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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.

2 Answers2

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The book states that

$$\frac{\sum \lambda_i a_i}{\sum \lambda_i b_i}=\lambda$$

while you seem to be trying to prove

$$\sum \frac{\lambda_i a_i}{\lambda_i b_i}=\lambda$$

Note where the summation sign is!

Now, to prove the original statement, rewrite $\frac{a_i}{b_i}=\lambda$ as $a_i = \lambda b_i$. Now, substitute $a_i$ into the sum:

$$\sum \lambda_i a_i = \sum \lambda_i (\lambda b_i) = \sum \lambda (\lambda_i b_i) = \lambda \sum \lambda_i b_i$$

Now, divide by $\sum \lambda_i b_i$ to get

$$\frac{\sum \lambda_i a_i}{\sum \lambda_i b_i}=\lambda$$

lisyarus
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$(a_i,b_i)$ 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 $$\,(a_i,b_i)\in V\ \Rightarrow\ \sum_i \lambda_i (a_i,b_i)\, =\, \left(\sum_i \lambda_i a_i,\, \sum_i \lambda_i b_i\right ) \in V$$

Bill Dubuque
  • 272,048