A projective space is not a vector space, correct? At a minimum, there is no additive identity in a projective space.
So can you even have a basis of a projective space? $[1 0 0], [0 1 0], [0 0 1]$ might work for $P^3$. Or are $[1 0 0]$ and $ [0 1 0]$ not linearly independent since they both lie on the line at infinity?
(is there a "too much" span problem with $[0 0 0]$ as well?)
If not, is there a roughly analogous concept for a projective space?
Perhaps it would be illustrative to demonstrate what happens with $\mathbb P^1$ (say over the real numbers). You might think about this as the set of lines in $\mathbb R^2$. Now, rather than thinking in terms of numbers, draw a picture or two. What could it reasonably mean to "add" two lines in $\mathbb R^2$ and expect to get a line back?
Perhaps a more convincing argument is this. Remember that a vector space is not just saying "hey I have a basis". It needs to remember that its a group. So in particular, you need an identity. You've thrown out $(0,0)$ remember, this is not an element of $\mathbb P^1$. But now anything you pick is going to run into the same problem that the comments as above described. You cant write down a meaningful statement like: $(a:b)+id = (a:b)$ and expect it to be true, because $id$ could be scaled to mess up your arithmetic.
See the comments above for other examples.