Let $\{v_1,v_2,\dots,v_n\}$ be a basis of vector space $V_n$,
$V_1=\operatorname{span}\{v_1+v_2+\dots+v_n\}$ and $V_2=\{v=\sum_{j=1}^nx_jv_j | \sum_{j=1}^nx_j=0,x_j\in R\}$.
Prove that $V=V_1\bigoplus V_2$.
Put this in $3\rm{D}$ scenario and let the basis be standard, then $V_2$ is the plane $x+y+z=0$ that perpendicular to $V_1$ which is spanned by the vector $(1,1,1)$
therefore their union fills $\mathbb{R}^3$
However, how do I explain this in general?
Take $v\in V$. Then\begin{align}v&=\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n\\&=\frac{\alpha_1+\alpha_2+\cdots+\alpha_n}n(v_1+v_2+\cdots+v_n)+\\&\quad +\left(\alpha_1-\frac{\alpha_1+\alpha_2+\cdots+\alpha_n}n\right)v_1+\cdots+\left(\alpha_n-\frac{\alpha_1+\alpha_2+\cdots+\alpha_n}n\right)v_n.\end{align}And this equality expresses $v$ as the sum of an element of $V_1$ with an element of $V_2$.
On the other hand, if $w\in V_1\cap V_2$ and if you write $w$ as $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n$, then $\alpha_1=\alpha_2=\cdots=\alpha_n$ and $\alpha_1+\alpha_2+\cdots+\alpha_n=0$; therefore, each $\alpha_k$ is equal to $0$. So, $w=0$, and therefore $V_1\cap V_2=\{0\}$.