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Using mathematical induction I am to prove:

$ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^n $ = $ \left( \begin{array}{ccc} F_{n+1} & F_n \\ F_n & F_{n-1} \end{array} \right) $

where $F_k$ represents the $k^{th}$ Fibonacci number.

my base case is $n =2$

LHS: $ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right) \times$ $ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right) $$=$ $ \left( \begin{array}{ccc} 2 & 1 \\ 1 & 1 \end{array} \right) $

RHS: $ \left( \begin{array}{ccc} F_3 & F_2 \\ F_2 & F_1 \end{array} \right) $$=$ $ \left( \begin{array}{ccc} 2 & 1 \\ 1 & 1 \end{array} \right) $

So $n = k + 1$

$ \left( \begin{array}{ccc} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{array} \right) $

So for my inductive step I did:

$ \left( \begin{array}{ccc} F_{k+1} & F_k \\ F_k & F_{k-1} \end{array} \right) $ $+$ $ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^{k+1} $

And now I'm not sure where to proceed from here. Can anyone point me in the right direction? Assuming my previous work is correct.

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  • $\begingroup$ Why are you summing the matrices? $\endgroup$ – user2345215 Feb 28 '14 at 11:13
  • $\begingroup$ I was under the assumption that when proving by induction you add the LHS where $n = k+1$ to the RHS where $n=k$ $\endgroup$ – user126959 Feb 28 '14 at 11:15
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    $\begingroup$ Yes --- if you're trying to prove something about a sum. That's not the case here. What you want to do in an induction proof, is write down the induction hypothesis at the top of the page, the thing to be proved at the bottom of the page, and then fill in a logical argument that leads from the top to the bottom. $\endgroup$ – Gerry Myerson Feb 28 '14 at 11:17
  • $\begingroup$ See also: math.stackexchange.com/questions/61997/… )and other posts linked there) $\endgroup$ – Martin Sleziak Nov 7 '15 at 19:57
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To prove it for $n=1$ you just need to verify that

$ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^1 $ = $ \left( \begin{array}{ccc} F_2 & F_1 \\ F_1 & F_0 \end{array} \right) $

which is trivial.

After you established the base case, you only need to show that assuming it holds for $n$ it also holds for $n+1$.

So assume

$ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^n $ = $ \left( \begin{array}{ccc} F_{n+1} & F_n \\ F_n & F_{n-1} \end{array} \right) $

and try to prove

$ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^{n+1} $ = $ \left( \begin{array}{ccc} F_{n+2} & F_{n+1} \\ F_{n+1} & F_n \end{array} \right) $

Hint: Write $ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^{n+1} $ as $ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right)^n $$ \left( \begin{array}{ccc} 1 & 1 \\ 1 & 0 \end{array} \right) $.

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  • $\begingroup$ Geez, I really managed to over-complicate this problem for myself! Thank you. $\endgroup$ – user126959 Feb 28 '14 at 11:23
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Inductive proof:

For $n=1$ is true, as the OP correctly observed.

Assume that it is true for $n=k$. Then $$ \left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^k=\left(\begin{matrix} F_{k+1}& F_k \\ F_k& F_{k-1}\end{matrix}\right) $$ Then for $n=k+1$ we have \begin{align} \left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^{k+1}&=\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^{k}\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right) =\left(\begin{matrix} F_{k+1}& F_k \\ F_k& F_{k-1}\end{matrix}\right)\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)= \left(\begin{matrix} F_k+F_{k+1}& F_{k+1} \\ F_{k-1}+F_{k}& F_k\end{matrix}\right)\\ &= \left(\begin{matrix} F_{k+2}& F_{k+1} \\ F_{k+1}& F_k\end{matrix}\right), \end{align} and hence it is true for $n=k+1$. Note that in the last equality above we used the recursive definition of Fibonacci sequence.

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