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Setup. Let $n\in\Bbb N$. Let $a_{1,1}, a_{1,2},\dots, a_{1,n}\in\Bbb R$ be a given sequence of real numbers that sum to $0$, i.e. $a_{1,n}=-(a_{1,1}+a_{1,2}+\dots+a_{1,n-1})$. For $i=2,\dots,n$ define $$a_{i,j}=a_{1,j}+a_{1,j+1}+\dots+a_{1,j+i-1}=\sum_{k=j}^{j+i-1} a_{1,k}\quad(\text{for } j=1,\dots,n-i+1).$$ The "half-matrix" $(a_{i,j})_{i,j}$ can be visualized as follows: $$ \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & \dots & a_{1,n-2} & a_{1,n-1} & -(a_{1,1}+a_{1,2}+\dots+a_{1,n-1}) \\ a_{1,1}+a_{1,2} & a_{1,2}+ a_{1,3} & a_{1,3}+a_{1,4} & \dots & a_{1,n-2} + a_{1,n-1} & -(a_{1,1}+a_{1,2}+\dots+a_{1,n-2}) \\ a_{1,1}+a_{1,2}+a_{1,3} & a_{1,2}+a_{1,3}+a_{1,4} & a_{1,3}+a_{1,4}+a_{1,5} & \dots & -(a_{1,1}+a_{1,2}+\dots+a_{1,n-3}) \\ \vdots & \vdots & ⋰& ⋰ \\ a_{1,1}+a_{1,2}+\dots+a_{1,n-1} & -a_{1,1} \\ 0 \end{pmatrix} $$

Now I have the following proposition:

Proposition. Let $n, a_{i,j}$ be as in the setup. Then there are at least $n$ distinct pairs $(i,j)$ with $i\in\{1,\dots, n\}$ and $j\in\{1,\dots,n-i+1\}$ such that

  • $a_{i,j}=0$ or
  • $j\le n-i$ and $a_{i,j}\cdot a_{i,j+1} < 0$.

More informally, the number of zeros of the $a_{i,j}$ plus the number of "sign switches" between adjacent $a_{i,j}$ in all rows is at least $n$.

My question: How can we prove this propostion?.

Context. Proving this proposition would enable me to solve another problem about zeroes of special continuous functions that I found on StackExchange.

Example ($n=4$). Consider \begin{pmatrix} 1 & \frac12 & -\frac14 & -\frac54 \\ \frac32 & \frac14 & -\frac32 \\ \frac54 & -1 \\ 0 \end{pmatrix}

Then $a_{1,2}\cdot a_{1,2}<0$; $a_{2,2}\cdot a_{2,3}<0$; $a_{3,1}\cdot a_{3,2}<0$ and $a_{4,1}=0$. So in our example we have exactly $n$ zeros/sign switches.

My work. I tried using induction over $n$ which didn't work.

Edit: A rigorous proof of the Proposition can be found on StackExchange. It is based on the very nice idea by Ilya Bogdanov from below.


The cyclic sequence $a_1, a_2, \dots, a_n, a_1$ contains two occurrences of (zero or sign change), as it sums up to $0$. Each zero corresponds to a zero in either $1$st or $(n-1)$th row, or in both. Each sign change corresponds to a sign change in exactly one of the two rows (note that a sign change $a_n, a_1$ appears in the $(n-1)$th row!). So those two rows contain in total at least two occurrences.

Similarly, each occurrence in $a_1+a_2, a_2+a_3, \dots, a_n+a_1, a_1+a_2$ leads to an occurrence either in the $2$nd, or in the $(n-2)$th row (or in both), and so on. All in all, this gives $n-1$ occurrences in the first $n-1$ rows, as desired

  • $\begingroup$ Hi Ilya, thank you very much for your nice answer! I translated your sketch of proof into a very rigorous framework here on StackExchange. $\endgroup$ – Maximilian Janisch Sep 25 '19 at 22:35

EDIT: my answer below is wrong, as shown by Ilya Bogdanov's comment.

If there is no zero and no sign change on row $i$ among $a_{i,1}, \cdots, a_{i,i-1}$, then they are of the same sign and $a_{i,i}$ is of the opposite sign. Therefore there is some zero or sign change on every row.

  • 2
    $\begingroup$ Start with row $(1,-2,1)$; the second row will get no sign change. $\endgroup$ – Ilya Bogdanov Sep 25 '19 at 21:24

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