# product of two Jacobi matrices

Let $$A = \begin{bmatrix} a_1 & b_1 \\ b_1 & a_2 & b_2 \\ & b_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-1} \\ & & & b_{n-1} & a_n \end{bmatrix},\ B = \begin{bmatrix} 1&0&0&\cdots\\ 0&a_2 & b_2 \\ 0&b_2& a_3 & \ddots & \\ \vdots&\ddots & \ddots & b_{n-1} \\ & & b_{n-1} & a_n \end{bmatrix},$$ where $$b_i \neq 0$$ and all entries are real. The following matrix is $$A^{-1} \times B$$ $$C = \begin{bmatrix} *&*&0&0&\cdots&0\\ *&*&0&0&\cdots&0 \\ *&*&1&0&\cdots&0 \\ *&*&0&1&\cdots&0 \\ \vdots&\vdots&&\ddots&\ddots&\\ *&*&0&0&\cdots&1 \\ \end{bmatrix},$$

As you see an identity matrix appears in $$C[2 \cdots n]$$ and rows 1, 2 from the third column are zero. I need to prove that it always happen. I tried this using syms in Matlab and it verified this result. I would be appreciated for any help.

## update

Fonseca 2001 in his paper Explicit inverses of some tridiagonal matrices gave the following relations for a symmetric tridiagonal matrix:

$$A^{-1} = \begin{bmatrix} u_1v_1&u_1v_2&u_1v_3&\cdots&u_1v_n\\ u_1v_2&u_2v_2&u_2v_2&\cdots&u_2v_2\\ u_1v_3&u_2v_3&u_3v_3&\cdots&u_3v_3\\ \vdots\\ u_1v_n&u_2v_n&u_3v_n&\cdots&u_nv_n\\ \end{bmatrix}$$

such that \begin{align} v_1&= \dfrac{1}{d_1}\\ v_k&= -\dfrac{b_{k-1}}{d_k}v_{k-1}, k=2, \cdots,n\\ d_n&=a_n\\ d_i&=a_i - \dfrac{b_ic_i}{d_{i+1}}, i=n-1, \cdots, 1\\ u_n &= \dfrac{1}{\delta_nv_n}\\ u_k&=-\dfrac{b_k}{\delta_k}u_{k+1}, k=n-1, \cdots, 1\\ \delta_1&=a_1\\ \delta_i&=a_i-\dfrac{b_{i-1}c_{i-1}}{\delta_{i-1}}, i=2, \cdots, n-1 \end{align}

I tried to replace these variables in their position to get the appropriate results, but I failed.

I would like to know is there any block version of computing inverse of a matrix? because two blocks of $$A, B$$ are the same, it may help.

You have $$B=A-D,$$ where $$D=\begin{bmatrix} a_1-1&b_1&0&0&\cdots&0\\ b_1&0&0&0&\cdots&0 \\ 0&0&0&0&\cdots&0 \\ 0&0&0&0&\cdots&0 \\ \vdots&\vdots&&\ddots&\ddots&\\ 0&0&0&0&\cdots&0 \\ \end{bmatrix}.$$ Then $$C=A^{-1}B=I-A^{-1}D.$$ Since $$D$$ has nonzero entries only at the first two columns, we have $$A^{-1}D= \begin{bmatrix} *&*&0&0&\cdots&0\\ *&*&0&0&\cdots&0 \\ *&*&0&0&\cdots&0 \\ *&*&0&0&\cdots&0 \\ \vdots&\vdots&&\ddots&\ddots&\\ *&*&0&0&\cdots&0 \\ \end{bmatrix},$$ and so $$C=\begin{bmatrix} 1&0&0&0&\cdots&0\\ 0&1&0&0&\cdots&0 \\ 0&0&1&0&\cdots&0 \\ 0&0&0&1&\cdots&0 \\ \vdots&\vdots&&\ddots&\ddots&\\ 0&0&0&0&\cdots&1 \\ \end{bmatrix}- \begin{bmatrix} *&*&0&0&\cdots&0\\ *&*&0&0&\cdots&0 \\ *&*&0&0&\cdots&0 \\ *&*&0&0&\cdots&0 \\ \vdots&\vdots&&\ddots&\ddots&\\ *&*&0&0&\cdots&0 \\ \end{bmatrix},$$ which is of the desired form.
• One problem, the $B \neq A-D$ because entry $(1,1)$ of $B$ is $1$ not $a_1$ Oct 13, 2018 at 15:18
• Infact $$D=\begin{bmatrix} a_1-1&b_1&0&0&\cdots&0\\ b_1&0&0&0&\cdots&0 \\ 0&0&0&0&\cdots&0 \\ 0&0&0&0&\cdots&0 \\ \vdots&\vdots&&\ddots&\ddots&\\ 0&0&0&0&\cdots&0 \\ \end{bmatrix}.$$ and the rest of the calculations are as you said Oct 13, 2018 at 16:11