Where is the contradiction? My linear algebra book states the following:
Let $$a_0 + a_1 z + \dots + a_m z^m = 0.$$
(Where $z$'s signify polynomials). If at least one of the coefficients was non-zero, then there would be at most $m$ distinct values of $z$ that would satisfy this equation. Hence, by contradiction, all coefficients are equal to zero. I do not understand the reasoning here, I know the coefficients must be equal to zero, but can someone explain this specific proof?
 A: Alright, I'm assuming you are proving linear independence of the polynomials $1, z, z^2, \ldots, z^m$ over any infinite field (such as $\mathbb{C}$ or $\mathbb{Q}$ or $\mathbb{R}$). The definition of linear independence is that if a linear combination of these is 0, the coefficients all must equal 0. So you assume that they are not linearly independent. This means you assume there is a linear combination $a_0 + \ldots + a_m z^m$ which is 0 in your vector space.
But remember that being 0 in the space of polynomials means that it is the 0 function, which is 0 for every input. A nonconstant polynomial of degree m has at most m zeros, which is something you probably know (and can be proven fairly easily). So it can't possibly be 0 everywhere, since it is only 0 at finitely many points and there are infinitely many in your field. Thus, the polynomial must be a constant polynomial, and hence it is the 0 linear combination, i.e. each $a_i=0$.
Saying that it is a contradiction isn't really accurate, but many authors and mathematicians and students overuse proof by contradiction. Really, this proof doesn't have a contradiction the way I've presented it. If you want to formulate it with a contradiction, then make the assumption that there is a nonzero linear combination which gives you the 0 polynomial, and then show that this contradicts the fact that any such polynomial has finitely many zeros. But I find this method more roundabout and worse in style than the above, which actually makes no assumptions or contradictions (it is a so-called direct proof).
