# Matrix whose Determinant is Fibonacci Number

## Theorem

The $n \times n$ determinant:

$D_n = \begin{vmatrix} 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix}$

evaluates to $F_{n + 1}$.

## Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $P \left({n}\right)$ be the proposition:

$D_n = F_{n + 1}$

### Basis for the Induction

$P \left({2}\right)$ is the case:

 $\ds D_2$ $=$ $\ds \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}$ $\ds$ $=$ $\ds \left({1 \times 1}\right) - \left({-1 \times 1}\right)$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds 2$ $\ds$ $=$ $\ds F_3$ Definition of Fibonacci Number

Thus $P \left({2}\right)$ is seen to hold.

$P \left({3}\right)$ is the case:

 $\ds D_3$ $=$ $\ds \begin{vmatrix} 1 & -1 & 0 \\ 1 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix}$ $\ds$ $=$ $\ds \left({1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} }\right) - \left({\left({-1}\right) \times \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} }\right) + \left({0 \times \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} }\right)$ Expansion Theorem for Determinants: expanding by $1$st row $\ds$ $=$ $\ds 1 \times D_2 + 1 \times \left({1 \times 1}\right)$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds 2 + 1$ from above $\ds$ $=$ $\ds F_3 + F_2$ Definition of Fibonacci Number $\ds$ $=$ $\ds F_4$ Definition of Fibonacci Number

Thus $P \left({3}\right)$ is seen to hold.

$P \left({2}\right)$ and $P \left({3}\right)$ together form the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ and $P \left({k - 1}\right)$ are true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:

 $\ds D_k$ $=$ $\ds F_{k + 1}$ $\ds D_{k - 1}$ $=$ $\ds F_k$

from which it is to be shown that:

$D_{k + 1} = F_{k + 2}$

### Induction Step

This is the induction step:

Recall the definition of $D_n$:

$D_{k + 1} = \underbrace {\begin{vmatrix} 1 & -1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \cdots & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix} }_{k + 1 \text { columns} }$

Expanding by $1$st row using the Expansion Theorem for Determinants requires the evaluation of $2$ cofactors:

$1 \times C_{1 1}$

and:

$\left({-1}\right) \times C_{1 2}$

where $C_{r s}$ denotes the determinant of order $k$ obtained from $D_{k + 1}$ by deleting row $r$ and column $s$.

By the construction of $D_{k + 1}$, it can be seen that $C_{1 1}$ is $D_k$.

The structure of $C_{1 2}$ is seen to be:

$C_{1 2} = \left({-1}\right) \times \underbrace {\begin{vmatrix} 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix} }_{k \text { columns} }$
$C_{1 2} = \left({-1}\right) \times \underbrace {\begin{vmatrix} 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -1 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix} }_{k - 1 \text { columns} }$

which equals $-D_{k - 1}$.

Putting this together, we have:

 $\ds D_{k + 1}$ $=$ $\ds 1 \times C_{1 1} + \left({-1}\right) \times C_{1 2}$ Expansion Theorem for Determinants $\ds$ $=$ $\ds C_{1 1} - C_{1 2}$ $\ds$ $=$ $\ds D_k - \left({-D_{k - 1} }\right)$ $\ds$ $=$ $\ds D_k + D_{k - 1}$ $\ds$ $=$ $\ds F_{k + 1} + F_k$ Induction Hypothesis $\ds$ $=$ $\ds F_{k + 2}$ Definition of Fibonacci Number

So $P \left({k}\right) \land P \left({k - 1}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\ge 2}: D_n = F_{n + 1}$

$\blacksquare$