# Deriving the solutions of a system of linear equations

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.

• Why are you interested in this question? – James Mar 25 at 17:31
• It is part of a proof I have to develop. – STF Mar 25 at 18:21
• (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. – Paul Sinclair Mar 26 at 4:21
• 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. – Paul Sinclair Mar 26 at 4:30