Suppose the quadratic form $f: \mathbb R^3\to \mathbb R$ with $$f(x_1,x_2,x_3) = x_1^2 - x_2^2 - 11x_3^2 - 2x_1x_2 + 4x_1x_3 + 8x_2x_3.$$ By using Lagrange's Reduction, we have the canonical expression of $f,$ $$g(y_1,y_2,y_3) = y_1^2 - 2y_2^2 + 3y_3^2,$$ where $$y_1 = x_1 - x_2 + 2x_3,\\ y_2 = x_2 - 3x_3,\\ y_3 = x_3.$$

My question is: How to find the sets $f(\mathbb R^3)$ and $f(\mathbb R^3\setminus \{(0,0,0)\})$?

Thank you for your help!


1 Answer 1


Since $f(x_1,0,0)=x_1^2$ and $f(0,x_2,0)=-x_2^2,$ it's clear that $f(\mathbb{R}^3)=\mathbb{R},$ and since $f(0,0,0)=0,$ the only question is whether $f(\mathbb{R}^3 \setminus \{(0,0,0)\})=\mathbb{R}$ or whether $f(\mathbb{R}^3 \setminus \{(0,0,0)\})=\mathbb{R} \setminus \{0\}.$ That is, does $f$ have any zeros other than $(0,0,0)?$

It should be easy to find another zero using your expression for $g$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .