Approximating the inverse of a diagonally dominant matrix with positive elements

TL;DR

This paper proposes a simple diagonal matrix approximation for the inverse of diagonally dominant matrices with positive elements, providing explicit error bounds that demonstrate high accuracy.

Contribution

The paper introduces a diagonal approximation method for matrix inversion and derives explicit error bounds, improving understanding of approximation accuracy for such matrices.

Findings

The approximation error is of order O(n^{-2}).

The diagonal matrix S closely approximates T^{-1}.

The method is effective for diagonally dominant matrices with positive elements.

Abstract

For an $n\times n$ diagonally dominant matrix $T=(t_{i,j})_{n\times n}$ with positive elements satisfying certain bounding conditions, we propose to use a diagonal matrix $S=(s_{i,j})_{n\times n}$ to approximate the inverse of $T$, where $s_{i,j}=\delta_{i,j}/t_{i,i}$ and $\delta_{i,j}$ is the Kronecker delta function. We derive an explicitly upper bound on the approximation error, which is in the magnitude of $O(n^{-2})$. It shows that $S$ is a very good approximation to $T^{-1}$.

Equations

$$t_{i,j}>0,t_{i,i}\ge \sum _{j=1,j\ne i}^n{t}_{i,j},i=1,\cdots ,n.$$

$$s_{i,j}=\frac{{\delta }_{i,j}}{t_{i,i}},$$

$$\Vert A\Vert :=\underset{i,j}{\max }|a_{i,j}|.$$

$$m:=\underset{1\le i<j\le n}{\min }{t}_{i,j},{\Delta }_i:=t_{i,i}-\sum _{j=1,j\ne i}^n{t}_{i,j},M:=\max \{\underset{1\le i<j\le n}{\max }{t}_{i,j},\underset{1\le i\le n}{\max }{\Delta }_i\}.$$

$$C(m,M)=\frac{2(n-2)m}{nM+(n-2)m}-\frac{(n-2)Mm}{[(n-2)m+M][(n-2)m+2M]}-\frac{M}{m(n-1)}.$$

$$\Vert T^{-1}-S\Vert \le \frac{M}{m^2(n-1{)}^{2}C(m,M)}.$$

$$F=T^{-1}-S,V=(v_{ij})=I_n-TS,W=(w_{ij})=SV.$$

$$F=T^{-1}-S=(T^{-1}-S)(I_n-TS)+S(I_n-TS)=FV+W.$$

$$v_{i,j}={\delta }_{i,j}-\sum _{k=1}^n{t}_{i,k}{s}_{k,j}={\delta }_{i,j}-\sum _{k=1}^n{t}_{i,k}\frac{{\delta }_{k,j}}{t_{j,j}}=({\delta }_{i,j}-1)\frac{t_{i,j}}{t_{j,j}},$$

$$w_{i,j}=\sum _{k=1}^n{s}_{i,k}{v}_{k,j}=\sum _{k=1}^n\frac{{\delta }_{i,k}}{t_{i,i}}[({\delta }_{k,j}-1)\frac{t_{k,j}}{t_{j,j}}]=\frac{({\delta }_{i,j}-1)t_{i,j}}{t_{i,i}{t}_{j,j}}.$$

$$0<\frac{t_{i,j}}{t_{i,i}{t}_{j,j}}\le \frac{M}{m^2(n-1{)}^2},$$

$$|w_{i,i}|=0,|w_{i,j}|\le \frac{M}{m^2(n-1{)}^2},$$

$$|w_{i,j}-w_{i,k}|\le \frac{M}{m^2(n-1{)}^2},|w_{i,i}-w_{i,k}|\le \frac{M}{m^2(n-1{)}^2}.$$

$$\max(|w_{i,j}|,|w_{i,j}-w_{i,k}|)\leq\frac{M}{m^{2}(n-1)^{2}},~{}~{}~{}~{}~{}\mbox{for all $i,j,k$}.$$

$$f_{i,j}=\sum _{k=1}^n{f}_{i,k}[({\delta }_{k,j}-1)\frac{t_{k,j}}{t_{j,j}}]+w_{i,j},j=1,\cdots ,n.$$

$$|f_{i,j}|\le \frac{M}{m^{2}C(M,m)(n-1{)}^2}.$$

$$\sum _{k=1}^n{f}_{i,k}{t}_{k,i}=\sum _{k=1}^n\left([T^{-1}{]}_{i,k}-\frac{{\delta }_{i,k}}{t_{i,i}}\right)t_{k,i}=1-1=0,$$

$$f_{i,\beta }\sum _{k=1}^n{t}_{k,i}\le \sum _{k=1}^n{f}_{i,k}{t}_{k,i}=0.$$

$$t_{\alpha ,\alpha }=-\{\sum _{k=1}^n[({\delta }_{k,\alpha }-1)t_{k,\alpha }-{\delta }_{k,\alpha }{\Delta }_{\alpha }]\},$$

