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:

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

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

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:

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} }$

By Determinant with Unit Element in Otherwise Zero Column:


 * $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:

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