Invertiblity of the Derivative Matrix ?! (to use Inverse function Theorem)

Let $$p(x):=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$$ be the given polynomial, and assume that $\alpha\in{\Bbb R}$ is a simple root of $p$. Consider the auxiliary function $$f:\quad{\Bbb R}^{n+1}\to{\Bbb R},\qquad (\xi, u_{n-1}, u_{n-2},\ldots, u_0)\mapsto \xi^n+u_{n-1} \xi^{n-1}+\ldots+u_1\xi +u_0\ .$$ One has $$f(\alpha,a_{n-1},\ldots, a_0)=p(\alpha)=0\ ;$$ furthermore $${\partial f\over\partial\xi}(\alpha,a_{n-1},\ldots, a_0)=p'(\alpha)\ne0\ .$$ By the implicit function theorem it then follows that there is a $C^1$-function $$\psi:\quad(u_{n-1},\ldots,u_0)\to \xi:=\psi(u_{n-1},\ldots,u_0)\ ,$$ defined in a neighborhood $U$ of $(a_{n-1},\ldots, a_0)$, such that $\psi(a_{n-1},\ldots, a_0)=\alpha$, and that $$f\bigl(\psi(u_{n-1},\ldots,u_0),u_{n-1},\ldots, u_0\bigr)\equiv 0$$ in $U$. But this is saying that when the coefficients of the polynomial $p$ are slightly perturbed the resulting polynomial $p_\epsilon$ still has a zero in the immediate neighborhood of $\alpha$.

This argument can be applied to any single real root $\alpha_j$ of $p$, whence we are done.

The determinant of your matrix, $$ p = \det D_{\alpha_!,\dots,\alpha_n} \psi = \det \begin{pmatrix} \frac{\partial\psi_1}{\partial\alpha_1}&\cdots&\frac{\partial\psi_1}{\partial\alpha_n}\\ \vdots&\ddots&\vdots\\ \frac{\partial\psi_n}{\partial\alpha_1}&\cdots &\frac{\partial\psi_n}{\partial\alpha_n} \end{pmatrix} $$ is an alternating polynomial in $\alpha_1,\dots,\alpha_n$, since $$ \frac{\partial \psi_k}{\partial \alpha_i}(\alpha_1,\dots,\alpha_n) = \frac{\partial \psi_k}{\partial \alpha_j}(\alpha_1,\dots,\alpha_n) \quad\text{when $\alpha_i=\alpha_j$}. $$ Hence, the polynomials $(\alpha_i-\alpha_j)$ for $i<j$ all divide $p$, in fact the Vandermonde determinant $$ v = \prod_{1\le i<j\le n} (\alpha_j-\alpha_i) $$ divides $p$. The Vandermonde determinant is a homogeneous polynomial of degree $\binom{n}{2}$. Since the entries of your matrix in every row are homogeneous polynomials of the same degree, the determinant is also a homogenous polynomial of degree $$ (n-1) + (n-2) + \cdots + 0 = \binom{n}{2}. $$ Since $p$ is not the zero-polynomial we conclude $p$ and $v$ can only differ by a factor of degree $0$, so $p(\alpha_1,\dots,\alpha_n)=0$ if and only if $\alpha_i=\alpha_j$ for some $i<j$.