I would like to calculate a basis for the factor (quotient) space of $\mathbb{R}^{4}/span(\vec{z})$ where $\vec{z}=(1,-1,-1,0)^{T}$.
What I have:
If x , y $\in \mathbb{R}^{4}$ are in the same class, we have that $\pi x = \pi y$ $~$ (where $\pi:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}/span(\vec{z})$ is the projection into the factor space). And so we have $\pi (x-y) = \bar{0}$ which imply that $x-y \in span(\vec{z})$. At this point I can infer that for 2 vectors $x,y$ to be in the same class, they have to have the same 4th component, this give me a vector for the basis, namely $b_1=(0,0,0,1)^{T}$. I know that the factor space has to have dimension 3 since $dim~(span(\vec{z})) + dim~(\mathbb{R}^{4}/span(\vec{z}))= 4$. I could give me any other 2 vectors l.i to $b_1$ and $\vec{z}$ and that should be theoretically enough.
Am I right? Is there a more clever way to do it? Could you pls give some feedback?
Cheers!
Whatever you have done is OK. But the null space of $(1,-1,-1,0)$ of dimension three would directly give you the basis of the factor space in your question. Thus, the vectors $(0\ 0\ 0\ 1)^T$, $(1\ 1\ 0\ 0)^T$ and $(1\ 0\ 1\ 0)^T$ would form a basis of your factor space.