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.

See also my paper “Fibonacci Identities via the Determinant Sum Property.”

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