# Calculating the product of tridiagonal matrix times its transpose

Denote by tridiag($a$,$b$,$c$) the tridiagonal matrix of size $n \times n$ with diagonal elements $b = (b_1, \ldots,b_n)$. Let $a = (a_1, \ldots, a_{n-1})$ and $c = (c_1,\ldots,c_{n-1})$ be the entries of the other two diagonals.

I am trying to compute $$\text{tridiag}(a,b,c)^{\intercal}\text{tridiag}(c,b,a)$$ where $a = (-1, \ldots, -1)$, a vector of $(n-1)$ ones, $b = (1,\ldots,1)$ and $c = (c_1,c_2,\ldots,c_{i-1},c_n)$ is given by $c_1 = -1-\theta$, $c_2 = c_3 = \ldots = c_{n-1} = -\theta k$ and $c_n = 1-\theta$. Here, $\theta$ and $k$ are real numbers.

I found a few articles that gave inverses of specific tridiagonal matrices, but none of this form. Any help would be greatly appreciated.

 To clarify, I'm trying to calculate

$$\left(\begin{array}{cccc} (-1-\theta) && 1 && \ldots \\ -1 && -\theta k && 1 &&\ldots\\ \vdots&&-1&&-\theta k && 1 &&\ldots\\ \vdots && && \ddots &&\ddots && \\ \cdots && && &&\cdots && 1 \\ \cdots && && && -1 && 1-\theta \end{array}\right)^T \left(\begin{array}{cccc} (-1-\theta) && 1 && \ldots \\ -1 && -\theta k && 1 &&\ldots\\ \vdots&&-1&&-\theta k && 1 &&\ldots\\ \vdots && && \ddots &&\ddots && \\ \cdots && && &&\cdots && 1 \\ \cdots && && && -1 && 1-\theta \end{array}\right)$$

where all the dots denote zeroes. Apologies for the broken lay-out, I don't know how to fix this.

I assume that $A^T$ denotes the transpose of $A$. There is a mistake. The considered product matrix (in EDIT) is not $tridiag(a,b,c)^T tridiag(c,b,a)$ but $tridiag(a,b,c)^T triadiag(a,b,c)$.
1. In the edit case, the result $U$ is the symmetric matrix ($\theta$ is denoted by $t$)
$u_{1,1}=t^2+2t+2,u_{n,n}=t^2-2t+2$, the other $u_{i,i}=k^2t^2+2$, $u_{1,2}=u_{2,1}=kt-t-1,u_{i,i+2}=u_{i+2,i}=-1,u_{n-1,n}=u_{n,n-1}=-kt+t-1$, the other $u_{i,j}$ being zero.
1. Now $tridiag(a,b,c)^T tridiag(c,b,a)=tridiag(c,b,a)^2$. Is that you want ?