# Closed form expression for particular determinant?

For a few days now I've been trying to find a closed form expression for the determinant of the following $$n\times n$$ tridiagonal matrix

$$\begin{pmatrix}c_1+b_1+a_1 & b_1 & 0 & \ddots & 0 \\ c_2 & c_2+b_2+a_2 & b_2 & \ddots & 0 \\ 0 & c_3 & c_3+b_3+a_3 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & b_{n-1}\\ 0 & ... & ... & c_{n} & c_{n}+b_n +a_n\end{pmatrix}$$

For the sequences $$c_n$$, $$b_n$$, and $$a_n$$. I've figured out closed form expression for special cases. Namely, when $$a_n=0$$, the determinant is $$\Big(\prod_{i=1}^nb_i\Big)\sum_{l=0}^n\prod_{k=1}^l\frac{c_{k}}{b_k}$$ When $$l=0$$ in the product series, that returns a $$1$$. Additionally, if $$c_1=0$$, then the determinant is simply $$\prod_{i=1}^nb_i.$$

I would really like to find an analogous formula in the case where $$a_n \neq 0$$. For your benefit I will list the first few determinants for small $$n$$ $$n=1:\quad\quad c_1+b_1+a_1$$ $$n=2:\quad\quad a_1a_2+b_1a_2+a_1b_2+b_1b_2+c_1a_2+c_1b_2+a_1c_2+c_1c_2$$ $$n=3:\quad\quad a_1a_2a_3+b_1a_2a_3+a_1b_2a_3+b_1b_2a_3+a_1a_2b_3+b_1a_2b_3+a_1b_2b_3+b_1b_2b_3+c_1a_2a_3+c_1b_2a_3+c_1a_2b_3+c_1b_2b_3+a_1c_2a_3+a_1c_2b_3+c_1c_2a_3+c_1c_2b_3+a_1a_2c_3+b_1a_2c_3+c_1a_2c_3+a_1c_2c_3+c_1c_2c_3$$

When you look at this, you may suspect that it is just the sum of every $$n$$th order product of $$a$$'s $$b$$'s and $$c$$'s with no subscript repeated, however this is not the case. For instance, $$b_1c_2$$ does not appear in the $$n=2$$ formula. Similarly there are $$6$$ terms which do not appear in the $$n=3$$ formula.

I would really appreciate anyones input on this!

• Perhaps something simplifies if you use the recursive formula for the determinant of a tridiagonal matrix: en.wikipedia.org/wiki/Tridiagonal_matrix#Determinant Oct 3, 2020 at 20:49
• I may have potentially found a nice simplification Oct 3, 2020 at 20:59
• @Buraian did you ever find the nice simplification? Oct 23, 2020 at 19:18
• I did but it was too big to be written as an answer :-) Oct 23, 2020 at 19:25

Your matrix is a general tridiagonal matrix, with $$d_i:=a_i+b_i+c_i$$ along the diagonal. If we denote the determinant of the $$n\times n$$-matrix by $$f_n$$, then we have the recurrence relation $$f_n=d_nf_{n-1}-b_{n-1}c_{n-1}f_{n-2}.$$ Not much more can be said for general sequences $$b_n$$, $$c_n$$ and $$d_n$$. For more information see Wikipedia.

I believe I have an explicit solution!

Using the case that I had already figured out (when $$a_k=0$$), we can Taylor expand around this solution. For finite $$n$$, this will be a finite expansion.

First I define the quantity $$\theta_{km}$$, with $$1\leq k,m\leq n$$, which satisfies the following recursive relations

$$\theta_{km}=(c_m+b_m+a_m)\theta_{k,m-1}-b_{m-1}c_m\theta_{k,m-2},\quad \theta_{kk}=c_k+b_k+a_k,\quad \theta_{k,k-1}=1$$ $$\theta_{km}=(c_k+b_k+a_k)\theta_{k+1,m}-b_{k}c_{k+1}s\theta_{k+2,m},\quad \theta_{mm}=c_m+b_m+a_m,\quad \theta_{m+1,m}=1$$ and $$\theta_{km}=0$$ when $$k> m+1$$ and $$m< k-1$$.

Note that this quantity combines the $$\theta_n$$ and $$\phi_n$$ which is defined in this Wikipedia article. And $$\theta_{1n}$$ is the determinant of the matrix.

When $$a_k=0$$, this quantity has an explicit solution:

$$\theta_{km}=\Big(\prod_{i=k}^mb_i\Big)\sum_{l=k-1}^m\prod_{j=k}^l\frac{c_{j}}{b_j}$$

Using the recursive relations, it can be shown that this quantity satisfies

$$\frac{d\theta_{km}}{da_j}=\theta_{k,j-1}\theta_{j+1,m}$$

Thus the general solution for nonzero $$a_k$$ is

$$\theta_{1n}+\sum_{k=1}^n\theta_{1k-1}a_k\theta_{k+1n}+\cdots+\sum_{k_1\cdots k_p=1}^n\theta_{1k_1-1}a_{k_1}\theta_{k_1+1,k_2-1}\cdots a_{k_p}\theta_{k_p+1,n}+\cdots+a_1\cdots a_n$$

Where all of the $$\theta$$'s in the above expression are for the case where $$a_k=0$$.

To tidy up the formula a bit more, one can note that $$(a\theta)_{nm}=a_n\theta_{n+1,m-1}$$ is a nilpotent upper triangular matrix. So this formula can actually be cast as

$$\Big(\theta(1-a\theta)^{-1}\Big)_{0n}$$

That's about as explicit as I can do for now.