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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?2011-09-18
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    @Arturo: sorry. z are polynomials of a certain degree, scalars are either real or complex (I think the book says it doesn't matter).2011-09-18
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    @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?2011-09-18
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    @anon: here's the precise quote: "If at least one of the coefficients $a_0, a_1, \dots, a_m$ were nonzero, then 2.3 [the equation] could be satisfied by at most $m$ distinct values of z; this contradiction shows that all the coefficients in 2.3 equal 0." I think it's just polynomials taking on values.2011-09-18
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    @linalconfused: Okay, so **what is it exactly** that "at most $m$ values" contradicts? That's the piece you aren't telling us.2011-09-18
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    @anon: that's what I am asking you, because I don't understand the contradiction. :)2011-09-18
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    @linalconfused: The problem is **context**: there is no context in what you are quoting. If you say the $z$s are polynomials, then this would be a polynomial expression in a *polynomial*, which seems unlikely. Is there any way you can post the *entire* context, or link to a scan of the page or something?2011-09-18
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    It would be very unusual to talk about "distinct values of z" if z is a polynomial. If z is a variable or indeterminate, this sounds like a proof that the powers of an indeterminate are linearly independent over the field of scalars, but it's really hard to tell without more context.2011-09-18
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    @Arturo: I wouldn't be able to scan right now, but initially the book introduces $p(z)=a_0+\dots+a_m z^m$ as a polynomial of a degree $m$, then talks about scalars in $\mathbb{F}$ (R or C) and gives the equation I presented above. The chapter is on span and linear independence (it's in the very beginning). So I guess z's are parts of the polynomial? Not sure how to say it in English.2011-09-18
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    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.2011-09-18
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    @linalconfused: The piece you weren't telling us was that the equation was, by hypothesis, supposed to hold for *all* $z$. :)2011-09-18

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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.2011-09-18