Let $f:\mathbb R^n \to \mathbb R^n$ such that $f$ maps roots of a polynomial to its coefficients.

Meaning: if $(x-x_1)(x-x_2)...(x-x_n)=x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_n$ then $f\begin{pmatrix} x_1 \\x_2\\x_3\\ \vdots \\x_n \end{pmatrix} =\begin{pmatrix} a_1\\a_2\\a_3 \\ \vdots \\a_n \end{pmatrix}$

Show that the rank of the differential of $f$ is equal to the number of different roots. $rank(Df)=cardinal\{x_1,...x_n\}$

  • $\begingroup$ Hi Oria. I edited my answer. The previous one was completely wrong. I believe the answer is ok now. $\endgroup$
    – Daniel
    Commented Dec 11, 2014 at 22:31

1 Answer 1


Let $g(x,x_1,\ldots,x_n)=(x-x_1)\ldots(x-x_n)=x^n+\sum_{i=1}^{n}a_ix^{n-i}$.

Now $\frac{\partial g}{\partial x_i}=-(x-x_1)\ldots(x-x_{i-1})(x-x_{i+1})\ldots(x-x_n)=\sum_{j=1}^{n}\frac{\partial a_j(x_1,\ldots,x_n)}{\partial x_i}x^{n-j}$. $(1)$

Now, let $r_1,\ldots,r_n\in\mathbb{R}$. Let $\alpha=\{1,x,x^2,\ldots,x^{n-1}\}$.

Now the column $i$ of the jacobian matrix of the function $f(x_1,\ldots,x_n)=(a_j(x_1,\ldots,x_n))_{j=1}^n$ on the point $(r_1,\ldots,r_n)$ is equal to $\left[\frac{\partial a_j(r_1,\ldots,r_n)}{\partial x_i}\right]_{j=1}^n$. By $(1)$, this column is the coordinates of the polynomial $-(x-r_1)\ldots(x-r_{i-1})(x-r_{i+1})\ldots(x-r_n)$ on the basis $\alpha$.

Now, let us assume that the cardinality of $\{r_1,\ldots,r_n\}$ is $s$ (there are only $s$ distinct $r_i$'s). Thus, in order to solve the problem we must prove that the polynomials $p_1(x),\ldots,p_n(x)$ span a $s-$dimensional vector space, where $p_i(x)=-(x-r_1)\ldots(x-r_{i-1})(x-r_{i+1})\ldots(x-r_n)$.

Without loss of generality suppose $r_1,\ldots,r_s$ are distinct. Thus, for every $j>s$ exists $i\leq s$ such that $r_j=r_i$. Let $p(x)=(x-r_1)\ldots(x-r_n)$ and notice that $p_i(x)=-\dfrac{p(x)}{(x-r_i)}$.

Thus, for every $j>s$ exists $i\leq s$ such that $p_j(x)=-\dfrac{p(x)}{(x-r_j)}=-\dfrac{p(x)}{(x-r_i)}=p_i(x)$. Therefore, we must prove that $p_1(x),\ldots,p_s(x)$ are l.i.

Now, $p(x)=(x-r_1)^{n_1}\ldots(x-r_s)^{n_s}$ and $p_i(x)=-(x-r_1)^{n_1}\ldots (x-r_{i})^{n_{i}-1}\ldots(x-r_s)^{n_s}$. Let $q(x)=(x-r_1)^{n_1-1}\ldots(x-r_s)^{n_s-1}$.

If $\sum_{i=1}^sc_ip_i(x)=0$ then $0=\displaystyle\lim_{x\rightarrow r_i}\sum_{i=1}^sc_i\dfrac{p_i(x)}{q(x)}=-c_i(r_i-r_1)\ldots (r_i-r_{i-1})(r_i-r_{i+1})\ldots(r_i-r_s)$. So $c_i=0$. Therefore, $p_1(x),\ldots,p_s(x)$ are l.i.


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