Show that $\mathbb R^n$ is the direct sum of the subspaces $W_1$ and $W_2$ Show that $\mathbb R^n$ is the direct sum of the subspaces
$W_1$ = {( $a_1$ , · · · , $a_n$) ∈ $\mathbb R^n $ : $a_1$ + $a_2$ + · · · + $a_n$ = $\mathbf 0$}
and
$W_2$ = {($a_1$, · · · , $a_n$) ∈ $\mathbb R^n$ : $a_1$ = $a_2$ = · · · = $a_n$}.
I have no idea as to how I can approach this aside from the Theorem which states that $\mathbb R^n$ is the direct sum of the subspaces $W_1$ and $W_2$ if and only if each $\mathbf v$ ∈ $\mathbb R^n$ can be uniquely written as $\mathbf x_1$ + $\mathbf x_2$ , where $\mathbf x_1$ ∈ $W_1$ and $\mathbf x_2$ ∈ $W_2$. Also, is there a way of solving this by showing that $W_1$ + $W_2$ = V and $W_1$ $\bigcap$ $W_2$ = {$\mathbf 0$}? Some help will be much appreciated. 
 A: Hint:
$W_2$ has dimension $1$ and is directed by vector $\vec u=(1,1,\dots,1)$. All you have to check is $W_1=W_2^{\bot}$ for the standard inner product.
Elementary argument:
$W_1$ has codimension $1$, i.e. dimension $n-1$ since it is defined by $1$ (non-trivial) linear equation, $W_2$ has dimension $1$, hence to prove $\mathbf R^n=W_1\oplus W_2$, it is enough to prove $W_1\cap W_2=\{0\}$.
But this is obvious: if $a_1=a_2=\dots=a_n=\lambda$, say,  and $a_1+a_2+\dots+a_n=0$, then $a_1+a_2+\dots+a_n=n\lambda=0$, so $\lambda=0$.
A: This is a fancy way of saying that a list of $n$ numbers can be re-centered, by subtracting the same amount from all the numbers, so that the average of the recentered list is $0$.  The direct sum decomposition says that knowing the average and the recentered list is equivalent (and in a linear way) to knowing the collection of numbers.
A: Consider the case in $\Bbb R^2$;
If we take $(a_1,a_2)=(b_1,b_2)+(c_1,c_2)$ then we have the relations $c_1=c_2;b_1+b_2=0$
Also $a_1=b_1+c_1;a_2=b_2+c_2$;
On solving we get that ;
Any $(a_1,a_2)\in \Bbb R^2$ can be uniquely expressed as 
$(a_1,a_2)= (\dfrac{a_1-a_2}{2},\dfrac{-a_1+a_2}{2})+(\dfrac{a_1+a_2}{2},\dfrac{a_1+a_2}{2})$
Do the same in $\Bbb R^n$ you get such relations but definitely more in number;Solve them you get the required result
