Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do you prove that: $$ \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n\\ F_{n} & F_{n-1} \end{pmatrix}$$

share|improve this question
11  
What about induction? –  O.L. May 7 at 8:17

2 Answers 2

up vote 13 down vote accepted

Let

$$A=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$

And the Fibonacci numbers, defined by

$$\begin{eqnarray} F_0&=&0\\ F_1&=&1\\ F_{n+1}&=&F_n+F_{n-1} \end{eqnarray}$$

Then, by induction,

$$A^1=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_2 & F_1 \\ F_1 & F_0 \end{pmatrix}$$

And if for $n$ the formula is true, then

$$A^{n+1}=A\,A^n=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1} \end{pmatrix}=\begin{pmatrix} F_{n+1}+F_n & F_{n}+F_{n-1} \\ F_{n+1} & F_{n} \end{pmatrix}=\begin{pmatrix} F_{n+2} & F_{n+1} \\ F_{n+1} & F_{n} \end{pmatrix}$$

So, the induction step is true, and by induction, the formula is true for all $n>0$.

share|improve this answer

$$\begin{align} F(n+1) &= 1\,F(n) + 1\,F(n-1)\\ F(n) &= 1\,F(n) + 0\,F(n-1)\\ \\ \begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} F(n) \\ F(n - 1) \end{bmatrix} \\ \begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(1) \\ F(0) \end{bmatrix} \\ \\ \text{as well as} \\ \begin{bmatrix} F(n) \\ F(n-1) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(0) \\ F(-1) \end{bmatrix} \\ \\ \text{from which it follows}\\ \begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(1) & F(0) \\ F(0) & F(-1) \end{bmatrix} \\ \\ \text{and choosing} \\ F(1) &= 1 \\ F(0) &= 0 \\ F(-1) &= 1 \end{align}$$

share|improve this answer
    
Shouldn't that be F(1)=1 , F(0)=1 , F(-1)=0 ? –  Raidri May 7 at 15:56
    
@Raidri If F(-1) + F(0) = F(1), what do you get? The negatives of the fibonacci form a pretty recognizable pattern actually ^_^ –  DanielV May 7 at 16:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.