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Let $K\in \mathbb{N}$ with $K\geq 2$ and consider the system of equations $$ \begin{cases} \sum_{j=1}^h y_j *x_{K+j-h}=\sum_{j=1}^h x_j *y_{K+j-h} & \text{ for }h=1,...,K-1\\ \sum_{j=1}^Ky_j=1\\ \end{cases} $$ that is $$ \begin{cases} y_1*x_K=x_1*y_K\\ y_1*x_{K-1}+y_2*x_K=x_1*y_{K-1}+x_2*y_K\\ y_1*x_{K-2}+y_2*x_{K-1}+y_3*x_K=x_1*y_{K-2}+x_2*y_{K-1}+x_3*y_K\\ ...\\ y_1*x_2+y_2*x_3+...+y_{K-1}*x_K=x_1*y_2+x_2*y_3+...+x_{K-1}*y_K\\ y_1+y_2+...+y_K=1\\ \end{cases} $$

For example, when $K=3$, the system becomes $$ \begin{cases} y_1*x_3=x_1*y_3\\ y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\\ y_1+y_2+y_3=1\\ \end{cases} $$ Note:

  • The unknowns of the system are $y_1,...,y_K$, while $x_1,...,x_K$ are treated as parameters.

  • I assume that $y_j>0,x_j>0$ for $j=1,...,K$.

  • We know that the parameters $x_1,...,x_K$ are such that $\sum_{j=1}^K x_j=1$


Question: I want to show that the solutions of the system above are $$ \begin{cases} (a) \text{ }y_j=x_j & \text{ for }j=1,...,K\\ \text{or }\\ (b) \text{ }y_j=y_{K+1-j} & \text{ for }j=1,...,\lfloor\frac{K+1}{2} \rfloor \end{cases} $$ where $\lfloor A \rfloor$ denotes the the largest integer less than or equal to $A$.


What I have tried (incomplete): I have the proof when $K=3$ (very simple).

$$ \begin{cases} y_1*x_3=x_1*y_3\\ y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\\ y_1+y_2+y_3=1\\ \end{cases} \Rightarrow \begin{cases} \frac{x_3}{x_1}=\frac{y_3}{y_1}\\ \frac{x_2}{x_1}+\frac{y_2}{y_1}\frac{x_3}{x_1}=\frac{y_2}{y_1}+\frac{y_3}{y_1}\frac{x_2}{x_1} \end{cases} $$ $$\Rightarrow \frac{x_2}{x_1}+\frac{y_2}{y_1}\frac{y_3}{y_1}=\frac{y_2}{y_1}+\frac{y_3}{y_1}\frac{x_2}{x_1}\Leftrightarrow \Big( \frac{y_2}{y_1}-\frac{x_2}{x_1}\Big)\Big(\frac{y_3}{y_1}-1 \Big)=0 $$ $$ \Leftrightarrow \text{ }y_3=y_1 \text{ or } \frac{y_2}{y_1}=\frac{x_2}{x_1} $$

If $y_3=y_1$ then we are in case (b) of the general claim above.

If $ \frac{y_2}{y_1}=\frac{x_2}{x_1}$, then the system becomes $$ \begin{cases} \frac{y_3}{y_1}=\frac{x_3}{x_1}\\ \frac{y_2}{y_1}=\frac{x_2}{x_1}\\ y_1+y_2+y_3=1\\ \end{cases} \Rightarrow y_1+\frac{x_2}{x_1}y_1+\frac{x_3}{x_1}y_1=1 $$ $$ \Leftrightarrow y_1(\frac{\overbrace{x_1+x_2+x_3}^{=1}}{x_1})=1 \Leftrightarrow y_1=x_1 \Rightarrow y_3=x_3, y_2=x_2 $$ that is case (a) of the general claim above. I am struggling to generalise this to any $K$ because I'm very weak in linear algebra. Any help from your side would be greatly appreciated.

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  • $\begingroup$ Why are you interested in this question? $\endgroup$ – James Mar 25 at 17:31
  • $\begingroup$ It is part of a proof I have to develop. $\endgroup$ – STF Mar 25 at 18:21
  • $\begingroup$ (b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half. $\endgroup$ – Paul Sinclair Mar 26 at 4:21
  • $\begingroup$ You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done. $\endgroup$ – Paul Sinclair Mar 26 at 4:30

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