TITLE: Find the span of a set of vectors
QUESTION [13 upvotes]: In general, what is the most straightforward way to find the span of a set of vectors? I'm trying to find the span of these three vectors:
$$\{[1, 3, 3], [0, 0, 1], [1, 3, 1]\}$$

REPLY [11 votes]: You're quite right that the span would be all vectors of the form $$[a+c,3a+3c,3a+b+c],$$ where $a,b,c$ are real. The question becomes how we can describe this using as few parameters as possible. The idea is to reduce your spanning set to a basis--that is, a spanning set that is linearly independent--by discarding superfluous vectors.
Let's suppose that $[a+c,3a+3c,3a+b+c]=[0,0,0]$ and see what we can determine about $a,b,c$. Well, since $a+c=0$, then we certainly have $c=-a$, and making the substitution $c=-a$ gives us $$[0,0,2a+b]=[0,0,0].$$ This holds exactly when $b=-2a$. Thus, $$a[1,3,3]-2a[0,0,1]-a[1,3,1]=[0,0,0],$$ regardless of what value of $a$ we choose, so your given spanning set is linearly dependent. Indeed, setting $a=1$, this means that $$[1,3,3]=2[0,0,1]+[1,3,1],$$ so the first vector is unnecessary to span the whole space, since it's a linear combination of the other two vectors. In particular, $$[a+c,3a+3c,3a+b+c]=(2a+b)[0,0,1]+(a+c)[1,3,1].$$
Thus, every vector in the generated space can be written in the form $$u[0,0,1]+v[1,3,1]=[v,3v,u+v]$$ for some real $u,v$. All that's left to do is confirm that $[0,0,1]$ and $[1,3,1]$ are linearly independent, which we can do by setting $[v,3v,u+v]=0$, and showing that this implies $u=v=0$.
One final step you can take is to prove that your span is the following: $$\bigl\{[x,y,z]\in\Bbb R^3:y=3x\bigr\}.$$