$$f_{i,\beta }=-\sum _{k=1}^n{f}_{i,\beta }\frac{({\delta }_{k,\alpha }-1)t_{k,\alpha }-{\delta }_{k,\alpha }{\Delta }_{\alpha }}{t_{\alpha ,\alpha }}.$$

$$f_{i,\beta }=-\sum _{k=1}^n{f}_{i,\beta }\frac{({\delta }_{k,\beta }-1)t_{k,\beta }-{\delta }_{k,\beta }{\Delta }_{\beta }}{t_{\beta ,\beta }}.$$

$$f_{i,\alpha }+f_{i,\beta }=\sum _{k=1}^n(f_{i,k}-f_{i,\beta })\frac{[({\delta }_{k,\alpha }-1)t_{k,\alpha }-{\delta }_{k,\alpha }{\Delta }_{\alpha }]}{t_{\alpha ,\alpha }}+w_{i,\alpha }.$$

$$2f_{i,\beta }=\sum _{k=1}^n(f_{i,k}-f_{i,\beta })\frac{[({\delta }_{k,\beta }-1)t_{k,\beta }-{\delta }_{k,\beta }{\Delta }_{\beta }]}{t_{\beta ,\beta }}+w_{i,\beta }.$$

$$\begin{array}[]{lll}&&f_{i,\alpha}-f_{i,\beta}\\ &=&\sum_{k=1}^{n}(f_{i,k}-f_{i,\beta})[(\delta_{k,\alpha}-1)\frac{t_{k,\alpha}}{t_{\alpha,\alpha}}-(\delta_{k,\beta}-1)\frac{t_{k,\beta}}{t_{\beta,\beta}}]\\ &&+w_{i,\alpha}-w_{i,\beta}-(\frac{\Delta_{\beta}}{t_{\beta,\beta}}-\frac{\Delta_{\alpha}}{t_{\alpha,\alpha}})f_{i,\beta}.\end{array}$$

$$\sum _{k=1}^n(f_{i,k}-f_{i,\beta })[({\delta }_{k,\alpha }-1)\frac{t_{k,\alpha }}{t_{\alpha ,\alpha }}-({\delta }_{k,\beta }-1)\frac{t_{k,\beta }}{t_{\beta ,\beta }}]$$

$$f(\lambda )$$

$$g(\lambda )$$

$$f^{\prime }(\lambda )$$

$$f^{\prime \prime }(\lambda )=-2(M-m)Mm(n-1)\left(\frac{1}{[\lambda M+(n-1-\lambda )m{]}^3}+\frac{1}{[(\lambda -1)m+(n-\lambda )M{]}^3}\right).$$

$$f(\frac{n}{2})$$

$$\underset{\lambda \in [0,n-1]}{\sup }f(\lambda )\le \frac{nM-(n-2)m}{nM+(n-2)m}.$$

Full text

Key words: Approximation error, Diagonally dominant, Inverse.

Mathematics Subject Classification: 15A09, 15B48.

1 Introduction

In this paper, we consider the approximate inverse of an $n\times n$ diagonally dominant matrices $T=(t_{i,j})_{n\times n}$ with positive elements satisfying certain bounding conditions, i.e.,

[TABLE]

It is easy to show that $T$ must be positive definite. We propose to use a diagonal matrix $S=(s_{i,j})_{n\times n}$ to approximate the inverse of $T$, where

[TABLE]

and $\delta_{i,j}$ is the Kronecker delta function. We obtain an explicitly upper bound on the approximation error in terms of $\max_{i,j}|(T^{-1}-S)_{ij}|$, which has the magnitude of $1/n^{2}$. This shows that $S$ is a very good approximation to $T^{-1}$.

The problems on inverses of nonnegative matrices have been extensively investigated; see Berman and Plemmons (1994); Loewy and London (1978); Egleston et al. (2004). It has applications to solving a large system of linear equations, in which a good approximate inverse of the coefficient matrix plays an important role in establishing fast convergence rates of iterative algorithms Axelsson (1985); Benzi (2002); Bruaset (1995); Zhang et al. (2009). Within statistics, Yan (2019) use the approximate inverse of $T$ to obtain a fast geometric rate of convergence of an iterative sequences for solving the estimate of parameters in the node-parameter network models with dependent structures. Further, it is used to derive the asymptotic representation of an estimator of the model parameter.

2 An explicit bound on the approximation error

For a general matrix $A=(a_{i,j})$, define the matrix maximum norm:

[TABLE]

We measure the approximation error of using $S$ to approximate $T^{-1}$ in terms of $\|T^{-1}-S\|$. Some notations are defined as follows:

[TABLE]

Note that $M\geq m>0$. Let

[TABLE]

The approximate error is formally stated below.

Theorem 1**.**

If $C(m,M)>0$, then for $n\geq 3$, we have

[TABLE]

Proof.

