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We have a simple graph with vertices $\{v_1, v_2, ... v_n\}$.

The adjacency matrix of this graph is $A= (a_{ij})$ so that

• $a_{ij}=1$ if $i+j$ belongs to the Fibonacci sequence;

• $a_{ij}=0$ if $i+j$ doesn't belong to the Fibonacci sequence.

We claim that the determinant of this matrix is $0$ when $n$ is odd. And that when $n$ is even, it is $1$, $-1$ or $0$.

How can we prove this claim?

Edit: on MSE, the OP added that $a_{ii}=0$ along the diagonal which is confirmed by the OP's observation that the determinant should be zero in the odd case (e.g. $n=1,3$ do yield $0$ then). So in particular, this is not a Hankel transform.

We have a simple graph with vertices $\{v_1, v_2, ... v_n\}$.

The adjacency matrix of this graph is $A= (a_{ij})$ so that

• $a_{ij}=1$ if $i+j$ belongs to the Fibonacci sequence;

• $a_{ij}=0$ if $i+j$ doesn't belong to the Fibonacci sequence.

We claim that the determinant of this matrix is $0$ when $n$ is odd. And that when $n$ is even, it is $1$, $-1$ or $0$.

How can we prove this claim?

Edit: on MSE, the OP added that $a_{ii}=0$ along the diagonal which is confirmed by the OP's observation that the determinant should be zero in the odd case (e.g. $n=1,3$ do yield $0$ then). So in particular, this is not a Hankel transform.

we have a simple graph that it's vertices are {v_1, v_2, ... v_n}.

The Adjacency matrix of this graph is A= (a_ij) so that;

(a_ij)=1 if i+j belongs to the Fibonacci sequence. (a_ij)=0 if i+j doesn't belong to the Fibonacci sequence.

We claim that the determinant of this matrix when n is odd is 0 and when n is even is 1 or -1 or 0. How can we prove this claim?

We have a simple graph with vertices $\{v_1, v_2, ... v_n\}$.

The adjacency matrix of this graph is $A= (a_{ij})$ so that

• $a_{ij}=1$ if $i+j$ belongs to the Fibonacci sequence;

• $a_{ij}=0$ if $i+j$ doesn't belong to the Fibonacci sequence.

We claim that the determinant of this matrix is $0$ when $n$ is odd. And that when $n$ is even, it is $1$, $-1$ or $0$.

How can we prove this claim?

Edit: on MSE, the OP added that $a_{ii}=0$ along the diagonal which is confirmed by the OP's observation that the determinant should be zero in the odd case (e.g. $n=1,3$ do yield $0$ then). So in particular, this is not a Hankel transform.