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 $\gamma$ : $$\boxed{(\gamma-\alpha)(\gamma-\beta)=\color{red}{\gamma^2}-(\alpha+\beta)\color{red}{\gamma}+\alpha\beta=0 \leftrightarrow \gamma=\alpha\vee\gamma=\beta\longrightarrow\bot}$$


2 Answers 2


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 $i<j$.

  • 1
    $\begingroup$ @FardadPouran I just learned about alternating polynomials and the Vandermonde determinant today, so this was a great coincidence ;-) $\endgroup$
    – Christoph
    Apr 17, 2015 at 15:35
  • $\begingroup$ Wow. Absolutely it's the amazing world of mathematics :) $\endgroup$ Apr 17, 2015 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.