Cassini's Identity/Proof 2

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Theorem

$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$


Proof

First we use this lemma:

$\forall n \in \Z_{>1}: \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$


Then the determinant of both sides is taken.

The left hand side follows directly from Determinant of Order 2:

$\begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix} = F_{n + 1} F_{n - 1} - F_n^2$


Now for the right hand side:

Basis for the Induction

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \paren {-1}^1$


Induction Hypothesis

For $k \in \Z_{>0}$, it is assumed that:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k = \paren {-1}^k$


It remains to be shown that:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k + 1} = \paren {-1}^{k + 1}$


Induction Step

The induction step follows from Determinant of Matrix Product:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = \paren {-1}^k \paren {-1} = \paren {-1}^{k + 1}$


Hence by induction:

$\forall n \in \Z_{>0}: \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^n = \paren {-1}^n$

$\blacksquare$


Source of Name

This entry was named for Giovanni Domenico Cassini.


Sources