# General inverse for a symmetric matrix of the form $\mathbf A_n = \left(a_{\min(i,j)}b_{\max(i,j)}\right)$

I am looking for a general inverse formula for a symmetric matrix of the form $$\mathbf A_n = \left(a_{\min(i,j)}b_{\max(i,j)}\right) = \pmatrix{ a_1b_1 & a_1b_2 & a_1b_3 & \cdots & a_1b_n \\ a_1b_2 & a_2b_2 & a_2b_3 & \cdots & a_2b_n \\ \vdots & & \ddots & & \vdots \\ a_1b_n & a_2b_n & & & a_nb_n}$$ where $$a_i,b_i$$ are real numbers. Empirically I can see that $$\mathbf A_n^{-1}$$ is tridiagonal so I suspect it has previously been studied. For example we have $$\mathbf A_5 = \left( \begin{array}{ccccc} \frac{a_2}{a_1 a_2 b_1-a_1^2 b_2} & \frac{1}{a_1 b_2-a_2 b_1} & 0 & 0 & 0 \\ \frac{1}{a_1 b_2-a_2 b_1} & \frac{a_1 b_3-a_3 b_1}{\left(a_2 b_1-a_1 b_2\right) \left(a_2 b_3-a_3 b_2\right)} & \frac{1}{a_2 b_3-a_3 b_2} & 0 & 0 \\ 0 & \frac{1}{a_2 b_3-a_3 b_2} & \frac{a_2 b_4-a_4 b_2}{\left(a_3 b_2-a_2 b_3\right) \left(a_3 b_4-a_4 b_3\right)} & \frac{1}{a_3 b_4-a_4 b_3} & 0 \\ 0 & 0 & \frac{1}{a_3 b_4-a_4 b_3} & \frac{a_3 b_5-a_5 b_3}{\left(a_4 b_3-a_3 b_4\right) \left(a_4 b_5-a_5 b_4\right)} & \frac{1}{a_4 b_5-a_5 b_4} \\ 0 & 0 & 0 & \frac{1}{a_4 b_5-a_5 b_4} & \frac{b_4}{a_5 b_4 b_5-a_4 b_5^2} \\ \end{array} \right)$$ which seems to be based on a bunch of 2x2 determinants...

Any help appreciated, thank you!

p.

Ok so after researching inversion of symmetric tridiagonal matrices, it appears that $$\mathbf A$$ is known as a "matrice factorisable" with tridiagonal inverse $$\mathbf A^{-1} = \pmatrix{ \alpha_1 & \beta_1 & & \\ \beta_1 & \alpha_2 & \beta_2 & \\ & \ddots & \ddots & \\ & & \beta_{n-1} & \alpha_n }$$ where: $$\alpha_1 = - \frac{a_2}{a_1 u_{1,2}}, \alpha_i = \frac{u_{i-1,i+1}}{u_{i-1,i}u_{i,i+1}}, \alpha_n = -\frac{b_{n-1}}{b_n u_{n-1,n}}, \beta_i = \frac{1}{u_{i,i+1}}, u_{i,j} = a_i b_j (v_j - v_i), v_i = \frac{b_i}{a_i}$$ subject to a's and b's being nonzero and $$v_i \neq v_{i+1}$$.
There is also an interesting factorization of the form $$\mathbf A = \mathbf {DL}^T\boldsymbol\Lambda\mathbf{LD}$$ where $$\mathbf D$$ is diagonal, $$\mathbf L$$ is lower triangular, and $$\boldsymbol\Lambda$$ is superdiagonal together with one last diagonal coefficient. I hope this helps anyone encountering a similar type of matrix!
• Your original expression for the inverse looks to me a lot more elegant. It would be good to have an explanation of why the denominators are connected with $a\wedge b$ in the way that they are! May 30, 2022 at 16:58