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Is there a formula to compute the determinant of a structurally symmetric $n$-banded matrix? I am specifically interested in the 5-banded matrix:

$$ \left[\begin{matrix} c_{0} & s_{0} & 0 & w_{0} & 0 & 0 & 0 & 0 & 0\\ n_{0} & c_{1} & s_{1} & 0 & w_{1} & 0 & 0 & 0 & 0\\ 0 & n_{1} & c_{2} & 0 & 0 & w_{2} & 0 & 0 & 0\\ e_{0} & 0 & 0 & c_{3} & s_{3} & 0 & w_{3} & 0 & 0\\ 0 & e_{1} & 0 & n_{3} & c_{4} & s_{4} & 0 & w_{4} & 0 & \dots \\ 0 & 0 & e_{2} & 0 & n_{4} & c_{5} & 0 & 0 & w_{5}\\ 0 & 0 & 0 & e_{3} & 0 & 0 & c_{6} & s_{6} & 0\\ 0 & 0 & 0 & 0 & e_{4} & 0 & n_{6} & c_{7} & s_{7}\\ 0 & 0 & 0 & 0 & 0 & e_{5} & 0 & n_{7} & c_{8} \\ & & & & \vdots & & & & & \ddots\end{matrix}\right] $$

which is defined by the five sequences:

$$ \left\lbrace w_0, w_1, \cdots \right\rbrace, \left\lbrace s_0, s_1, \cdots \right\rbrace, \left\lbrace c_0, c_1, \cdots \right\rbrace, \left\lbrace n_0, n_1, \cdots \right\rbrace, \left\lbrace e_0, e_1, \cdots \right\rbrace $$

All other elements are zero.

It's not clear to me whether formulas for tridiagonal matrices can be extended straightforwardly to compute the determinant of the above matrix.

Notes:

  • In my special case some terms $ n_2 = n_5 = n_8 = \cdots = 0 $, $ s_0 = s_3 = s_6 = \cdots = 0 $, but I'm interested in the solution for arbitrary sequences $w_i$, $s_i$, $c_i$, $n_i$, $e_i$.
  • All off-diagonal elements are nonnegative, and all diagonal elements are negative.
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    $\begingroup$ Can you not obtain a recursive formula (as is done for tridiagonal forms), based on the size? $\endgroup$ – David Handelman Jan 20 '18 at 15:44
  • $\begingroup$ This paper: ac.inf.elte.hu/Vol_002_1979/013.pdf describes a quick way to calculate determinants of a 5-banded matrix if the matrix is actually symmetric rather than just symmetrically structured. $\endgroup$ – Josiah Park Jan 21 '18 at 2:40
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This determinant can be calculated using the block transfer matrix method of Molinari (Linear Algebra and its Applications 429 (2008) 2221-2226, https://arxiv.org/abs/0712.0681), if $w_i \neq 0$. You have $3 \times 3$ blocks that read, after renumbering, $$ \mathbf A_k = \begin{pmatrix} c_{k,0} & s_{k,0} & 0 \\ n_{k,0} & c_{k,1} & s_{k,1} \\ 0 & n_{k,1} & c_{k,2} \end{pmatrix}, \,\, \mathbf B_k = \begin{pmatrix} w_{k,0} & 0 & 0 \\ 0 & w_{k,1} & 0 \\ s_{k,2} & 0 & w_{k,2} \end{pmatrix}, \,\, \mathbf C_k = \begin{pmatrix} e_{k,0} & 0 & n_{k,2} \\ 0 & e_{k,1} & 0 \\ 0 & 0 & e_{k,2} \end{pmatrix}. $$

Defining the block transfer matrix $$ \mathbf T_k = \begin{bmatrix} -\mathbf B_k^{-1} \mathbf A_k & -\mathbf B_k^{-1} \mathbf C_k \\ \mathbf 1 & \mathbf 0 \end{bmatrix}, $$ the determinant $\Delta$ of the $K \times K$ block matrix (with $3 \times 3$ blocks) above then reads $$ \Delta = \det \langle \mathbf 1,\mathbf 0| \mathbf T_K \mathbf T_{K-1} \cdots \mathbf T_1 |\mathbf 1,\mathbf 0\rangle \prod_{k=1}^{K} \det(-\mathbf B_{k}) . $$ Note that I used bra-ket notation, where the $2 \times 1$ block ket vector $|\mathbf 1,\mathbf 0\rangle$ has total size $6 \times 3$, such that the bra-ket pair picks the (1,1)-block of the block matrix product. For a similar application, see

Hucht, Alfred, The square lattice Ising model on the rectangle. I: Finite systems, J. Phys. A, Math. Theor. 50, No. 6, Article ID 065201, 23 p. (2017). ZBL1357.81154, https://arxiv.org/abs/1609.01963 .

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