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In the Analysis2 midterm exam, we had the following problem:

Let the equation $a_nx^n+\cdots+a_1x+a_0=0$ has $n$ simple real roots (distinct) $\{\alpha_1,\cdots,\alpha_n\}$. Prove that the above equation has still $n$ distinct real roots when the change in coefficients is small enough !

I'm pretty sure that $(a_1,\cdots,a_n,\alpha)\mapsto a_n\alpha^n+\cdots+a_1\alpha+a_0$ plus implicit function theorem will work. But it didn't came to my mind.

Instead, I thought that the coefficients are $C^\infty$ function of the roots by Vietta Theorem. So I hoped the map $\psi:(\alpha_1,\cdots,\alpha_n)\mapsto(a_0,\cdots,a_{n-1})$ has a full rank derivative at the current roots and start to apply Inverse function Theorem to conclude that, locally, $(\alpha_1,\cdots,\alpha_n)$ is $C^\infty$ diffeomorphism map of $(a_1,\cdots,a_n)$.

Hence, I conjectured the following proposition :

Conjecture. We know by Vietta's Theorem that : $$\left\{\begin{array}{ll} \psi_1=a_0=(-1)^n \alpha_1\cdots\alpha_n\\ \psi_2=a_1=(-1)^{n-1} \displaystyle\sum_{r=1}^n \alpha_1\cdots\hat{\alpha_r}\cdots\alpha_n\\ \vdots\\ \psi_{n-1}=a_{n-2}=\displaystyle\sum_{i,j=1}^n \alpha_i\alpha_j\\ \psi_{n}=a_{n-1}=-(\alpha_1+\cdots+\alpha_n) \end{array}\right.$$

Then the matrix $$D_{(\alpha_1,\cdots,\alpha_n)}\psi=\left[\begin{matrix} \frac{\partial\psi_1}{\partial\alpha_1}&\cdots&\frac{\partial\psi_1}{\partial\alpha_n}\\ \vdots&\ddots\\ \frac{\partial\psi_n}{\partial\alpha_1}&&\frac{\partial\psi_n}{\partial\alpha_n} \end{matrix}\right]= \left(\begin{matrix} (-1)^n\alpha_2\cdots\alpha_{n}&(-1)^n\alpha_1\alpha_3\cdots\alpha_{n}&\color{red}{\cdots}&(-1)^n\alpha_1\cdots\alpha_{n-1}\\ \color{red}{\vdots}&\color{red}{\ddots}&\color{red}{\vdots}\\ \alpha_2+\cdots+\alpha_{n}&\alpha_1+\alpha_3+\cdots+\alpha_{n}&\color{red}{\cdots}&\alpha_1+\cdots+\alpha_{n-1}\\ -1&-1&\color{red}{\cdots}&-1 \end{matrix}\right)$$ is Invertible, whenever $\alpha_j$s are pairwise distinct.

For case $n=2$ and $n=3$, I prove that $\det\big( D_{(\alpha_1,\cdots,\alpha_n)}\psi\big)=0$ if and only if $\alpha_i=\alpha_j$ for some $i\neq j$.

But I don't know what to do for general $n$. Is it a famous matrix ? Is this conjecture correct for general $n$?


Proof for $n=2$ and $n=3$ :

n=2 :$\quad D_{(\alpha,\beta)}\psi=\left[\begin{matrix} \beta&\alpha\\ -1&-1 \end{matrix}\right]$. So, $\boxed{\det(D_{\alpha,\beta}\psi)=0 \leftrightarrow \alpha=\beta\rightarrow\bot}$


n=3$ :\quad D_{(\alpha,\beta,\gamma)}\psi=\left[\begin{matrix} -\beta\gamma&-\alpha\gamma&-\alpha\beta\\ \beta+\gamma&\alpha+\gamma&\alpha+\beta\\ -1&-1&-1 \end{matrix}\right] $. Now by computing determinant respect to the last row:

$\begin{align} \det(D_{(\alpha,\beta,\gamma)}\psi)= +\big[-\alpha^2\gamma-\alpha\beta\gamma+\alpha^2\beta+\alpha\beta\gamma\big]&-\big[-\alpha\beta\gamma-\beta^2\gamma+\alpha\beta^2+\alpha\beta\gamma\big]\\ &-\big[-\alpha\beta\gamma-\beta\gamma^2+\alpha\beta\gamma+\alpha\gamma^2\big] \end{align}$. $$\Rightarrow\det(D_{(\alpha,\beta,\gamma)}\psi)= [\alpha-\beta]\color{red}{\gamma^2}+[\beta^2-\alpha^2]\color{red}{\gamma}+[\alpha\beta(\alpha-\beta)] $$ Now the equation $\det(D_{(\alpha,\beta,\gamma)}\psi)=0$, while $\alpha\neq\beta$, becomes the following quadratic equation respect to $\gamma$ : $$\boxed{(\gamma-\alpha)(\gamma-\beta)=\color{red}{\gamma^2}-(\alpha+\beta)\color{red}{\gamma}+\alpha\beta=0 \leftrightarrow \gamma=\alpha\vee\gamma=\beta\longrightarrow\bot}$$

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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 $i<j$.

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  • 1
    $\begingroup$ @FardadPouran I just learned about alternating polynomials and the Vandermonde determinant today, so this was a great coincidence ;-) $\endgroup$ – Christoph Apr 17 '15 at 15:35
  • $\begingroup$ Wow. Absolutely it's the amazing world of mathematics :) $\endgroup$ – Fardad Pouran Apr 17 '15 at 15:36
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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.

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