Hankel Determinant

Let $F_n$ denote the $n$-th Fibonacci number. Prove that the determinant of the Hankel matrix defined as:

$$\mathcal{H}=\begin{bmatrix} F_n &F_{n+1}  &F_{n+2}  &\cdots  &F_{2n-1} \\  F_{n+1}&F_{n+2}  & F_{n+3} &\cdots  &F_{2n} \\  F_{n+2}&F_{n+3}  &F_{n+4}  &\cdots  &F_{2n+1} \\  \vdots & \vdots  &\vdots   & \ddots  & \vdots \\  F_{2n-1}&F_{2n}  &F_{2n+1}  &\cdots  & F_{3n-2} \end{bmatrix}$$

is equal to zero.

Solution



 For the Fibonacci sequense holds: \(F_{n+1}=F_{n}+F_{n-1}\,,\quad n\in\mathbb{N}\,.\) Using this on the elements of the third \((\dagger)\) column we have

$$\begin{aligned}
|H_n|&=\begin{vmatrix}
F_n &F_{n+1}  &F_{n+2}  &\cdots  &F_{2n-1} \\  
F_{n+1}&F_{n+2}  & F_{n+3} &\cdots  &F_{2n} \\  
F_{n+2}&F_{n+3}  &F_{n+4}  &\cdots  &F_{2n+1} \\  
\vdots & \vdots  &\vdots   & \ddots  & \vdots \\  
F_{2n-1}&F_{2n}  &F_{2n+1}  &\cdots  & F_{3n-2}
\end{vmatrix}\\\\
 &=\begin{vmatrix}
F_n      &  F_{n+1}  & F_{n+1}+F_n  &\cdots  &F_{2n-1} \\  
F_{n+1}  &  F_{n+2}  & F_{n+2}+F_{n+1} &\cdots  &F_{2n} \\
F_{n+2}  &  F_{n+3}  & F_{n+3}+F_{n+2}  &\cdots  &F_{2n+1} \\  
\vdots & \vdots  &\vdots   & \ddots  & \vdots \\  
F_{2n-1} &  F_{2n}  &  F_{2n}+F_{2n-1}  &\cdots  & F_{3n-2}
\end{vmatrix}\\\\
 &\stackrel{C_{3}\to C_3-C_2-C_1}{=\!=\!=\!=\!=\!=\!=\!=\!=}
\begin{vmatrix}
F_n      &  F_{n+1}  & 0  &\cdots  &F_{2n-1} \\  
F_{n+1}  &  F_{n+2}  & 0 &\cdots  &F_{2n} \\
F_{n+2}  &  F_{n+3}  & 0  &\cdots  &F_{2n+1} \\  
\vdots & \vdots  &\vdots   & \ddots  & \vdots \\  
F_{2n-1} &  F_{2n}  &  0  &\cdots  & F_{3n-2}
\end{vmatrix}\\\\
&=0\,.
\end{aligned}$$

\((\dagger)\) We treat the cases for \(n\times n\) matrices with \(n\geqslant3\).  For \(n=1,2\) the determinants are not \(0\) because \(|H_1|=|F_1|=|1|=1\) and \(|H_2|=F_2F_4-F_3^2=-1\)