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We know that $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix},$ of which there is a simple proof by induction. However, since matrix multiplication is associative, we should be able to split $k$ into $a + b$ so that, $$ \begin{pmatrix}F_{a + 1} & F_a \\ F_a & F_{a - 1}\end{pmatrix} \begin{pmatrix}F_{b + 1} & F_b \\ F_b & F_{b - 1}\end{pmatrix} = \begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix} .$$ The matrix multiplication gives, $$ \begin{pmatrix}F_{a+1}F_{b+1} + F_aF_b & F_{a+1}F_b + F_aF_{b-1} \\ F_aF_{b+1} + F_{a-1}F_b & F_aF_b + F_{a-1}F_{b-1} \end{pmatrix}.$$ Is there a proof using more elementary Fibonacci identities that this is equal to $\begin{pmatrix}F_{k + 1} & F_k \\ F_k & F_{k - 1}\end{pmatrix}?$

Mike Smith
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  • The identity you have quoted is a proof of this fact. Are you looking for a different proof? – Carl Schildkraut Jul 08 '22 at 01:55
  • Yes, that is how I know this is correct. I just want to know why $F_{a+1}F_{b+1} + F_aF_b = F_{k+1},$ etc. The matrix multiplication was to show why this must be true. – Mike Smith Jul 08 '22 at 01:56
  • The identity is among the many listed on the wikipedia page. You can probably verify it using Binet's formula directly, but the most natural/convincing derivation is still the one using the matrix form. – dxiv Jul 08 '22 at 02:26
  • There is also some known combinatorial proofs of this identity, but your phrasing makes it confusing what you're asking: "Here is a proof of this identity. Is there a proof of this identity?" – Brian Moehring Jul 08 '22 at 02:37
  • See, for instance, https://math.stackexchange.com/q/11477 – Brian Moehring Jul 08 '22 at 02:41
  • @BrianMoehring my bad, I meant is there a proof using more elementary Fibonacci identities – Mike Smith Jul 08 '22 at 18:14
  • I still cannot figure out your request. Which part of the proof using matrices you've shown do you find advanced? The inductive proof in the beginning or the associativity of matrix multiplication? My absolute best guess is you were given $\begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix}^k = \begin{pmatrix}F_{k + 1} & F_k \ F_k & F_{k - 1}\end{pmatrix}$ as something to prove rather than discovering it? If that were the case, then it may seem to be more complicated than it actually is, which is just the natural encoding of $F_{k+1} = F_k + F_{k-1}$, $F_0=0$ and $F_1=F_2 = 1$. – Brian Moehring Jul 08 '22 at 20:58

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I can't honestly call this proof "more elementary" than the proof you've given in the question because I find that proof to be quite elementary, but perhaps you'll find it more palatable.

Decomposing sequences defined by linear recurrences:

Let $A_k, B_k$ denote two sequences that satisfy the same recursive relation as $F_k$ but with the initial conditions $A_2 = B_1 = 1, A_1=B_2 = 0$.

Then for any $a$, the sequence $G_k = F_{a+1}A_k + F_aB_k$ satisfies $G_{k+1} = G_k + G_{k-1}$, $G_2 = F_{a+1}$ and $G_1 = F_a$, so that $G_k = F_{k+a-1}$. That is, we've shown $$F_{k+a-1} = F_{a+1}A_k+F_aB_k$$

Now we consider the identity of the $A,B$ sequences:

  • $A$ satisfies the defining recurrence for the Fibonacci sequence. Also, $A_2 = 1=F_1$ and $A_3 = A_2+A_1 = 1+0 = 1=F_2$. That is, $A$ is just a re-indexed Fibonacci sequence, specifically $A_k = F_{k-1}$.
  • $B$ satisfies the defining recurrence for the Fibonacci sequence. Also, $B_3 = B_2+B_1 = 0+1=F_1$ and $B_4 = B_3+B_2 = 1+0 = F_2$. That is, $B$ is another re-indexed Fibonacci sequence, specifically, $B_k = F_{k-2}$.

Put together, this shows $$F_{k+a-1} = F_{a+1}F_{k-1}+F_aF_{k-2}$$ which is exactly the identity we were looking for. To make it look more alike, we can set $k=b+2$ to find $$F_{a+b+1} = F_{a+1}F_{b+1}+F_aF_b$$ and changing $a$ to $a-1$ or $b$ to $b-1$ explicitly gives all the remaining identities.

Brian Moehring
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