So I have this:
$$ F = \{(x_1, x_2, x_3) : x_1 - x_2 + x_3 = 0\} $$
I'm having trouble finding out the vectors whose linear combination generate the given subspace...
Basically I need something like: $x_1(a_1,b_1,c_1) + x_2(a_2,b_2,c_2) + x_3(a_3,b_3,c_3)$ so that given any value for $x_1, x_2$ and $x_3$ the equation gives me a vector in $F$ and so that each of the vectors above is in $F$
I can't find any ...
On
$F$ is the solution space of the set of linear equations $$x_1-x_2+x_3=0.$$
We always solve sets of linear equations by Gauss elimination, because it is guaranteed to find the complete set of solutions. In this case there is no work to be done, the equations -- the equation ! -- is already in row reduced form. So we can read off the solutions, they are $$x_3=\lambda,\ x_2=\mu,\ x_1=-\lambda+\mu\ \text{ for arbitrary} \lambda,\mu$$ and hence $$(x_1,x_2,x_3)=\lambda(-1,0,1)+\mu(1,1,0).$$
Gauss elimination guarantees not only do these two vectors $$(-1,0,1),\ \text{and}\ (1,1,0)$$span $F$, they actually form a basis of $F$.
On
we have $F=\{(x_1, x_2, x_3) : x_1 - x_2 + x_3 = 0\}=\{(x_1, x_2, x_3) : x_1 + x_3 =x_2 \}$ it means: $$\begin{align} x \in F&\leftrightarrow x=(x_1,x_1+x_3,x_3)\\ &\leftrightarrow x =(x_1,x_1,0)+(0,x_3,x_3) \\ &\leftrightarrow x=x_1(1,1,0) + x_3(0,1,1) \\ &\leftrightarrow x \in \langle(1,1,0),(0,1,1) \rangle\end{align}$$ therefore the vectors $(1,1,0)$ and $(0,1,1)$ generate $F$
Note that if you know $x_1$ and $x_2$, then $x_3$ is uniquely determined. (So, this is a two-dimensional space.)
You can write $x_3=x_2-x_1$. Therefore, any such vector can be written as $$ \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}x_1\\x_2\\x_2-x_1\end{bmatrix}=\begin{bmatrix}x_1\\0\\-x_1\end{bmatrix}+\begin{bmatrix}0\\x_2\\x_2\end{bmatrix}=x_1\begin{bmatrix}1\\0\\-1\end{bmatrix}+x_2\begin{bmatrix}0\\1\\1\end{bmatrix} $$ So, a reasonable choice of basis is the two vectors $\langle 1,0,-1\rangle^T$ and $\langle0,1,1\rangle^T$.