# Determinant of an “almost cyclic” matrix

Let $$n\geq 3$$, let $$Z$$ be the matrix of the cyclic shift (the companion matrix of $$X^n-1$$), and for $$\mathbf{d}\in \mathbb{C}^n$$ let $$\operatorname{diag}(\mathbf{d})$$ be the diagonal matrix with $$\mathbf{d}$$ on the diagonal, and $$M_\mathbf{d}:=\operatorname{diag}(\mathbf{d})+Z+Z^2$$.

Notation: call a subset of $$[n]:=\{1,\ldots,n\}$$ separated if does not contain two (cyclically) adjacent elements, let $$S_j(n):=\{ M\in [n]\;:\, |M|=j\}$$ denote the set of separated subsets of $$[n]$$ with $$j$$ elements.

For $$n\leq 12$$ the determinant is given by: $$\det(M_{\mathbf{d}})=\sigma_n(d) +\sum_{j=0}^{n-1} (-1)^{n-j-1} \sigma_j(\mathbf{d})$$ where $$\sigma_0(\mathbf{d})=\det(Z+Z^2)=\bigg\{\begin{array}{cr} 2 & \text{ for odd } n\\ 0 & \text{ for even } n\end{array}$$

$$\sigma_1(\mathbf{v})=d_1+\ldots+d_n$$, $$\sigma_n(\mathbf{v})=\prod_{i=1} d_i$$, and for $$2\leq j \leq n$$ $$\sigma_j(\mathbf{v})= \sum_{(k_1,\ldots,k_j)\in S_j(n)}\prod_{i=1}^j d_{k_i}$$

Does the formula hold for all $$n\geq 3$$?

Motivation: this (and related) determinants appear in a combinatorial problem arising in homological algebra, see Classification of algebras of finite global dimension via determinants of certain 0-1-matrices

• Given that the conjecture is true, can one count the number of $v$ for $w=3$ with finite global dimension as in the other thread? – Mare Mar 21 at 21:22
• Is it possible to contact you via email for a question @esg ? – Mare Jun 3 at 6:06
• I'll send you a mail. – esg Jun 4 at 17:58
• ok, do you have my mail? – Mare Jun 4 at 18:42
• I sent a mail to the email given in your paper with Rubey and Stump – esg Jun 5 at 14:33

For a length $$n-1$$ vector $$a=(a_1,\dots,a_{n-1})$$, define the $$n$$-by-$$n$$ tridiagonal matrix $$T(a)$$ with $$1$$'s on its main diagonal and subdiagonal and $$a$$ on its superdiagonal. The first two determinants are $$\det T(\emptyset)=1$$ and $$\det T(a_1)=1-a_1$$. By cofactor expansion on the last row we have $$\det T(a_1,\dots,a_{n-1})=\det T(a_1,\dots,a_{n-2})-a_{n-1}\det T(a_1,\dots,a_{n-3})$$.

Define $$\tilde S_j(n)$$ in a similar way but with linear (not cyclic) adjacency. Define also $$\tilde \sigma_j(a)=\sum_{(k_1,\ldots,k_j)\in \tilde S_j(n-1)}\prod_{i=1}^j a_{k_i}$$ for $$0\le j\le n-1$$. It's not hard to prove that $$\det T(a_1,\dots,a_{n-1})=\sum_{j=0}^{n-1} (-1)^j\tilde\sigma_j(a)$$ by showing both sides have the same initial conditions and satisfy the same recurrence.

I'll take $$Z_{ij}=\delta_{i,j+1}$$ (subscripts taken cyclically). $$M$$ is a lower triangular matrix $$L$$ plus three $$1$$'s in the upper right corner at $$(1,n-1)$$, $$(1,n)$$, and $$(2,n)$$.

The terms in the expansion of $$\det M$$ which do not contain any of these $$1$$'s contribute $$\det(L)=\sigma_n(d)$$.

The terms which contain exactly one of the three $$1$$'s contribute $$(-1)^{n-1}\det T(d_2,\dots,d_{n-1})$$, $$(-1)^nd_n\det T(d_2,\dots,d_{n-2})$$, and $$(-1)^nd_1\det T(d_3,\dots,d_{n-1})$$ from the minors at $$(1,n)$$, $$(1,n-1)$$, and $$(2,n)$$ respectively.

The final terms contain the $$1$$'s at $$(1,n-1)$$ and $$(2,n)$$ and contribute $$1$$ to the determinant, since the minor is upper triangular with $$1$$'s on its diagonal.

It's straightforward to show that these five terms add up to your formula.