Why is the coefficient of the $(n-1)$st degree term of a polynomial of degree $n$ the sum of its roots (up to sign)?

Question

My professor gave the following fact:

$$p(z)=(z-\zeta_0)\cdots(z-\zeta_{n-1})=z^n+(-1)^n(\zeta_0+\cdots+\zeta_{n-1})z^{n-1}+\cdots$$

It is fairly easy to see this for polynomials of a particular degree by expanding its factorization as a product of its roots. For example, for a cubic: $$(z-\zeta_0)(z-\zeta_1)(z-\zeta_2)=(z^2-\zeta_0z-\zeta_1z+\zeta_0\zeta_1)(z-\zeta_2)$$ $$=z^3-\zeta_0z^2-\zeta_1z^2-\zeta_0\zeta_1z-\zeta_2z^2+\zeta_0\zeta_2z+\zeta_1\zeta_2z-\zeta_0\zeta_1\zeta_2$$ $$=z^3-(\zeta_0+\zeta_1+\zeta_2)z^2+\cdots$$

However, I am not sure how to prove this for an arbitrary degree $n$. Any hints? I would like to use this fact as a lemma to show that the roots of the cyclotomic polynomial $z^n-1$ sum to zero.

Answer 1

Consider literally factoring the polynomial where $r_i$ are it's roots:

$$ P = (x-r_1)(x-r_2)...(x-r_n)$$

What happens when you take its product? We we end up with a sum of an $x^n$ term an $x^{n-1}$ term etc... down to a constant term.

Now here is a procedure for expanding the product: For each term $(x-r_i)$ make a decision of one of the two choices "$x$" or "root". Then form a product of $x$ and the negative of the roots for each combination of decisions that you made. The sum of ALL these "sub products" IS then equal to the expanded version of the polynomial.

Let's try a concrete example:

$$ P = (x-1)(x-2)$$

I can consider $(x-1)$-"x", $(x-2)$-"x" so my product is $x \times x = x^2$

I can consider $(x-1)$-"x", $(x-2)$-"root" so my product is $x \times (-2) = -2x$

I can consider $(x-1)$-"root", $(x-2)$-"x" so my product is $(-1) \times x = -x$

I can consider $(x-1)$-"root", $(x-2)$-"root" so my product is $(-1) \times (-2) = 2$

The sum of all the sub products calculated thus far is: $x^2 + (-2x) + (-x) + (2) = x^2 - 3x +2$. Confirm by hand using FOIL that $(x-1)(x-2) = x^2 - 3x + 2$.

So in general when you have a product of $n$ terms of the form $(x-r_i)$ that is $P = (x-r_1)(x-r_2)...(x-r_n)$ then there are $2^n$ possible ways to a unique list of "root,x,root,root,x,x,x...." to form sub products with. Don't worry if you can't prove the $2^n$ fact its just a counting exercise but not key to the insight.

Now the beautiful thing if you made it this far, is the final $x^{n-1}$ term must be a sum of the all the sub-products you can make using the procedure above that contain an $x^{n-1}$ term.

Concretely the final $-3x$ in $x^2 - 3x + 2$ from expanding $(x-1)(x-2)$ is made by considering the sum $(-2x) + (-x)$.

But these $x^{n-1}$ sub products can only be made in one way, that is by making the decision:

"root", "x", "x", "x" ... "x" (rest are x)

or

"x", "root", "x", "x" ... "x" (rest are x)

or

"x", "x", "root", "x" ... "x" (rest are x)

That is: by making decisions of ALL x's and ONE root.

So your subproducts will be of the form $-x^{n-1}r_i$ and collecting all these subproducts yields

$$ -{r_1}x^{n-1} - {r_2}x^{n-1} - {r_3}x^{n-1} ... = -(r_1 + r_2 + r_3 ... r_n) x^{n-1}$$

after simplifying.

And that's intuitively whats going on. If you found this interesting I recommend checking out:

https://en.wikipedia.org/wiki/Vieta%27s_formulas

Answer 2

It follows from the Fundamental Theorem of Algebra that any monic polynomial over $\Bbb C$ (or any algebraically closed field) can be factored as $$z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = (z - \zeta_0) \cdots (z - \zeta_{n - 1}) ,$$ where $\zeta_1, \ldots, \zeta_{n - 1}$ are the roots, including multiplicity. If we expand the right-hand side the terms that will contribute to the $z^{n - 1}$ term are precisely those with $n - 1$ factors of $z$ and $1$ factor $\zeta_i$, namely, those of the form $-\zeta_i z^{n - 1}$, and there is one of these for each $i$. So, $$z^n + a_{n - 1} z^{n - 1} + \cdots = z^n + (-\zeta_0 - \cdots - \zeta_{n - 1}) z^{n - 1} + \cdots ,$$ and comparing coefficients gives that $$a_{n - 1} = -(\zeta_0 + \cdots + \zeta_{n - 1})$$ as claimed.

Remark Instead comparing constant terms shows, for example, that $$a_0 = (-1)^n \zeta_0 \cdots \zeta_{n - 1} ,$$ that is, up to sign the constant term of the polynomial is the product of the roots. For a general coefficient $a_i$, we get (up to sign) the elementary symmetric polynomials in the roots. The identities for $a_{n - 1}, \ldots, a_0$ we produce this way are Vieta's Formulas.

Answer 3

If you want you can prove it easily by induction:

$(x-a)\left(x^r-bx^{r-1}+p_{r-2}(x)\right)=x^{r+1}-(a+b)x^r+p_{r-1}(x)$ where $p_k(x)$ is a polynomial of degree at most $k$.

The base case is obviously true.

The $(-1)^n$ factor does not appear for this coefficient - but it is required for the constant coefficient, which is numerically the product of the roots.

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Of course the roots may not be elements of the field under consideration, but the field can (it turns out) always be extended to a context in which the roots do exist. The coefficients, which are symmetric functions of the roots, are all elements of the original field. It therefore makes sense to reason about the roots and their functions.

Answer 4

Consider the polynomial: $$p(z)=(z-\zeta_0)\cdot(z-\zeta_1)\cdot(z-\zeta_2)\cdots(z-\zeta_{n-3})\cdot(z-\zeta_{n-2})\cdot(z-\zeta_{n-1})$$ The easiest way to calculate $z^{n-1}$ is multiply $n-1$ terms $z$ and a $\zeta_k$, obtaining a polynomial in the form $z^{n-1}\zeta_k$. You can do this in $n-1$ different ways, obtaining $n-1$ polynomial: $$\sum_{i=0}^{n-1}z^{n-1}\zeta_i$$ Picking up a $z^{n-1}$, I obtain: $$z^{n-1}\sum_{i=0}^{n-1}\zeta_i$$