Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My problem is the following:

Let $p$ be a non constant polynomial over $\mathbb {R}$ and define $F(x,y)=(p(x+y),p(x-y))$.Show that $DF(x,y)$ is invertible in a dense and open subset of $\mathbb {R^2}$.

I've been thinking a lot on this one, but couldn't get far... I'm really stuck... any help is much appreciated!

share|cite|improve this question
up vote 3 down vote accepted

The Jacobi matrix of $F$ at $(x,y)$ is $$DF(x,y)\left(\begin{matrix} p'(x+y) & p'(x-y)\\ p'(x+y) & -p'(x-y) \end{matrix}\right)$$ and its determinant is given by $-2p'(x+y) p'(x-y)$. So $DF$ is not invertible at $(x,y)$ if and only if $x+y \in Z$ or $x-y \in Z$ where $Z$ is the set of roots of $p'$. Since $p$ is non-constant, $p'$ is not the zero polynomial, so $Z$ is finite. The set of points where $DF$ is not invertible is therefore given by $$\bigcup_{a \in Z} \left(\operatorname{Graph}(y = a+x) \cup \operatorname{Graph}(y = a-x)\right),$$ a finite union of lines in $\mathbb R^2$. Clearly, the complement of this is open and dense.

share|cite|improve this answer
Thank you very Much!very helpful! – HipsterMathematician Nov 19 '12 at 19:08

Very carefully write out the matrix $DF(x,y)$ using the chain rule. You know that the matrix is invertible if and only if its determinant is nonzero, so write out an expression for the determinant and set it equal to zero. You should end up with something like $$p'(x+y)p'(x-y) = 0 .$$ When is this possible? Remember that $p'$ is itself just a polynomial over $\mathbb{R}$, so think about the set of zeroes of that polynomial and how they related to the zeroes of the expression $p'(x+y)p'(x-y).$

share|cite|improve this answer
Thank you very much! – HipsterMathematician Nov 19 '12 at 19:07
"know that the matrix is invertible if and only if its determinant is zero" isn't it non-zero? – HipsterMathematician Nov 19 '12 at 19:31
Yeah, made a typo. Should be fixed now. :) – Zach L. Nov 20 '12 at 0:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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