# Inverse of a symmetric tridiagonal matrix

I have a symmetric $$n\times n$$ matrix $$\mathbb A$$ with entries:

$$A_{ij} = (a_i + a_{i-1})\delta_{ij} - a_i\delta_{i,j-1}-a_{j}\delta_{i-1,j}$$

where $$a_0,\dots,a_n$$ are given positive numbers.

Is there an analytical formula for the inverse of $$\mathbb A$$?

Old: I've found numerically that $$\mathbb A^{-1}$$ can also be tridiagonal. But I have not been able to prove this.

Edit: $$\mathbb A^{-1}$$ is not tridiagonal in general, as pointed out by a simple counterexample by Jean-Claude in the comments. But I'm still interested in a closed-form formula for $$\mathbb A^{-1}$$, if it exists.

• It is not nice to replace the original question with something completely different while others might working on an answer to the original question, which then becomes meaningless. If you have two questions, then please open a new question for the second one and do not replace the first with the second. – Reinhard Meier May 16 at 21:43
• Counterexample: $[2,-1,0 ; -1,2,-1 ; 0,-1,2]$ with $a_0=a_1=a_2=a_3=1$. – Stop hurting Monica May 16 at 21:54
• @ReinhardMeier I'm sorry! I did not realize you were working on this. I will revert the edit and open a new question. – becko May 16 at 22:00
• @ReinhardMeier I had to modify notation to avoid automatic marking as duplicate, but here it is: math.stackexchange.com/q/3228940/10063. Again I apologize. – becko May 16 at 22:04
• @Jean-ClaudeArbaut Can we at least find an analytical inverse? – becko May 16 at 22:26

First of all it is convenient to index the matrices from $$0$$.
I will indicate with $${\mathbf{X}_{ \, h} }$$ a square matrix with indices in $$[0,h]^2$$.

Then it is convenient to put that $$a_n = 0 \; | \, n < 0$$ , and keeping your definition, starting from $$n_0$$, then the matrix $$\bf A$$ becomes, e.g. for $$h=3$$,
$${\bf A}_{\,3} = \left( {\matrix{ {a_{\,0} } & { - a_{\,0} } & 0 & 0 \cr { - a_{\,0} } & {a_0 + a_{\,1} } & { - a_{\,1} } & 0 \cr 0 & { - a_{\,1} } & {a_{\,1} + a_{\,2} } & { - a_{\,2} } \cr 0 & 0 & { - a_{\,2} } & {a_{\,2} + a_{\,3} } \cr } } \right)$$ We can see that the lower diagonal block contains the matrix as you defined it.

It is not difficult to demonstrate that the determinant is simply
$$d(h) = \left| {\;{\bf A}_{\,h} \;} \right| = \prod\limits_{0\, \le \,k\, \le \,h} {a_{\,k} }$$ while that of the matrix defined by you is $$d_1 (h) = \left| {\;{\bf A}_{\,1 \ldots h} \;} \right| = \sum\limits_{0\, \le \,j\, \le \,h} {\prod\limits_{0\, \le \,k\, \ne \;j\, \le \,h} {a_{\,k} } } = \left( {\prod\limits_{0\, \le \,k\, \le \,h} {a_{\,k} } } \right)\sum\limits_{0\, \le \,j\, \le \,h} {{1 \over {a_{\,j} }}}$$

The eigenvalues are however complicated, and so is the Jordan decomposition..

Trying instead the LU decomposition for the lowest value of $$h$$ we get the hint that it might be quite straight and simple.
We get $${\bf A}_{\,h} = {\bf L}_{\,h} \,{\bf U}_{\,h} = {\bf L}_{\,h} \,{\bf D}_{\,h} \;\overline {{\bf L}_{\,h} }$$ where the overbar denotes the transpose, and where we adopt the following notation \eqalign{ & {\bf D}_{\,h} = \left( {a_{\,n} \circ {\bf I}_{\,h} } \right)\quad \left| {\quad \left( {f(n) \circ {\bf I}} \right)_{\,n,\,m} = f(n)\;\delta _{\,n,\,m} } \right. \cr & {\bf L}_{\,h} = {\bf I}_{\,h} - {\bf E}_{\,h} \quad \left| {\quad {\bf E}_{\,n,\,m} = \;\delta _{\,n,\,m + 1} } \right. \cr}

