Q:

General Componendo and Dividendo rule.

I have a book that states in the footnotes

$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... =\frac{a_n}{b_n}= \lambda \implies \frac{\sum\lambda_ia_i}{\sum \lambda_ib_i} = \lambda $$
  for any $\lambda_i$.

I can prove the "usual" rule given this name. (the first one that comes up when googling it.) 
But I can not seem to show this. 
If I proceed as follows, in the case n = 2
$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \lambda$$
$$\implies 2\frac{a_1}{b_1} = \frac{a_2}{b_2} + \frac{a_1}{b_1}= \lambda+\frac{a_1}{b_1} $$
$$\implies \frac{b_1a_2+b_2a_1}{b_2b_1+\lambda^{-1}a_1b_2} = \lambda$$
but these are very specific coefficients! 
I have a feeling like this is going to be an inductive argument,so I think it would be sufficient to understand the case n = 2, as it will be used to complete the inductive phase. 

A:

$(a_i,b_i)$ are solutions of $\, L(a,b) =  (1,-\lambda)\cdot (a,b) = a-b\lambda = 0.\,$ Since $\,L\,$ is linear, the solutions form a vector space $V$, i.e solutions are closed under addition and scalings, therefore $$\,(a_i,b_i)\in V\ \Rightarrow\ \sum_i \lambda_i (a_i,b_i)\, =\, \left(\sum_i \lambda_i  a_i,\, \sum_i \lambda_i b_i\right ) \in V$$

