatrix}}\,,}$

where the equality in the last row rests on the condition

${\displaystyle {}\sum _{i=1}^{n}v_{i}=0\,.}$