Linear Algebra Subspaces Proving Let $V_1$ and $V_2$ be subspaces of $\mathbb R^n$ defined by
$$
V_1 = \{(x_1, x_2, ..., x_n) | x_1 + x_2 + \cdots + x_n = 0\}
$$
$$
V_2 = \{(x_1, x_2, ..., x_n) |  x_1 = x_2 = \cdots = x_n\}.
$$
Prove that any vector $v \in\mathbb R^n$ can be uniquely expressed as $v = v_1 + v_2$ such that $v_1 \in V_1$ and $v_2 \in V_2$.
I have no idea how I should go about starting this proof. I can't see a way for me to use the conditions of $V_1$. Also, how do I go about proving the uniqueness of the expression?
 A: If $(x_1,x_2,\ldots,x_n)\in\mathbb R^n$ and if $m=\frac1n(x_1+x_2+\cdots+x_n)$, then$$(x_1,x_2,\ldots,x_n)=\overbrace{(m,m,\ldots,m)}^{\phantom{V_2}\in V_2}+\overbrace{(x_1-m,x_2-m,\ldots,x_n-m)}^{\phantom{V_1}\in V_1}.$$And if $m',y_1,y_2,\ldots,y_n\in\mathbb R$ are such that $y_1+y_2+\cdots+y_n=0$ and that\begin{align}(x_1,\ldots,x_n)&=(m',m',\ldots,m')+(y_1,y_2,\ldots,y_n)\\&=(m'+y_1,m'+y_2,\ldots,m'+y_n),\end{align}then\begin{align}m&=\frac1n(x_1+x_2+\cdots+x_n)\\&=\frac1n(m'+y_1+m'+y_2+\cdots+m'+y_n)\\&=m'\end{align}and therefore$$(y_1,y_2,\ldots,y_n)=(x_1-m,x_2-m,\ldots,x_n-m).$$
A: You can also view the problem somewhat geometrically. We can express
\begin{align*}
V_1 &= \{x \in \Bbb{R}^n : x \cdot (1, 1, \ldots, 1) = 0\} \\
V_2 &= \{t(1, 1, \ldots, 1) : t \in \Bbb{R}\}.
\end{align*}
Looking at it this way, $V_2$ is a line through $\Bbb{R}^n$ and $V_2$ is the hyperplane of points perpendicular to $V_1$.
This helps us get the decomposition too. Essentially we need to orthogonally project $v$ onto the line, decomposing $v$ into a vector along the line $V_2$, and a perpendicular vector necessarily in $V_1$. We even have a formula for the projection of one vector onto another. Let $w = (1, 1, \ldots, 1)$. Then,
$$\operatorname{proj}_w(v) = \frac{v \cdot w}{w \cdot w} w = \frac{v_1 + v_2 + \ldots + v_n}{n}(1, 1, \ldots, 1)$$
is the component in $V_2$. The component in $V_1$ is therefore
$$v - \operatorname{proj}_w(v),$$
which is perfectly in line with Jose's answer.
This shows that $V_1 + V_2 = \Bbb{R}^n$. To show the sum is direct, try showing that their intersection is trivial. If $v \in V_1 \cap V_2$, then $v = tw$ for some $t$, but $v \cdot w = 0$. Thus,
$$(tw) \cdot w = 0 \implies t (w \cdot w) = 0 \implies tn = 0 \implies t = 0 \implies v = tw = 0.$$
A: Hint: 
Define $f:\mathbb R^n\to \mathbb R $ via $f(x)=x_1+\cdots +x_n$, where  $x=(x_1,\ldots, x_n) $. It is easy to verify that  $f $ is a linear functional.  Let $a=(1,1,\ldots ,1)\in\mathbb R^n$. Observe that $f (a)\neq0$.  Let $v\in\mathbb R^n $ be arbitrary. Consider  $$v_1:=v-\frac{f (v)}{f (a)}a.$$ Note that  $f(v_1)=0$ and $\dfrac{f (v)}{f (a)}a\in V_2$. 
