Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

GIven an equation such as $ax^2+by^2+cz^2+dxy+exz+fyz=g$ where $a,b,c,d,e,f,g\in \mathbb R$, How can we tell if the surface described is a bounded one without explicitly plotting a graph?

share|cite|improve this question
up vote 2 down vote accepted

The surface is bounded iff the eigenvalues of the matrix $\begin{pmatrix} a & d/2 & e/2 \\ d/2 & b & f/2 \\ e/2 & f/2 & c \\ \end{pmatrix}$ have the same sign.

One way to show this is to compute the extrema of function $$(x,y,z)\mapsto x^2+y^2+z^2$$ restricted to the surface. You can do this using Lagrange multipliers very easily.

Alternatively, it is not difficult to see that there is a linear change of variables which takes your function to one of the form $$\alpha X^2+\beta Y^2+\gamma Z^2$$ for which the question is very easy.

share|cite|improve this answer
+1 Clever method, I think I'll have to use it sometime. – Alex Becker May 22 '12 at 23:24

With $$ g > 0: $$ It is bounded if and only if $$ a > 0 $$ and $$ 4 ab - d^2 > 0 $$ and $$ 4 abc + def - a f^2 - b e^2 - c d^2 > 0. $$

This is called Sylvester's Criterion. See also TWEETY. And SYLVESTER.

share|cite|improve this answer
Haha! Nice associations...! – Miserable May 23 '12 at 8:10

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.