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

Say that a vector $x=(x_1,x_2, \ldots ,x_n)\in {\mathbb R}^n$ is nondecreasing if $x_1 \leq x_2\leq \ldots \leq x_n$. Can anyone show or find a counterexample to the following : if three nondecreasing vectors are mutually orthogonal, one of them is the zero vector.

This question is the “discrete” version of that recent question.

What I have achieved so far : I can show the result when one of the vectors has sum $0$. Indeed, the following stronger result holds in this case :

Theorem If two nondecreasing vectors $u$ and $v$ are orthogonal and the “integral” (sum of coordinates) of $u$ is zero, then either $u$ is the zero vector or $v$ is “constant” (all its coordinates are equal).

Proof Suppose that $u=(u_1,u_2, \ldots ,u_n)$ is orthogonal to $v=(v_1,v_2, \ldots, v_n)$ and that the sum of $u$ is $0$ but $u$ is not the zero vector. The set $\Omega=\lbrace i | u_i \lt 0\rbrace$ is nonempty, call $r$ its largest element. Then we can decompose $u=(-g_r,-g_{r-1}, \ldots ,-g_2,-g_1,0, \ldots, 0, h_1,h_2, \ldots ,h_s)$ where $s \leq n-r$ and $0 \lt g_1 \leq g_2 \leq \ldots \leq g_r$ and $0 \lt h_1 \leq h_2 \leq \ldots \leq h_s$, and $\sum_{i=1}^r g_i=\sum_{j=1}^r h_j$ (since the integral of $u$ is zero) ; call $S$ this common value. Then, the orthogonality of $u$ and $v$ implies that

$$ Sv_r=\sum_{i=1}^r g_i v_r \geq \sum_{i=1}^r g_i v_i = \sum_{j=1}^s h_j v_{n+1-s+j} \geq \sum_{j=1}^s h_j v_{n+1-s}=Sv_{n+1-s} $$

Now $v_r \leq v_{n+1-s}$ because $v$ is nondecreasing, so all the inequalities above must be equalities and $v$ is indeed constant.

share|cite|improve this question
up vote 0 down vote accepted

This question has been cross-posted at MO, and promptly answered there

share|cite|improve this answer

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.