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I have a standard proof for the theorem:

$$\sum_{}^n f_1+f_3+f_5+...+f_{2n-1} = f_{2n}$$

$$f_i$$ refers to the Fibonacci numbers for future reference.

It involves setting p(k) as p(k+1) and proving it through weak induction, it has been graded by my professor and is correct.

However, I have recently came across the fibonacci matrix formulation from here:

How to prove Fibonacci sequence with matrices?

I am curious how I would go about solving this theorem with matrices.

I have tried using the product operator: $$\Pi$$

but I am not experienced enough to correctly formulate it so that they equal each other, for example:

$$\Pi_{i=1}^n \begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}^{2n-1} = \begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}^{2n}$$

Using the product operator may be completely pointless, but I honestly just don't know since I have never really used them before.

Any idea on how the original theorem is shown in matrices?

Thank you for any help in advance.

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  • $\begingroup$ Yes, $$f_i$$ refers to the fibonacci numbers, i'll edit that in. Thank you. $\endgroup$
    – Antas Santas
    Commented Sep 10, 2018 at 19:25
  • $\begingroup$ There is no $i$ in the summands of your sum $\sum_i$. $\endgroup$
    – Dietrich Burde
    Commented Sep 10, 2018 at 19:27
  • $\begingroup$ @DietrichBurde That is a completely different Fibonacci identity. $\endgroup$
    – Mike Earnest
    Commented Sep 10, 2018 at 19:31
  • $\begingroup$ Ooops, will be corrected soon (this one is the right identity., but not yet with matrices). $\endgroup$
    – Dietrich Burde
    Commented Sep 10, 2018 at 19:31
  • $\begingroup$ One nice setup for the matrix proof is to begin by noting that $$ \pmatrix{f_1\\f_0} + \pmatrix{f_3\\f_2} + \cdots + \pmatrix{f_{2n-1}\\f_{2n-2}} = \\ \pmatrix{1\\0} + F^2 \pmatrix{1\\0} + \cdots + F^{2n-2}\pmatrix{1\\0} $$ where $F = \pmatrix{1&1\\1&0}$ $\endgroup$
    – Ben Grossmann
    Commented Sep 10, 2018 at 19:32

1 Answer 1

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Letting $F=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, so $F^2=F+I$, then the finite geometric series formula gives $$ I + F^2 + F^4 + \dots + F^{2n-2} = (F^{2n}-I)(F^2-I)^{-1} = (F^{2n}-I)F^{-1}=F^{2n-1}-F^{-1}. $$ Since $F^{-1}=\begin{bmatrix}0&1\\1&-1\end{bmatrix}$, conclude by examining the upper left entry of the equation $$ I + F^2 + F^4 + \dots + F^{2n-2} = F^{2n-1}-F^{-1}. $$ and recalling that $F^n=\begin{bmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{bmatrix}$.

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  • $\begingroup$ Also, $F^{-1}=F-I$. $\endgroup$
    – lhf
    Commented Oct 18, 2018 at 0:37

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