Uniqueness of the roots of a polynomial equation

[maverick280857]

Hi,

I have a question, which seems deceptively simple to me, but when I thought about it, I couldn't really come up with a rigorous proof. Here goes,

Are the roots of a polynomial equation unique?

Suppose we have a general monic polynomial equation:

$$z^{n} + c_{1}z^{n-1} + c_{2}z^{n-2} + \ldots + c_{n} = 0$$

where $z \in \mathbb{C}$ and $c_{1}, c_{2}, \ldots$ are real coefficients. Now let S be the set of all roots of this equation (counted according to their multiplicities so that if x is a root of multiplicity n, then S contains n occurrences of x. Strictly S is not a set, but rather a list, but you see my point.)

Can we find some $S' \neq S$ such that every member $z^{'} \in S'$ is a root of the above equation (and $z^{'} \notin S$)? Stated differently, does this equation have two different sets of roots?

The picture I have in mind for real roots is as follows: if the roots were not unique, the zero crossings of the function on the left hand side of the equation would not be unique, which is impossible. This reasoning does not work for complex roots :-(

Any ideas?

Thanks in advance.
-Vivek

 

[rasmhop]

Hi,
Can we find some $S' \neq S$ such that every member $z^{'} \in S'$ is a root of the above equation (and $z^{'} \notin S$)? Stated differently, does this equation have two different sets of roots?

No. If it had then just take an element z of S'. z is a root so it must be in S since S is the set of all roots.

In general if you let a set T be all elements with some property, then you can't find an element with the property, but not in T.

One small note is that if you allow the empty set, then the answer is yes since $\emptyset \not= S$, but I'm assuming this isn't what you meant by a set S'.

 

[maverick280857]

No. If it had then just take an element z of S'. z is a root so it must be in S since S is the set of all roots.

In general if you let a set T be all elements with some property, then you can't find an element with the property, but not in T.

One small note is that if you allow the empty set, then the answer is yes since $\emptyset \not= S$, but I'm assuming this isn't what you meant by a set S'.

Right. I get your point. Now, how does one prove that the roots of a n-th order polynomial equation (with say, real coefficients) are unique? Is my set theoretic statement of the problem equivalent to this question? If not, how does one sketch a proof?

 

[rasmhop]

I'm not really sure what you don't understand. Given a polynomial $f \in \mathbb{C}[x]$ a complex number z is a root in f iff f(z)=0. There is no ambiguity. Thus two sets of roots can differ if they don't contain all roots. For instance for the polynomial (x-2)(x-3)(x-4) we may let S={2} and S'={3,4} and these are different sets of roots, but if we form the set of all roots S''={2,3,4}, then we can't find another set of all roots because it would have to contain all elements of S'' and can't contain any more.

As for your actual statement, then yes we can find such an S' because we can let $S'=\emptyset$. You can find no other S' because you said S consists of exactly all roots, so you can't find a different set that consists of exactly all roots. If we had a non-empty set S' as you state, then $z \in S'$ would imply f(z) =0 since S' consists of roots, and since S consists of all roots $z \in S$.

I'm sorry if this wasn't what you wanted, but I find your question a bit confusing. It seems to me that your question is of the same kind as:
Is the set {1,2} unique.
Since it's specified completely, then of course yes. In the same sense the set of roots is specified completely so of course it's unique.

 

[elibj123]

Perhaps you are trying to prove the fundamental theorem of algebra, and you don't explain yourself correctly?

 

[maverick280857]

Yes, I realized its not phrased properly. Someone asked me the following question (I'm citing an example to make things simple):

Suppose we're given the equation $x^{2}-5x+6 = 0$. We know the roots are x = 2 and x = 3. The question is: can this equation have some other set of 2 roots? Of course, this particular example is absolutely trivial because the "uniqueness" is proved by writing

$x^{2}-5x+6 = (x-2)(x-3)$

For a general scenario, the question is that for an arbitrary positive integer n, if

$$x^{n} + c_{1}x^{n-1} + \ldots + c_{n} = (x-\alpha_{1})(x-\alpha_{2})\cdots(x-\alpha_{n})$$

so that $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ are the roots, then is it possible to find some set $\beta_{1}, \beta_{2}, \ldots, \beta_{n}$ of numbers such that

$$x^{n} + c_{1}x^{n-1} + \ldots + c_{n} = (x-\beta_{1})(x-\beta_{2})\cdots(x-\beta_{n})$$

and none of the $\beta$'s are from $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$.

