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Let $F_n$ denote the $n$th Fibonacci number, adopting the convention $F_1=1$, $F_2=1$ and so on. Consider the $ntimes n$ matrix defined by

$$mathbf M_n:=begin{bmatrix}F_1&F_2&dots&F_n\F_{n+1}&F_{n+2}&dots&F_{2n}\vdots&vdots&ddots&vdots\F_{n^2-n+1}&F_{n^2-n+2}&dots&F_{n^2}end{bmatrix}.$$

I have the following conjecture:

Conjecture. For all integers $ngeq3$, $detmathbf M_n=0$.

I have used some Python code to test this conjecture for $n$ up to $9$, but I cannot go further. Note that $detmathbf M_1=detmathbf M_2=1$. Due to the elementary nature of this problem I have to assume that it has been discussed before, perhaps even on this site. But I couldn't find any reference on it, by Googling or searching here. Can someone shed light onto whether the conjecture is true, and a proof of it if so?

Here's a hint: what's the relationship between $F_{k+1}+F_{k+2}$ and $F_{k+3}$? What does that say about the 1st, 2nd, and 3rd columns of this matrix?

The resolution is remarkably simple (many thanks to obscurans' answer for the hint!) By the definition of the Fibonacci numbers, $F_k+F_{k+1}=F_{k+2}$ for all $k$. If $ngeq3$ then these numbers are going to be in the first three columns of every row. Hence the first three rows are linearly dependent, so the determinant is $0$. It follows from this that any such sequence following a linear recurrence (of the form $F_{n}=aF_{n-1}+bF_{n-2}$, $a,b$ are constant), with possibly different starting terms, also satisfies the stated conjecture. In fact, this shows that all such matrices have rank $2$, with the only two linearly independent columns being the first two. If the linear recurrence is of higher order, say $m$, then the determinant is $0$ when $n>m$, and the rank of the matrix will be $m$.