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I'm having difficulty with the following, problem 5.3.7 from Gilbert Strang's "Linear Algebra and its Application".

The numbers $\lambda_1^k$ and $\lambda_2^k$ satisfy the Fibonacci rule $F_{k+2}=F_{k+1} + F_k$ :

$\lambda_1^{k+2}=\lambda_1^{k+1} + \lambda_1^k$ and $\lambda_2^{k+2}=\lambda_2^{k+1} + \lambda_2^k$

Prove this by using the original equation for the $\lambda's$ (Multiply it by $\lambda^k$)

Then any combination of $\lambda_1^k$ and $\lambda_2^k$ satisfies the rule.

The combination $F_k=(\lambda_1^k-\lambda_2^k)/(\lambda_1-\lambda_2)$ gives the right start of $F_0=0$ and $F_1=1$.

I'm not sure what "the original equation for the $\lambda's$" is, or what I'm supposed to prove. Can someone please help?

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    The author is Gilbert Strang.2012-11-11

2 Answers 2

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It looks to me like $\lambda_1$ and $\lambda_2$ were defined earlier, probably as eigenvalues of the matrix $\left( \begin {matrix} 1&1\\1&0 \end {matrix} \right)$ so $\lambda_1=\frac 12(1+\sqrt 5)$ and $\lambda_2=\frac 12(1-\sqrt 5)$. You are now supposed to prove the sentence just under "Fibonacci Rule".

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    @Ross Millikan I know that it is late of an answer but right now I was also trying to solve the same exercise. So is it possible to verify the validity of my answer? Thanks.2017-02-20
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For me there are three parts on this exercise.

Part A: Prove $\lambda_2^{k+2}=\lambda_2^{k+1} + \lambda_2^k$ and $\lambda_1^{k+2}=\lambda_1^{k+1} + \lambda_1^k$.

To do that we need

  1. $F_{k+2} = F_{k+1}+F_{k}$ and
  2. $F_{k+2} = c_1\lambda_1^k + c_2\lambda_2^k$

Replacing 2. in 1. we have:

$c_1\lambda_1^{k+2} + c_2\lambda_2^{k+2} = c_1[\lambda_1^{k+1}+\lambda_1^{k}]+c_2[\lambda_2^{k+1}+\lambda_2^{k}]$

Based on the last equation, we may say that $\lambda_2^{k+2}=\lambda_2^{k+1} + \lambda_2^k$ and $\lambda_1^{k+2}=\lambda_1^{k+1} + \lambda_1^k$ is true.

Part B: Prove that Fibonacci rule $F_{k+2} = c_1\lambda_1^k + c_2\lambda_2^k$ is true for any $c_1$ and $c_2$.

From Part A we may have:

$c_1\underbrace{[\lambda_1^{k+2}-(\lambda_1^{k+1}+\lambda_1^{k})]}_{zero} = c_2\underbrace{[\lambda_2^{k+2}-(\lambda_2^{k+1}+\lambda_2^{k})]}_{zero}$

The later equation is true for any $c_1$ and $c_2$.

Part C: The combination $F_k = (\lambda_1^k-\lambda_2^k)/(\lambda_1-\lambda_2)$ gives the right start of $F_0$ and $F_1$.

Easily, using the later combination for $n=0$ and $n=1$ we have the Fibonacci start $F_0=0$ and $F_1=1$.