Find a basis for the subspace of $\mathbb{R}^4$ spanned by the given vectors. 
(1,1,-4,3), (2,0,2,-2), (2,-1,3,2) 

My question is. how do you know how to make these row vectors and find the rowspace or make them column vectors and find the columspace? Will you get the correct answer regardless? The book made these row vectors.  
Is it something to do with having more unknowns than equations? or is it simply choosing these as row vectors makes less rows than if we made column vectors and so there will be less row operations to do and, hence less working out. 
I tried doing it as column vectors to see if it gives the same result using column space and I didn't get the same answer as the book finding rowspace of these three row vectors. As  you probably expect because the columnspace would be using the original columns.
Thanks.
 A: If you’re using row-reduction to find these row and column spaces, you should expect to get different results. If you use these vectors as the columns of a matrix, the pivots in the rref will give you a linearly-independent subset of these vectors that spans the same space. If you use them as rows, you will generally get a different set of linearly-independent vectors that span the same space.  
In this case, using them as columns we get $$\left[\begin{array}{rrr}1&2&2\\1&0&-1\\-4&2&3\\3&-1&2\end{array}\right]\to\left[\begin{array}{rrr}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}\right]$$ so the vectors are linearly independent and so form a basis for their span. Using them as rows, we get $$\left[\begin{array}{rrr}1&1&-4&3\\2&0&2&-1\\2&-1&3&2\end{array}\right]\to\left[\begin{array}{rrr}1&0&0&\frac98\\0&1&0&-\frac{37}8\\0&0&1&-\frac{13}8\end{array}\right]$$ which gives us $(8,0,0,9)^T$, $(0,8,0,-37)^T$ and $(0,0,8,-13)^T$ as a basis of the span of the original vectors.  
Clearly, either way will produce an answer, but to get the one in the book, you’ll have to do it whichever way is explained in the book.
A: The vectors themselves form a base. All you need to do is to show that they are linearly independent. Begin with the equation:
$$
a(1,1,-4,3) + b(2,0,2,-2) + c(2,-1,3,2) = (0,0,0,0)
$$
If the only solution to this equation is $a = b = c = 0$, then the vectors are linearly independent. Notice the 0 in the second component of the second vector. This gives us the equation: $1a+0b-1c=0 \iff a = c$. So replace the $c$ with $a$ in the first equation:
$$
a(1,1,-4,3) + b(2,0,2,-2) + a(2,-1,3,2) \\
= a((1,1,-4,3) + (2,-1,3,2)) + b(2,0,2,-2) \\
= a(3,0,-1,5) + b(2,0,2,-2) = (0, 0, 0, 0)
$$
Now the third component of the LHS vectors gives us $-a+2b=0$  and the fourth $5a-2b=0$. It is trivial to see that the only solution to those two equations are $a = b = 0$. Ergo, your vector set is a base.
