# Determinant of structurally symmetric $n$-banded matrix?

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.
• Can you not obtain a recursive formula (as is done for tridiagonal forms), based on the size? Jan 20 '18 at 15:44
• 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. Jan 21 '18 at 2:40

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