## Cassini’s Identity without Matrix Powers

Cassini’s identity for Fibonacci numbers says that $F_{n+1} F_{n-1} - F_n^2 = (-1)^n$.  The classic proof of this shows (by induction) that $\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n$.  Since $\det \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = -1$, Cassini’s identity follows.

In this post I’m going to give a different proof involving determinants, but one that does not use powers of the Fibonacci matrix.

Let’s start with the $2 \times 2$ identity matrix, which we’ll call $A_0$:

$\displaystyle A_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

To construct $A_1$, add the second row to the first and then swap the two rows.  This gives us

$\displaystyle A_1 = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$.

Continue this process of adding the second row to the first and then swapping the two rows.  This yields

$\displaystyle A_2 = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}, A_3 = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, A_4 = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}, \ldots$.

Since $A_1 = \begin{bmatrix} F_0 & F_1 \\ F_1 & F_2 \end{bmatrix}$ and $F_{n+1} = F_n + F_{n-1}$, the fact that we’re adding rows each time means that $A_n = \begin{bmatrix} F_{n-1} & F_n \\ F_n & F_{n+1} \end{bmatrix}$.

Since adding a row to another row doesn’t change the determinant, and swapping two rows changes only the sign of the determinant, we have

$\displaystyle F_{n+1} F_{n-1} - F_n^2 = \det A_n = (-1)^n \det A_0 = (-1)^n,$

which is Cassini’s identity.