How to show that $\mathbb{R}^3$ is the direct sum of $W_1={\rm span}\{(1,1,1)\}$ and $W_2={\rm span}\{(1,0,0),(1,1,0)\}$? We write linear combination  of two spans as $a(1,1,1)+b(1,0,0)+c(1,1,0)$.
Any vector $v$ in $\mathbb{R}^3$ can be expressed as $(a+b+c, a+b, a)$. Also linear combination of $W_2$ cannot form vector in $W_1$ other than $(0,0,0)$, $W_1\cap W_2=(0,0,0)$, and therefore direct sum is $\mathbb{R}^3$.  I am not sure if my method is correct, so could anyone help? 
 A: Note that you need to show that every vector in $\mathbb{R}^3$ can be expressed as
$$ (a + b + c, a + c, a) $$
and not the expression you wrote. Other then that, your method is correct (and you of course need to justify why $W_1 \cap W_2 = \{ (0,0,0) \}$). If you know the theorem that relates the dimension of the intersection $W_1 \cap W_2$ with the dimension of the sum $W_1 + W_2$, then it is enough to show that $W_1 \cap W_2 = \{ (0,0,0) \}$ as this shows that $W_1 + W_2$ is three-dimensional.
A: Since $\dim W_1=1$ and $\dim W_2=2$, if $\mathbb{R}^3=W_1+W_2$ the dimension formula gives
$$
\dim(W_1\cap W_2)=\dim\mathbb{R}^3-\dim W_1-\dim W_2=0
$$
So you just need to show that
$$
\mathbb{R}^3=\operatorname{Span}\{(1,1,1),(1,0,0),(1,1,0)\}
$$
The matrix
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 0
\end{bmatrix}
$$
clearly has rank $3$.
A: One simple way is the classical method to put vectors $$(1,1,1),(1,0,0),(1,1,0)$$ as column in a matrix $A$ and verify that $$\forall b \in \mathbb{R^3}$$
$$Ax=b$$
has solution.
That is $\iff$ $A$ is not a singular matrix.
Thus your consideration is correct as $W_1$ and $W_2$ are independent subspaces and $dim(W_1)+dim(W_2)=3$ thus they span $\mathbb{R^3}$.
