This is a cross-posting of a MSE question (which did not receive any feedback there so far).

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.


Consider the nondecreasing vector $v=(1,1,\ldots,1)$. It's not too hard to show that two nondecreasing nonzero vectors orthogonal to $v$ must have a positive inner product. Here is a proof.

Let $a=(a_1,\ldots, a_n)$ and $b=(b_1,\ldots, b_n)$ be orthogonal to $v$. Suppose that $a_j$ is the last negative entry of $a$ and $b_k$ is the last negative entry of $b$, for $k>j$. Then

\begin{equation}\sum_{i=1}^j |a_ib_i| \ge \sum_{i=1}^j |a_i b_{j+1}| \ge \sum_{i=j+1}^k |a_ib_{j+1}|\ge \sum_{i=j+1}^k |a_ib_i|\end{equation}

But the sum defining the inner product is

\begin{equation}\sum_{i=1}^j |a_ib_i| - \sum_{i=j+1}^k |a_ib_i| + \sum_{i=k+1}^n |a_ib_i|\end{equation}

So this is nonzero as long as $a_n$ and $b_n$ are nonzero; since $a$ and $b$ are orthogonal to $v$, $a_n$ is zero if and only if $a$ is zero, and likewise for $b.$

Now, a generic nondecreasing vector is of the form $av+u$ for $u$ nondecreasing and orthogonal to $v$.

Assume we have three mutually orthogonal nonzero nondecreasing vectors $v_i= a_i v +u_i$ for $i\in\{1,2,3\}$. No more than one $u_i$ can be zero; suppose $u_1$ and $u_2$ are nonzero. Then $a_1$ and $a_2$ must have opposite signs for $v_1$ and $v_2$ to be orthogonal (in particular, neither can be zero). Now $u_3$ cannot be zero because then $v_3$ would not be orthogonal to $v_1$ or $v_2$. Then by the same argument, $a_3$ must have a sign opposite to both $a_1$ and $a_2$, a contradiction.

  • $\begingroup$ I don’t quite follow your claim that “a generic nondecreasing vector is of the form $v+u$ for $u$ orthogonal to $v$”. When $n=3$, for example, take $u=(2,3,-5)$. Then $u$ is orthogonal to $v$, but $u+v=(3,4,-4)$ is not nondecreasing. $\endgroup$ – Ewan Delanoy Mar 7 '13 at 14:05
  • $\begingroup$ There was a typo in what I wrote, fixed now (I left out the scalar multiple on $v$), but I'm not sure it was the source of your confusion. You can always use orthogonal projection to write a vector $w$ as a sum of a scalar multiple of $v$ and a vector orthogonal to $v$. If $w$ is nondecreasing and you add or subtract any scalar multiple of $v$, you get a nondecreasing vector. $\endgroup$ – Gabriel C. Drummond-Cole Mar 7 '13 at 16:53
  • $\begingroup$ I also added the word "nondecreasing" to $u$, which was critical for the argument in the following paragraph. I didn't say that every vector of the form $av+u$ was nondecreasing for $u$ orthogonal to $v$. I said that every nondecreasing vector was of the form $av+u$ for $u$ orthogonal to $v$. That is true and your example is not a counterexample to that statement. Furthermore, any such $u$ will itself be nondecreasing. I used this fact to claim that $a_1$ and $a_2$ had opposite signs, but inadvertently left it out of the answer. $\endgroup$ – Gabriel C. Drummond-Cole Mar 7 '13 at 16:58
  • $\begingroup$ Right. It seems also that your argument generalizes nicely to the continuous version of this question, where vectors are replaced by continuous functions and the inner sum is replaced by the $L^2$ product. $\endgroup$ – Ewan Delanoy Mar 7 '13 at 17:20

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