[See also StackExchange](https://math.stackexchange.com/q/3368742/631742).


**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](https://math.stackexchange.com/q/3366262/631742).


**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](https://math.stackexchange.com/a/3370078/631742). It is based on the very nice idea by [Ilya Bogdanov](https://mathoverflow.net/users/17581/ilya-bogdanov) from below.



Number of zeroes and sign switches in a constructed zero-sum double sequence
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 


