Proving that two subspaces are complementary? Pardon me if I have the terminology wrong here, I'm not totally sure if this is what it's called in English.
I want to show that two sets are complementary in $\mathbb R^4$. From what I understand, this is equivalent to show that two sets are direct sum of $\mathbb{R}^4$. Is that true?
Here are my subspaces:
$H = \{(a,b,c,d)\in \mathbb{R}^4 \mid a + b + c + d = 0\}$
$\operatorname{span}((1,1,1,1))$
For starters, I proved that $H \cap \operatorname{span}((1,1,1,1))= \{(0,0,0,0)\}$
by saying that $\operatorname{span}((1,1,1,1))$ can be described as $\{(x,y,z,t)\in \mathbb{R}^4 \mid x = y = z = t\}$. Thus, $H \cap \operatorname{span}((1,1,1,1))= \{(x,y,z,t) \mid x + y + z + t = 0, x = y = z = t\}=\{(x,x,x,x)\mid 4x = 0\} = \{(0,0,0,0)\}$ 
Is this the right way to show this?
Now, to show that $H + \operatorname{span}((1,1,1,1)) = \mathbb{R}^4$:
I can say that
$$
H + \operatorname{span}((1,1,1,1)) = \{(a,b,c,d)\in \mathbb{R}^4 \mid a + b + c + d = 0\} + \{(x,y,z,t)\in \mathbb{R}^4 \mid x = y = z = t\}.
$$
I sort of have an idea of what I should be doing at this point, but I'm not too sure I'm even doing this properly. 
How do I go about finishing my proof?
Furthermore, is showing that two subspaces are complementary equivalent to showing that their sum creates a basis?
Sorry for the poor notation. Any help is appreciated. Thank you.
 A: Yes, you have to prove that


*

*$H\cap\operatorname{span}((1,1,1,1))=\{(0,0,0,0)\}$

*$H+\operatorname{span}((1,1,1,1))=\mathbb{R}^4$


You can simplify the proof of the first fact by taking $a(1,1,1,1)=(a,a,a,a)\in\operatorname{span}((1,1,1,1))$ and imposing the condition that $(a,a,a,a)\in H$, that is,
$$
a+a+a+a=0
$$
which of course implies $a=0$.
The second part can be done in several ways. One is to take $(x,y,z,w)\in\mathbb{R}^4$ and trying to express it in the form
$$
(x,y,z,w)=(a,b,c,d)+r(1,1,1,1)
$$
where $a+b+c+d=0$. This becomes a linear system
$$
\begin{cases}
a+r=x \\
b+r=y \\
c+r=z \\
d+r=w \\
a+b+c+d=0
\end{cases}
$$
The matrix is
$$
\left[\begin{array}{ccccc|c}
1 & 0 & 0 & 0 & 1 & x \\
0 & 1 & 0 & 0 & 1 & y \\
0 & 0 & 1 & 0 & 1 & z \\
0 & 0 & 0 & 1 & 1 & w \\
1 & 1 & 1 & 1 & 0 & 0
\end{array}\right]
$$
and a simple elimination brings it in the form
$$
\left[\begin{array}{ccccc|c}
1 & 0 & 0 & 0 & 1 & x \\
0 & 1 & 0 & 0 & 1 & y \\
0 & 0 & 1 & 0 & 1 & z \\
0 & 0 & 0 & 1 & 1 & w \\
0 & 0 & 0 & 0 & -4 & -x-y-z-w
\end{array}\right]
$$
which shows that the system has a (unique) solution.
Another way is to notice that $\dim H=3$ and $\dim\operatorname{span}((1,1,1,1))=1$. Since $H\cap\operatorname{span}((1,1,1,1))=\{(0,0,0,0)\}$, we know that
$$
\dim\bigl(H+\operatorname{span}((1,1,1,1))\bigr)=3+1=4
$$
so the sum is $\mathbb{R}^4$. Why is $\dim H=3$? Because a basis is
$$
\{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\}
$$
