We are given a vector space $W=(x,y,z)\in\Bbb{R^3}:x+y+z=0$. I claim that the vectors $$u:=(0,-1,1)$$$$v:=(1,1,-2)$$
form a set which is a basis for $W$. To prove this, I believe it should be sufficient to show the equation ( $\alpha$ and $\beta$ are scalars ) $$\alpha u+\beta v=0$$
has only the trivial solution. This means the vectors are linearly independent. Now the confusion begins. My reasoning for this basis, is that since $u$ and $v$ are both in $W$ and they are $LI$, then certainly they span $W$ – that is : we can generate each and every vector of $W$ from these two vectors. Is this really enough to show these vectors span $W$? If so, then why is this the case? I don't really understand the concept of spanning. Also, my choice of vectors was completely arbitrary ( apart from the property of LI ) – how should I choose vectors in general when forming a basis of some vector space?
Best Answer
To begin, the span of a set of vectors is just the set of linear combinations of them. It's also the smallest vector (sub-)space that contains all of the vectors in question. A set of vectors spanning W need not be linearly independent (for example, using the vectors you defined in the question, $u, v$ and $u+v$ together also span $W$). Linearly independent sets of vectors that span a subspace form a basis for it, that for the most part, behaves very similarly to the standard basis for, say, $\mathbb{R}^n$: you can express any vector in the subspace in question as a unique linear combination of the basis vectors.
Essentially, selecting a basis is like selecting coordinates for the vector space. The axes you choose don't have to be perpendicular to one another, but they still allow you to uniquely represent every point in your space. It's very convenient if the axes for your coordinate system are also perpendicular to one another, and even more so if your basis vectors also have length 1, because then things look exactly like euclidean spaces... but that's not necessary for the question at hand (it's just to maybe help you develop some intuition).
There's a lot of freedom to choose a basis for a vector space. For example, in $\mathbb{R}^2$, any two non-zero vectors that are not parallel constitute a basis. Similarly, in your example, you could choose many other basis sets, but the ones you chose are perfectly fine!
I think the piece you're missing to show that the vectors you chose, $u$ and $v$, form a basis is just the statement that $\dim W = 2$ (if $\dim W=n$, you need $n$ linearly independent vectors to span it). Here's how the logic plays out: Since $u$ and $v$ are two vectors in $W$, they span a subspace of $W$ (i.e., ${\rm span} \{u,v\}\subset W$). Since they are two linearly independent vectors, the dimension of this subspace is 2. Lastly, because $\dim W=2$, the only subspace of $W$ that has dimension 2 is $W$ itself, so ${\rm span}\{u,v\} = W$.
An elementary way to see that $\dim W = 2$ is to note that $\dim W \ge2$ because you found two linearly independent vectors in it (namely, $u$ and $v$), but that there are vectors in $\mathbb{R}^3$ that are not in $W$ (e.g., $(1,0,0)$), so $\dim W < \dim \mathbb{R}^3 = 3$. This implies $\dim W = 2$.
Another, more involved, way to see that $\dim W =2$ is to use the rank-nullity theorem. To do that, we want to build a linear map $T: U\to V$ such that its kernel is $W$. $W$ is a subspace of $\mathbb{R}^3$, so the map will go from $U = \mathbb{R}^3$. Here's a A simple one with $V=\mathbb{R}$: \begin{equation} T\begin{pmatrix} x \\ y \\ z \end{pmatrix} = x + y + z\,. \end{equation} By the very definition of $W$, $\ker T = W$. Because $T$ is not the zero map, its rank must be no smaller than 1, and it cannot be greater than $\dim V = 1$, so ${\rm rank\,} T =1$. The rank-nullity theorem tells us that ${\rm rank\,} T + \dim \ker T = \dim U = 3$, so $\dim W = \dim \ker T=2$.
Hope this helps!