In fact \eqalign{ & {\bf A}_{\,h} = {\bf L}_{\,h} \,{\bf D}_{\,h} \;\overline {{\bf L}_{\,h} } = \left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)\left( {a_{\,n} \circ {\bf I}_{\,h} } \right)\left( {{\bf I}_{\,h} - \overline {{\bf E}_{\,h} } } \right) = \cr & = \left( {a_{\,n} \circ {\bf I}_{\,h} } \right) - {\bf E}_{\,h} \left( {a_{\,n} \circ {\bf I}_{\,h} } \right) - \left( {a_{\,n} \circ {\bf I}_{\,h} } \right)\overline {{\bf E}_{\,h} } + {\bf E}_{\,h} \left( {a_{\,n} \circ {\bf I}_{\,h} } \right)\overline {{\bf E}_{\,h} } = \cr & = \left( {a_{\,n} \circ {\bf I}_{\,h} } \right) + \left( {a_{\,n - 1} \circ {\bf I}_{\,h} } \right){\bf E}_{\,h} \overline {{\bf E}_{\,h} } - {\bf E}_{\,h} \left( {a_{\,n} \circ {\bf I}_{\,h} } \right) - \left( {a_{\,n} \circ {\bf I}_{\,h} } \right)\overline {{\bf E}_{\,h} } = \cr & = \left( {\left( {a_{\,n} + \left[ {1 \le n} \right]a_{\,n - 1} } \right) \circ {\bf I}_{\,h} } \right) - {\bf E}_{\,h} \left( {a_{\,n} \circ {\bf I}_{\,h} } \right) - \left( {a_{\,n} \circ {\bf I}_{\,h} } \right)\overline {{\bf E}_{\,h} } \cr} which is the definition of $${\bf A}$$
(the square brackets denote the Iverson bracket ).

Since the inverse of $$\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)$$ is the "Summing" matrix $${\bf S}_{\,h}$$ $$\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)^{ - \,{\bf 1}} = {\bf S}_{\,h} \quad \left| {\;S_{\,n,\,m} = \left[ {m \le n} \right]} \right.$$ then we conclude that \eqalign{ & {\bf A}_{\,h} ^{ - \,{\bf 1}} = \left( {{\bf I}_{\,h} - \overline {{\bf E}_{\,h} } } \right)^{ - \,{\bf 1}} \left( {1/a_{\,n} \circ {\bf I}_{\,h} } \right)\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)^{ - \,{\bf 1}} = \cr & = \overline {{\bf S}_{\,h} } \left( {1/a_{\,n} \circ {\bf I}_{\,h} } \right){\bf S}_{\,h} \cr} that is \eqalign{ & \left( {{\bf A}_{\,h} ^{ - \,{\bf 1}} } \right)_{\,n,\,m} = \sum\limits_{0\, \le \,j\, \le \,h} {\sum\limits_{0\, \le \,k\, \le \,h} {\left[ {n \le k} \right]{{\left[ {k = j} \right]} \over {a_{\,k} }}\left[ {m \le j} \right]} } = \cr & = \sum\limits_{0\, \le \,k\, \le \,h} {\left[ {n \le k} \right]{1 \over {a_{\,k} }}\left[ {m \le k} \right] = \sum\limits_{\max \left( {n,m} \right)\, \le \,k\, \le \,h} {{1 \over {a_{\,k} }}} } \cr}

From here, by partitioning $$\bf A$$ into four blocks, enucleating the first row and the first column, and applying the Inversion by Blocks method, we can deduce the inverse of the matrix as defined by you.

with the convention now of indexing the matrices from $$1$$ to $$h$$

\eqalign{ & {\bf A}_{\,h} = \left( {\matrix{ {a_{\,0} + a_{\,1} } & { - a_{\,1} } & 0 & \cdots \cr { - a_{\,1} } & {a_{\,1} + a_{\,2} } & { - a_{\,2} } & \ddots \cr 0 & { - a_{\,2} } & {a_{\,2} + a_{\,3} } & \ddots \cr \vdots & \ddots & \ddots & \ddots \cr } } \right) = \cr & = \left( {\left( {a(n) + a(n - 1)} \right) \circ {\bf I}_{\,h} } \right) - \left( {a(n - 1) \circ {\bf I}_{\,h} } \right){\bf E}_{\,h} - \overline {{\bf E}_{\,h} } \left( {a(n) \circ {\bf I}_{\,h} } \right) \cr & \cr}

The determinant now is $$d (h) = \left| {\;{\bf A}_{\,h} \;} \right| = \sum\limits_{0\, \le \,j\, \le \,h} {\prod\limits_{0\, \le \,k\, \ne \;j\, \le \,h} {a_{\,k} } } = \left( {\prod\limits_{0\, \le \,k\, \le \,h} {a_{\,k} } } \right)\sum\limits_{0\, \le \,j\, \le \,h} {{1 \over {a_{\,j} }}}$$ and we conventionally put $$d(0)=1$$.

