Fibonacci numbers as determinants

[Grigorios Kostakos]

Let \(\{{F_{n}}\}_{n=1}^{\infty}\) be the Fibonacci sequence defined as \[F_{n}=F_{n-1}+F_{n-2}\,,\;n\geqslant3\,, \quad F_1=F_2=1\,.\] Prove for every \(n\in\mathbb{N},\,n\geqslant2\) that the \(n\)-th term \(F_n\) of the Fibonacci sequence it is given by the determinant of the \((n-1)\times(n-1)\)-matrix : \[\left[{\begin{array}{rrrrrrrr} 1 & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ -1 & 1 & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & -1 & 1 & 1 & \cdots & 0 & 0 & 0\\ 0 & 0 & -1 & 1 & \cdots & 0 & 0 & 0\\ \vdots &\vdots & \vdots& \vdots & \ddots &\vdots & \vdots& \vdots\\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & \cdots & -1 & 1 & 1\\ 0 & 0 & 0 & 0 & \cdots & 0 & -1 & 1 \end{array}}\right]\]

[Demetres]

Let us write $D_n$ for the determinant of the $n \times n $ matrix. It is immediate that $D_1=1$ and $D_2=2$.

Taking cofactor expansion along the first column we have

\[D_{n+2} = \left|{\begin{array}{rrrrrrr} 1 & 1 & 0 & \cdots & 0 & 0 & 0\\ -1 & 1 & 1 & \cdots & 0 & 0 & 0\\ 0 & -1 & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots& \vdots & \ddots &\vdots & \vdots& \vdots\\ 0 & 0 & 0 & \cdots & 1 & 1 & 0\\ 0 & 0 & 0 & \cdots & -1 & 1 & 1\\ 0 & 0 & 0 & \cdots & 0 & -1 & 1 \end{array}}\right| - (-1)\left|{\begin{array}{rrrrrrrr} 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ -1 & 1 & 1 & \cdots & 0 & 0 & 0\\ 0 & -1 & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots& \vdots & \ddots &\vdots & \vdots& \vdots\\ 0 & 0 & 0 & \cdots & 1 & 1 & 0\\ 0 & 0 & 0 & \cdots & -1 & 1 & 1\\ 0 & 0 & 0 & \cdots & 0 & -1 & 1 \end{array}}\right|\]

where the determinants in the right hand side are of $(n+1) \times (n+1)$ matrices. The first determinant by definition is equal to $D_{n+1}$. By expanding along the first row we see that the second determinant is equal to $D_n$.

Thus $D_{n+2} = D_{n+1} + D_n$ for every $n \geqslant 1$. From the initial conditions we get $D_n = F_{n+1}$ as required.