Take the 2-minute tour ×
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.

For arbitrary $d \in \mathbb{R}^n$, I am interested in the set of $x \in \mathbb{R}^n$ such that $d \cdot \langle x_1^2, \dots, x_n^2\rangle = 0$. If $d > 0$, then $x = 0$ is the unique solution. If $d \ge 0$, then the solution set is a $k$-dimensional flat where $x_i$ can be anything iff $d_i = 0$ (otherwise, $x_i = 0$).

What about when $d$ contains both positive and negative elements? I suspect it's a hyperplane, on the observation that if $x$ is a solution, then $\lambda x$ is a solution for any $\lambda \in \mathbb{R}$. But does the figure always have enough dimension to be a hyperplane, or is it sometimes lower-dimensional?

share|improve this question
2  
What you have in general is a type of cone, not a hyperplane. Consider the case $n=3$ with $d=\langle 1,1,-1\rangle$ where you have the cone $x^2 + y^2 = z^2$. –  Robert Israel Sep 23 '12 at 19:00
add comment

1 Answer

If $d$ has both positive and negative elements, this variety always contains a manifold of dimension $n-1$, because the Jacobian of $F(x_1,\ldots,x_n) = \sum_j d_j x_j^2$ is $(2 d_1 x_1, \ldots, 2 d_n x_n)$, which is nonzero when some $d_i x_i \ne 0$. Use the Implicit Function Theorem.
There are always solutions with some $d_i x_i \ne 0$, e.g. if $d_1 > 0 > d_2$ take $x_1 = \sqrt{-d_2}$, $x_2 = \sqrt{d_1}$, $x_i = 0$ otherwise.

share|improve this answer
add comment

Your Answer

 
discard

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.