Find a subspace $T \leq \mathbb{R}^3$ such that $\mathbb{R}^3 = S \oplus T$ Let
$$S= \left\{\left.
\left[
\begin{array}{c}
a-b+c \\
a+b-c \\
a
\end{array}
\right]
\,\middle|\,  (a,b,c) \in \mathbb{R}^3 \right.\right\}$$
Find a subspace $T\leq\mathbb{R}^3$ such that $\mathbb{R}^3 = S \oplus T$
I already proved that $S$ is a subspace of $\mathbb{R}^3$ and that $\operatorname{dim}(S)=2$.
I know also that $\operatorname{dim}(T)=1$ and $S \cap T= {0}$.
How should I find $T$?
 A: So you have found a basis for the S given by the set $\{[1,1,1]^T,  [-1,1,0]^T\}$. You need a subspace $T$ which  complements $S$.
The most elementary way to do this, would be the following : find a linear transformation that has $S$ in it's kernel, but is not completely zero. This can be done by simply looking at a general element of $S$.
The elements of $S$ are of the form $[a-b+c,a+b-c,a]$ for all $a,b,c$. Now, it doesn't take too long to realize the following : $S_1 + S_2 =2S_3$, where $S_i$ is the $i$th component of $S$. 
Hence, define a linear transformation $T : \mathbb R^3 \to \mathbb R$, given by $T([x,y,z]) = x+y-2z$. By what we know, the kernel of $T$ contains $S$. But is $T$ just the zero transformation? Of course not. For example, $T([1,1,0]) = 2$.
Let $W$ be the subspace generated by $[1,1,0]$. That is, every element is of the form $[a,a,0]$ for some $a$. I claim that $W \oplus S = \mathbb R^3$.
To do this, first prove that $W \cap S$ is trivial i.e. consists only of the zero element. Then, prove that the bases(plural of basis) of $S$ and $W$, when put together, form a basis for $\mathbb R^3$. This is enough to show that $W \oplus S = \mathbb R^3$.
First, we show that their intersection is trivial. Suppose that $v \in W \cap S$. Then, we know that the components of $v$, which I call $v_1,v_2,v_3$ look like $[a+b-c,a-b+c,a]$ for some $a,b,c$. Furthermore, it also looks like $[d,d,0]$ for some $d$, since it belongs in $W$. 
Comparing the vectors $[a+b-c,a-b+c,a]$ and $[d,d,0]$, we get that $a=0$ by comparing the third components. By comparing the first two components, $a+b-c = d  = a-b+c$, so $b-c = c-b$, hence $c=b$, which gives $d=0$. Since $v = [d,d,0]$, we conclude that $v = 0$.
Finally, to see that any element of $\mathbb R^3$ is a sum of vectors from $W$ and $S$, note that:
$$
[r_1,r_2,r_3] = \left[r_3 + \frac{r_1}2 - \frac{r_2}2,r_3 - \frac{r_1}2 + \frac{r_2}2,r_3\right] + \left[\frac{r_1+r_2}{2} - r_3,\frac{r_1 + r_2}2 - r_3,0\right]
$$
where the first component is in $S$, with $a,b,c$ demarcated clearly, and the second component is in $W$. Hence, the direct sum follows.
Note that $W$ is not unique, however, as $T$ above could be non-zero on many different vectors, each of which generates a vector space complementary to $S$.
A: We note that all vectors in  $S$ may be expressed in the form 
$\begin{pmatrix} a - b +c \\ a + b - c \\ a \end{pmatrix} = a \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + (b - c) \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}; \tag{1}$
also, the two vectors occurring on the right of (1) are orthogonal,viz.:
$(1, 1, 1) \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = 1(-1) + 1(1) = 0; \tag{2}$
therefore, an easy way to find a one-dimensional subspace $T$ such that $\Bbb R^3 = S \oplus T$ is to choose a vector $v$ orthogonal to both $(1, 1, 1)^T$ and $(-1, 1, 0)^T$; I suggest taking their cross product:
$v = (1, 1, 1)^T \times (-1, 1, 0)^T = (-1, -1, 2)^T; \tag{3}$
we now set 
$T = \Bbb R v, \tag{4}$
and we are done!
