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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?

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There's a lot of context missing! What are $z$? What is the context of this statement? – Arturo Magidin Sep 18 '11 at 5:04
@Arturo: sorry. z are polynomials of a certain degree, scalars are either real or complex (I think the book says it doesn't matter). – linalconfused Sep 18 '11 at 5:05
@Arturo: Based on the flow of reasoning I'd say the context is that there are not "at most m distinct values of $z$ that would satisfy this equation," which probably means the hypothesis is that the polynomial is always zero on some interval or for an infinite number of $z$ or something. / Edit: linalconfused - wait, in the OP you say $z$ is a value and in your comment you say there are more than one $z$ and that they are polynomials. Care to clarify? – anon Sep 18 '11 at 5:08
@anon: that's what I am asking you, because I don't understand the contradiction. :) – linalconfused Sep 18 '11 at 5:13
Thanks to Google, I found that linalconfused is working from Linear Algebra Done Right, page 24. For the future, linalconfused, it would have been easier to help you if you had copied the entire paragraph or provided a book name and page number. – Adam Saltz Sep 18 '11 at 5:22
up vote 5 down vote accepted

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).

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This made perfect sense, thanks so much! Of course, 0 maps everything to 0, and this thing can map only $m$ things at most. – linalconfused Sep 18 '11 at 5:19

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