The LU decomposition,gives hints to that $${\bf A}_{\,h} = {\bf L}_{\,h} \,{\bf U}_{\,h} = {\bf L}_{\,h} \,{\bf D}_{\,h} \;\overline {{\bf L}_{\,h} }$$ with $$\left\{ \matrix{ {\bf D}_{\,h} = \left( {{{d(n)} \over {d(n - 1)}} \circ {\bf I}_{\,h} } \right) \hfill \cr {\bf L}_{\,h} = {\bf I}_{\,h} - {\bf E}_{\,h} \left( {a(n) \circ {\bf I}_{\,h} } \right)\left( {{{d(n - 1)} \over {d(n)}} \circ {\bf I}_{\,h} } \right) = \hfill \cr = {\bf I}_{\,h} - {\bf E}_{\,h} \left( {a(n){{d(n - 1)} \over {d(n)}} \circ {\bf I}_{\,h} } \right) = \hfill \cr = {\bf I}_{\,h} - {\bf E}_{\,h} \left( {a(n) \circ {\bf I}_{\,h} } \right){\bf D}_{\,h} ^{ - \,{\bf 1}} \hfill \cr} \right.$$

Since \eqalign{ & {\bf I}_{\,h} - \left( {f(n - 1) \circ {\bf I}_{\,h} } \right){\bf E}_{\,h} = {\bf I}_{\,h} - {\bf E}_{\,h} \left( {f(n) \circ {\bf I}_{\,h} } \right)\quad \left| {\;0 \ne f(n)} \right.\;\left| {\;n = 1 \ldots h} \right.\quad = \cr & = \left( {\left( {\prod\limits_{1\, \le k\, \le \,n - 1} {f(k)} } \right) \circ {\bf I}_{\,h} } \right)\;\,\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)\,\;\left( {\left( {\prod\limits_{1\, \le k\, \le \,n - 1} {f(k)} } \right) \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \cr} then \eqalign{ & {\bf L}_{\,h} = {\bf I}_{\,h} - {\bf E}_{\,h} \left( {a(n){{d(n - 1)} \over {d(n)}} \circ {\bf I}_{\,h} } \right) = \cr & = \left( {\left( {{1 \over {d(n - 1)}}\prod\limits_{1\, \le k\, \le \,n - 1} {a(k)} } \right) \circ {\bf I}_{\,h} } \right)\;\,\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)\,\;\left( {\left( {{1 \over {d(n - 1)}}\prod\limits_{1\, \le k\, \le \,n - 1} {a(k)} } \right) \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} = \cr & = \left( {\left( {{{a_{\,0} } \over {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} }}} \right) \circ {\bf I}_{\,h} } \right)\;\,\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)\,\;\left( {\left( {{{a_{\,0} } \over {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} }}} \right) \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} = \cr & = \left( {\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} } \right) \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \;\,\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)\,\;\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} \circ {\bf I}_{\,h} } \right) \cr}

and it is clear the path reach to the conclusion, i.e.

\bbox[lightyellow] { \eqalign{ & {\bf A}_{\,h} ^{\,{\bf - }\,{\bf 1}} = \overline {{\bf L}_{\,h} } ^{\,{\bf - }\,{\bf 1}} \,\;{\bf D}_{\,h} ^{\,{\bf - }\,{\bf 1}} \;{\bf L}_{\,h} ^{\,{\bf - }\,{\bf 1}} \; = \cr & = \left( {\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} } \right) \circ {\bf I}_{\,h} } \right)\;\,\left( {{\bf I}_{\,h} - \overline {{\bf E}_{\,h} } } \right)^{\,{\bf - }\,{\bf 1}} \,\, \cdot \cr & \cdot \;\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \left( {{{\left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {a_{\,k} } } \right)\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} } \over {\left( {\prod\limits_{0\, \le \,k\, \le \,n} {a_{\,k} } } \right)\sum\limits_{0\, \le \,j\, \le \,n} {{1 \over {a_{\,j} }}} }}} \right)\left( {\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} } \right) \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \;\,\, \cdot \cr & \cdot \,\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)\,\;\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} \circ {\bf I}_{\,h} } \right) = \cr & = \left( {\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} } \right) \circ {\bf I}_{\,h} } \right)\;\,\overline {{\bf S}_{\,h} } \;\left( {\left( {a_{\,n} \sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} \sum\limits_{0\, \le \,k\, \le \,n} {{1 \over {a_{\,k} }}} } \right) \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \;{\bf S}_{\,h} \,\;\left( {\sum\limits_{0\, \le \,j\, \le \,n - 1} {{1 \over {a_{\,j} }}} \circ {\bf I}_{\,h} } \right) \cr} }

• Fantastic! Thank you. – becko May 18 at 22:47
• @becko: glad for the appreciation. Does the matrix constructed as above meet you requirements ? or do you exactly need the inverse of the matrix that you specified? – G Cab May 18 at 22:52
• Honestly the blockwise inversion gets a bit messy. I think you are missing a $-1$ in the sum limits in the final expression for the inverse. Should be? $$(A_h^{-1})_{i, j} = \sum_{\max (i, j) - 1 \leq k \leq n} \frac{1}{a_k}$$ – becko May 19 at 7:58
• @becko: the formula I gave is correct and checked, when indexing from $0$ to $h$ – G Cab May 19 at 10:49
• Ah ok. On a first reading I thought you were indexing the matrices from 0, not the entries themselves. – becko May 19 at 10:51