Question

Prove Cassini's identity using the matrix representation of the Fibonacci sequence

Answer 1

oh is that all? easy! one question, whats cassini's identity? To google!

Answer 2

Heh, I can open questions all over the shop now.....:-)

Answer 3

http://planetmath.org/ProofofCassinisIdentity.html

come across this which answeres your question, but certainly doesnt answer mine.

Answer 4

out of interest. whats the formula for the fibonacci sequence

Answer 5

Where's the matrix?

Answer 6

\[F_n = F_{n-2} + F_{n-1}\] ?

Answer 7

oh sorry yes your right there is none there

Answer 8

You can represent the Fibonacci sequence as:
\[\left[\begin{matrix}F_{n+1} & F_n \\ F_n &F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]

Answer 9

This is fairly easy to prove by induction

Answer 10

define F_0 as 0, F_1 as 1, and F_2 as 1

Answer 11

If the first equation is true, then\[\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^{n+1}=\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\]:
\[= \left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right]*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\]
\[= \left[\begin{matrix}F_{n+1}+F_n & F_{n+1} \\ F_n+F_{n-1} & F_n\end{matrix}\right]=\left[\begin{matrix}F_{n+2} & F_{n+1} \\ F_{n+1} & F_n\end{matrix}\right]\]

Answer 12

So:
\[\left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n \forall n \in Z_+\]

Answer 13

now, take the determinant of both sides:
\[F_{n+1}F_{n-1}-F_n^2=\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]

Answer 14

The determinant of the products of square matrices is the products of the determinants of the matrices:
\[= (\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right])^n\]

Answer 15

=\[(0-1)^n=-1^n\]