We are interested in investigating the function $f: \mathbb R^n \to \mathbb R^n$ defined by $f(a_1, a_2, \dots , a _n) = (c_1, c_2, \dots, c_n)$ where $\prod_{i=1}^{n}(x-a_i) = x^n + \sum_{i=1}^{n}c_ix^{n-i}$.

In simple language, $f$ maps the roots of a monic $n$ degree polynomial to its coefficients.

Specifically we want to:

1) Show that $\text{rank } (D_f(a_1, \dots ,a_n))$ equals the number of distinct values among the $a_i$.

2) Calculate the Jacobian $\det (D_f(a _1, \dots , a _n))$

Hint: Start with $n = 2$ and $n = 3$.

I mentioned inverse function theorem at the title because this exercise showed up right after the proof of the IFT, so it makes sense they are related somehow, but I couldn't find a direct connection myself.

What I did:

I followed the tip. In the case of $n=2$ we have $(x-a_1)(x-a_2)= x^2+x(-a_1-a_2)+a_1 a_2$, so $f(a_1, a_2) = (-a_1 - a_2, a_1 a_2)$

And the Jacobi matrix is $\begin{pmatrix}-1 & -1 \\ a_2 & a_1\end{pmatrix}$ and its determinant the Jacobian is $a_2 - a_1$. Indeed it is full rank iff $a_1 \neq a_2$ so the rank is the number of distinct roots.

Did the same thing with $n=3$. Got Jacobi matrix $\begin{pmatrix}-1 & -1 & -1 \\a_2 + a _3 & a_1+a_3 &a_1 + a_2 \\ -a_2a_3 &-a_1a_3 &-a_1a_2\end{pmatrix}$ and Jacobian $(a_1^2+a_2a_3)(a_2-a_3)$ hopefully no miscalculation.

How do we continue from here? and how do we extrapolate this to $n$ dimensions? How can we use IFT?


Proposition 1. $Jac((a_i)_i)=\pm\Pi_{i<j}(a_i-a_j)$.

Proof. $Jac$ is a homogeneous polynomial of degree $1+2+\cdots+n=n(n+1)/2$ (to construct any term of a determinant, we choose an entry in each row). Clearly, each $(a_i-a_j)$ is a factor of $Jac$ (if $a_i=a_j$ then $Jac=0$); we deduce that $Jac=\lambda \Pi_{i<j}(a_i-a_j)$ where $\lambda$ is a constant.

We consider the leader term in $a_1$: in RHS, it's $\pm\lambda a_1^{n-1}\Pi_{1<i<j}(a_i-a_j)$; in LHS, it suffices to consider the principal $2,\cdots,n$ submatrix and to keep the terms that contain $a_1$; we obtain $\pm a_1^{n-1}Jac(a_2,\cdots,a_n)$. Reasoning by recurrence, we conclude that $\lambda=\pm 1$.


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