Matrix whose Determinant is Fibonacci Number

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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:

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


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:

\(\ds D_{k + 1}\) \(=\) \(\ds 1 \times C_{1 1} + \left({-1}\right) \times C_{1 2}\) 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:

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

$\blacksquare$


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