# Show that a system of linear equations has a unique solution

Suppose that I have the following system of $$K+1$$ linear equations with $$K\geq 3$$ $$\begin{cases} \lambda_j\times \lambda_h'=\lambda'_j\times \lambda_h & \text{ for K different pairs (j,h) taken from A}\\ \lambda_1+...+\lambda_K=1 \end{cases}$$

where

• the unknowns are $$(\lambda_1,...,\lambda_K)$$

• $$(\lambda'_1,...,\lambda_K')$$ are known parameters such that $$\lambda'_1+...+\lambda'_K=1$$

• $$\lambda_1>0,...,\lambda_K>0$$ and $$\lambda'_1>0,...,\lambda'_K>0$$

• $$A\equiv \{(1,2),(1,3),...,(1,K),(2,K),(3,K),...,(K-1,K)\}$$ with cardinality $$2K-3$$.

Question: could you help me to show that this system has a unique solution that is $$\lambda_1=\lambda'_1,...,\lambda_K=\lambda'_K$$

For example, when $$K=5$$, we have that $$A\equiv \{(1,2),(1,3),(1,4),(1,5), (2,5),(3,5),(4,5)\}$$ and a specification of the system could be $$\begin{cases} \lambda_1\times \lambda_5'=\lambda'_1\times \lambda_5\\ \lambda_2\times \lambda_5'=\lambda'_2\times \lambda_5\\ \lambda_1\times \lambda_4'=\lambda'_1\times \lambda_4\\ \lambda_1\times \lambda_3'=\lambda'_1\times \lambda_3\\ \lambda_3\times \lambda_5'=\lambda'_3\times \lambda_5\\ \lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=1 \end{cases}$$

which implies $$\begin{cases} \lambda_2=\lambda_1\times \frac{\lambda_2'}{\lambda_1'}\\ \lambda_3=\lambda_1\times \frac{\lambda_3'}{\lambda_1'}\\ \lambda_4=\lambda_1\times \frac{\lambda_4'}{\lambda_1'}\\ \lambda_5=\lambda_1\times \frac{\lambda_5'}{\lambda_1'}\\ \end{cases}$$ Then, replacing this in $$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=1$$ we get $$\lambda_1\times \Big(\overbrace{1+\frac{\lambda_2'}{\lambda_1'}+ \frac{\lambda_3'}{\lambda_1'}+ \frac{\lambda_4'}{\lambda_1'}+ \frac{\lambda_5'}{\lambda_1'}}^{=1/\lambda_1'}\Big)=1$$ which implies $$\lambda_1=\lambda_1'$$ and hence $$\lambda_2=\lambda_2',\lambda_3=\lambda_3',\lambda_4=\lambda_4',\lambda_5=\lambda_5'$$.

I am unable to generalise this procedure to any $$K$$ elements from the set $$A$$. Could you help?

We can rewrite $$\lambda_j\times \lambda_h'=\lambda'_j\times \lambda_h$$ as $$\frac{\lambda_j}{\lambda_j'} = \frac{\lambda_h}{\lambda_h'}$$. If we can show from these $$K$$ equations that $$\frac{\lambda_1}{\lambda_1'} = \frac{\lambda_2}{\lambda_2'} = \dots = \frac{\lambda_K}{\lambda_K'}$$ then there is a constant $$C$$ such that $$\lambda_i = C \lambda_i'$$ for all $$i$$; from knowing that $$\lambda_1 + \dots + \lambda_k = \lambda_1'+ \dots + \lambda_K'= 1,$$ we deduce that $$C=1$$ and therefore $$\lambda_i = \lambda_i'$$ for all $$i$$.
However, we cannot necessarily conclude that all the ratios $$\frac{\lambda_j}{\lambda_j'}$$ are equal. This depends on the $$K$$ specific equations we chose. Let $$G$$ be the graph with vertex set $$\{1,\dots,K\}$$ and an edge $$hj$$ whenever we choose the pair $$(h,j)$$ to form an equation. The requirement to have a unique solution is that $$G$$ must be connected. If so, for any $$a,b \in \{1,\dots,K\}$$ there is a path from $$a$$ to $$b$$ in $$G$$, and we get $$\frac{\lambda_a}{\lambda_a'} = \dots = \frac{\lambda_b}{\lambda_b'}$$ by transitivity along that path.
But here is an example (for $$K=5$$) without a unique solution. Choose the $$5$$ equations \begin{align} \lambda_1 \lambda_2' &= \lambda_1'\lambda_2 \\ \lambda_1 \lambda_3' &= \lambda_1'\lambda_3 \\ \lambda_1 \lambda_5' &= \lambda_1'\lambda_5 \\ \lambda_2 \lambda_5' &= \lambda_2'\lambda_5 \\ \lambda_3 \lambda_5' &= \lambda_3'\lambda_5 \ \end{align} Together, these equations are equivalent to $$\frac{\lambda_1}{\lambda_1'} = \frac{\lambda_2}{\lambda_2'} = \frac{\lambda_3}{\lambda_3'} = \frac{\lambda_5}{\lambda_5'}$$, but they leave out $$\lambda_4$$ entirely. So all $$5$$-tuples $$\lambda$$ with $$(\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5) = (A \lambda_1', A\lambda_2', A\lambda_3', B\lambda_4', A\lambda_5')$$ satisfy these $$5$$ equations, and if $$A(\lambda_1'+\lambda_2'+\lambda_3'+\lambda_5') + B \lambda_4'= 1$$, then $$\lambda_1 + \dots + \lambda_5 = 1$$ also holds. For any $$0 < A < \frac{1}{\lambda_1'+\lambda_2'+\lambda_3'+\lambda_5'}$$, we can set $$B = \frac{1 - A(\lambda_1'+\lambda_2'+\lambda_3'+\lambda_5')}{\lambda_4'}$$ and get a valid solution this way.
1. For every $$i = 2,\dots,K-1$$, either $$\lambda_1 \lambda_i' = \lambda_1'\lambda_i$$ or $$\lambda_i\lambda_K' = \lambda_i'\lambda_K$$ is an equation, forcing $$\frac{\lambda_i}{\lambda_i'}$$ to be equal to either $$\frac{\lambda_1}{\lambda_1'}$$ or $$\frac{\lambda_K}{\lambda_K'}$$.
2. To get $$K$$ equations, we need to either include both such equations for some $$i$$, or else include the equation $$\lambda_1 \lambda_K' = \lambda_1'\lambda_K$$. In either case, we can conclude that $$\frac{\lambda_1}{\lambda_1'} = \frac{\lambda_K}{\lambda_K'}$$. Therefore all $$\frac{\lambda_i}{\lambda_i'}$$ are equal.