Cassini's Identity/Proof 2

Theorem
Let $F_k$ be the $k$th Fibonacci number.

Then $F_{n+1}F_{n-1} - F_n^2 = \left({-1}\right)^n$.

This is also sometimes reported (slightly less elegantly) as $F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$

Proof
First this identity is proved:


 * $\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.

Basis for the Induction

 * $\begin{bmatrix}

F_2 & F_1     \\ F_1 & F_0 \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^1$

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


 * $\begin{bmatrix}

F_{k+1} & F_k     \\ F_k     & F_{k-1} \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^k$

It remains to be shown that:


 * $\begin{bmatrix}

F_{k+2} & F_{k+1}     \\ F_{k+1} & F_k \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^{k+1}$

Induction Step
The induction step follows from conventional matrix multiplication:

So by induction:
 * $\begin{bmatrix}

F_{n+1} & F_n     \\ F_n     & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^n$

Determinant
Now we calculate the determinants:

The LHS follows directly from the order 2 determinant:


 * $\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 RHS:

Basis for the Induction

 * $\begin{vmatrix}

1    & 1    \\  1     & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \left({-1}\right)^1$

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


 * $\begin{vmatrix}

1    & 1    \\  1     & 0 \end{vmatrix}^k = \left({-1}\right)^k$

It remains to be shown that:


 * $\begin{vmatrix}

1    & 1    \\  1     & 0 \end{vmatrix}^{k+1} = \left({-1}\right)^{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} = \left({-1}\right)^k \left({-1}\right) = \left({-1}\right)^{k+1}$

Hence by induction:


 * $\forall n \in \Z_{>0}: \begin{vmatrix}

1    & 1    \\  1     & 0 \end{vmatrix}^n = \left({-1}\right)^n$