would any one explicitly and precisely help me to understand how the map from the set consists of monic complex polynomials with degree n to the set consists of n complex numbers without order by mapping each polynomial to its roots is continuous? if I understand this fact then i will be able to prove this one : Let X be a compact subset of $GL_n(\mathbb{C})$ and Y= set of all eigenvalues of matrices in X, then Y is compact in $\mathbb{C}$
A problem to prove a map is continuous
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0I dont understand anything, waiting patiently for answer. – 2012-04-20
1 Answers
Here's an approach that you may not like, ’cause it’s much more algebraic than analytic, and it works only for the open set in the domain corresponding to $n$-tuples of distinct complex numbers.
Consider the reverse map $(\rho_1,\rho_2,\cdots,\rho_n)\mapsto (S_1(\rho),S_2(\rho),\cdots,S_n(\rho))$, where the polynomials $S_i$ are the elementary symmetric functions, $S_j$ is the sum of all the possible products of $j$ of the roots $\rho_i$. In particular, $S_1$ is the sum of all the roots, and $S_n$ is the product of them all. So you know that the polynomial whose roots are the $\lbrace\rho_i\rbrace$ is $X^n - S_1X^{n-1}+\cdots+(-1)^nS_n$.
Now look at this as a map from $n$-space to $n$-space, defined by polynomials, and thus as differentiable as you could like. What is the Jacobian determinant of this map? This requires a proof, but up to sign, it’s the square root of the discriminant, namely \prod_{i
This has, in fact, nothing to do with real or complex analysis: I first made my own use of this in the context of $p$-adic analysis.
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0@Makuasi, the pleasure is mine. I hope that you are familiar with the connection between the roots of a polynomial and the elementary symmetric polynomials. This should be described in any basic Algebra text. If you haven’t seen this relationship, try it out for a quadratic polynomial (I know you've seen this!) and for a cubic. That will likely persuade you of the truth of the general fact, before you work out a proof for yourself. I should also say that the set of unordered $n$-tuples of complex numbers is definitely *not* ${\mathbb{C}}^n$: this latter is the set of ordered $n$-tuples. – 2012-04-22