Let $I_{n}$ be the $n\times n$ identity matrix. Define

[TABLE]

Then, we have the recursion:

[TABLE]

A direct calculation gives that

[TABLE]

and

[TABLE]

Recall that $m\leq t_{i,j}\leq M$ and $(n-1)m\leq t_{i,i}\leq nM$. When $i\neq j$, we have

[TABLE]

such that for three different subscripts $i,j,k$,

[TABLE]

It follows that

[TABLE]

We use the recursion (3) to obtain a bound of the approximate error $\|F\|$. By (3) and (4), for any $i$, we have

[TABLE]

Thus, to prove Theorem 1, it is sufficient to show that for any $i,j$,

[TABLE]

Define $f_{i,\alpha}=\max\limits_{1\leq k\leq n}f_{i,k}$ and $f_{i,\beta}=\min\limits_{1\leq k\leq n}f_{i,k}$.

First, we will show that $f_{i,\beta}\leq 0$. Since for any fixed $i$,

[TABLE]

we have

[TABLE]

It follows that $f_{i,\beta}\leq 0$. With similar arguments, we have that $f_{i,\alpha}\geq 0$.

Recall that $\Delta_{\alpha}=t_{\alpha,\alpha}-\sum_{k=1,k\neq\alpha}^{n}t_{k,\alpha}$. Since

[TABLE]

we have the identity

[TABLE]

Similarly, we have

[TABLE]

By combining (7) and (9), where we set $i=\alpha$ in (7), it yields that

[TABLE]

Again, by combining (7) and (10), we have

[TABLE]

By subtracting (12) from (11), we get

[TABLE]

Let $\Omega=\{k:(1-\delta_{k,\beta})t_{k,\beta}/t_{\beta,\beta}\geq(1-\delta_{k,\alpha})t_{k,\alpha}/t_{\alpha,\alpha}\}$ and define $\lambda:=|\Omega|$. Note that $1\leq\lambda\leq n-1$. Then,

[TABLE]

We will obtain the maximum value of the expression in the above bracket through dividing it into two functions $f(\lambda)$ and $g(\lambda)$ of $\lambda$, where

[TABLE]

We first derive the maximum value of $f(\lambda)$. There are two cases to consider the maximum value of $f(\lambda)$ in the range of $\lambda\in[1,n-1]$.

Case I: When $M=m$, it is easy to show $f(\lambda)=1/(n-1)$.

Case II: $M\neq m$. A direct calculation gives that

[TABLE]

and

[TABLE]

Since $f^{\prime\prime}(\lambda)\leq 0$ when $\lambda\in[1,n-1]$, $f(\lambda)$ is a convex function of $\lambda$ ($\in[1,n-1]$) such that $f(\lambda)$ takes its maximum value at $\lambda=n/2$ when $1\leq\lambda\leq n-1$. Note that

[TABLE]

So we have

[TABLE]

Next, we obtain the maximum value of $g(\lambda)$. Since

[TABLE]

$g^{\prime}(\lambda)>0$ when $1\leq\lambda\leq n-1$. So $g(\lambda)$ is an increasing function on $\lambda$ such that

[TABLE]

By combining (15) and (16), we have

[TABLE]

where $I(\cdot)$ is an indictor function. By combining (13), (LABEL:yyy) and (17), we have

[TABLE]

Since $f_{i,\alpha}\geq|f_{i,\beta}|$ and $f_{i,\beta}\leq 0$, we have

[TABLE]

Recall the definition of $C(m,M)$ in (2). By combining (18) and (19), it yields

[TABLE]

Consequently,

[TABLE]

This completes the proof. ∎

We discuss the condition $C(m,M)>0$. $C(m,M)$ can be represented as

[TABLE]

So if $M/m=o(n)$, then for large $n$

[TABLE]

Then we immediately have the corollary.

Corollary 1**.**

If $M/m=o(n)$, then for large $n$,

[TABLE]

3 Discussion

The bound on the approximation error in Theorem 1 depends on $m$, $M$ and $n$. When $m$ and $M$ are bounded by a constant, all the elements of $T^{-1}-S$ are of order $O(1/n^{2})$ as $n\to\infty$, uniformly. Therefore we conjecture that $T$ may belong to inverse $M$-matrices. The interested readers can refer to Berman and Plemmons (1994); Foregger (1990).

We illustrate by an example that the bound on the approximation error in Theorem 1 is optimal in the sense that any bound in the form of $K(m,M)/f(n)$ requires $f(n)=O(n^{2})$ as $n\to\infty$. Assume that the matrix $T$ consists of the elements: $t_{i,i}=(n-1)M,i=1,\cdots,n-1;t_{n,n}=(n-1)m$ and $t_{i,j}=m,i,j=1,\cdots,n;i\neq j$, which satisfies (1). By the Sherman-Morrison formula, we have

[TABLE]

In this case, the elements of $S$ are

[TABLE]

It is easy to show that the bound of $||T^{-1}-S||$ is $O(\frac{1}{(n-1)^{2}m})$. This suggests that the rate $1/(n-1)^{2}$ is optimal. On the other hand, there is a gap between $1/m$ and $O(M^{2}/m^{3})$ which implies that there might be space for improvement. It is interesting to see if the bounds in Theorem 1 can be further relaxed.