-1

Let $A=\bigl(\begin{smallmatrix} 1&1\\1&0 \end{smallmatrix}\bigr)$. Then $A^n~?$

I saw $A^2=\left(\begin{smallmatrix} 2&1\\1&1 \end{smallmatrix}\right)$. Is there any similarity or any recurrence relation ?

Sam
  • 107
  • Yes, eigenvalues of $A$ are ${1+\sqrt 5\over 2}$ and ${1-\sqrt 5\over 2}$ – Sam Jan 12 '21 at 14:40
  • 1
    This should remind you of the fibonacci recurrence. $\begin{bmatrix}1&1\1&0\end{bmatrix}\begin{bmatrix}f_2\f_1\end{bmatrix}=\begin{bmatrix}f_2+f_1\f_2\end{bmatrix}$ – JMoravitz Jan 12 '21 at 14:41
  • By approaching from the other direction of diagonalization of the matrix and taking the relation to the Fibonacci recurrence into account, one can then arrive at one of the elementary methods of proving the closed form to the Fibonacci sequence. – JMoravitz Jan 12 '21 at 14:45
  • DietrichBurde: I think your suggested link assumes OP must know the Fibonacci sequence, which only provides one approach to the question. But it does certainly not require any a priori knowledge in that sequence. –  Jan 12 '21 at 15:13
  • @mrsamy The duplicate answers the question of what $A^n$ is, and that is the above question (see title). So $F_n$ appears naturally. – Dietrich Burde Jan 12 '21 at 15:49
  • @DietrichBurde: I may misread something, but your link explicitly says "How to prove Fibonacci sequence with matrices". –  Jan 12 '21 at 15:50
  • @mrsamy Yes, you are right, that's what the link says. But the post itself (and the linked one) proves a formula for $A^n$, and this is what the OP wants. For $n=2$ he found it himself. – Dietrich Burde Jan 12 '21 at 15:51
  • @DietrichBurde there may be a better duplicate post anyway. Without context, nobody knows what $F_n$ means. But I agree, yes, this certainly relates to the Fibonacci sequence and it worths knowing that. :-) –  Jan 12 '21 at 15:54
  • @mrsamy Yes, I agree that there may be other duplicates. But I find it good enough that an explicit formula is proved for $A^n$. That's the important point for me. And it leads in addition to Fibonacci numbers, and that's also good. – Dietrich Burde Jan 12 '21 at 15:55
  • @DietrichBurde: There are various pedagogical approaches. Point taken :-) –  Jan 12 '21 at 15:57

2 Answers2

1

Find the eigenvalues of $A$ and write $A$ as $SJS^{-1}$ where $J$ is some nice matrix.

Then find $(SJS^{-1})^n$.

The point is that it is very easy to find a "recursive" relation in $J^n$.

0

$A^n=\bigl(\begin{smallmatrix} F_{n+1}& F_n\\F_n & F_{n-1}\end{smallmatrix}\bigr)$ where $F_n=n-$th fibonacci number.

Notice this is true for $n=1$. Now use Induction to prove my statement. You should read abot the matrix representation of fibonacci number [here] (htttps://en.wikipedia.org/wiki/Fibonacci_number)

Saikai Prime
  • 1,133