In other words, given the sum of roots taken one at a time, two at a time, three at a time, and so on.., does the following system of equations have a unique solution in $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$?.

$$\sum_{i=1}^{n}\alpha_{i} = -c_{1}$$
$$\sum_{i\neq j}^{n}\alpha_{i}\alpha_{j} = c_{2}$$
$$\cdots$$
$$\sum_{i_{1}\neq i_{2}\neq i_{3}\cdots\neq i_{n}}^{n}\alpha_{i_{1}}\alpha_{i_{2}}\ldots\alpha_{i_{n}}= (-1)^{n}c_{n}$$

I want to be able to give a rigorous proof to the uniqueness of the solution to this nonlinear system.

 

[rasmhop]

Ok suppose,
$$f(x) = (x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n) = (x-\beta_1)(x-\beta_2)\cdots(x-\beta_n)$$
and that $\alpha_1 \notin \{\beta_1,\ldots,\beta_n\}$. Now evaluate $f(\alpha_1)$.
$$f(\alpha_1) = (\alpha_1-\alpha_1)(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0$$
Thus we have,
$$0 = f(\alpha_1) = (\alpha_1-\beta_1)(\alpha_1-\beta_2)\cdots(\alpha_1-\beta_n)$$
We know that in $\mathbb{C}$ a product is 0 if and only if one of the factors is 0. Let i be an integer in {1,...,n} such that the factor $(\alpha_1-\beta_i)$ is 0 (we know that such an i exists since at least one factor is 0). Then we get $\alpha_1 = \beta_i$ which is a contradiction.

 

[maverick280857]

Ok suppose,
$$f(x) = (x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n) = (x-\beta_1)(x-\beta_2)\cdots(x-\beta_n)$$
and that $\alpha_1 \notin \{\beta_1,\ldots,\beta_n\}$. Now evaluate $f(\alpha_1)$.
$$f(\alpha_1) = (\alpha_1-\alpha_1)(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0$$
Thus we have,
$$0 = f(\alpha_1) = (\alpha_1-\beta_1)(\alpha_1-\beta_2)\cdots(\alpha_1-\beta_n)$$
We know that in $\mathbb{C}$ a product is 0 if and only if one of the factors is 0. Let i be an integer in {1,...,n} such that the factor $(\alpha_1-\beta_i)$ is 0 (we know that such an i exists since at least one factor is 0). Then we get $\alpha_1 = \beta_i$ which is a contradiction.

Thats neat. Thanks...so stupid of me not to get it. :-(

 

[HallsofIvy]

I don't want to embarass your further but I think it is even more fundamenental than that! (And it is true of all equations, not just polynomial equations.)

"Roots of an equation", f(x)= 0, means numbers, a, such that f(a)= 0. If you have two different purported sets of roots such that a is in one but not the other, one of them must be wrong. Either f(a)= 0 or not! If f(a) is equal to 0, then the set that does not include a is NOT a "set of roots". If f(a) is not equal to a, then the set that does include it is NOT a "set of roots".

But I think I see the cause of your confusion- there are many different ways of solving equations and you were, I suspect, thinking of "roots" as "what you get using this particular method" so, conceivably, different methods might give you different "roots".

Hopefully, if your methods are correct, that won't happen because "roots" are NOT just "what you get using this method."

Many years ago, I had a student in a basic algebra class complain bitterly that I had marked her wrong on a particular test problem just be she "used a different method" than I had taught in class. I don't believe I ever convinced her that I marked her wrong becuase her solution did not satisfy the equation!