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# 数学代写|数值分析代写Numerical analysis代考|STAT360 Monotone matrices

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## 数学代写数值分析代写Numerical analysis代考|Monotone matrices

If a positive real number $a$ is increased by $\varepsilon>0$ to $a+\varepsilon$, then its reciprocal $a^{-1}$ decreases to $(a+\varepsilon)^{-1}$. It is not usually true, however, that if we increase some or all of the elements of a nonsingular matrix $A \in \mathbb{R}^{n \times n}$, then the elements of the inverse $A^{-1} \in \mathbb{R}^{n \times n}$ will decrease. This useful property holds for the class of monotone matrices defined below.

The discussion in this section is not related to Gaussian elimination and LU factorisation, but it is of relevance in the iterative solution of systems of linear equations with monotone matrices which arise in the course of numerical approximation of boundary value problems for certain ordinary and partial differential equations.

Definition 3.5 The nonsingular matrix $A \in \mathbb{R}^{n \times n}$ is said to be monotone if all the elements of the inverse $A^{-1}$ are nonnegative.

Example 3.3 Suppose that $a$ and $d$ are positive real numbers, and $b$ and $c$ are nonnegative real numbers such that $a d>b c$. Then,
$$A=\left(\begin{array}{rr} a & -b \ -c & d \end{array}\right)$$
is a monotone matrix. This is easily seen by considering the inverse of the matrix A,
$$A^{-1}=\frac{1}{a d-b c}\left(\begin{array}{ll} d & b \ c & a \end{array}\right),$$
and noting that all elements of $A^{-1}$ are nonnegative.
Next we introduce the concept of ordering in $\mathbb{R}^n$ and $\mathbb{R}^{n \times n}$.

## 数学代写|数值分析代写Numerical analysis代考|Simultaneous iteration

Let $D$ be a nonempty closed subset of $\mathbb{R}^n$ and $\boldsymbol{f}: D\left(\subset \mathbb{R}^n\right) \rightarrow \mathbb{R}^n$ a continuous function defined on $D$. We shall be concerned with the problem of finding $\boldsymbol{\xi} \in D$ such that $\boldsymbol{f}(\boldsymbol{\xi})=\mathbf{0}$. If such $\boldsymbol{\xi}$ exists, it is called a solution to the equation $\boldsymbol{f}(\boldsymbol{x})=\mathbf{0}$ (in $D$ ). When written in componentwise form, $\boldsymbol{f}(\boldsymbol{x})=\mathbf{0}$ becomes
$$f_i\left(x_1, \ldots, x_n\right)=0, \quad i=1, \ldots, n,$$
a system of $n$ simultaneous nonlinear equations for $n$ unknowns, where $f_1, \ldots, f_n$ are the components of $\boldsymbol{f}$.

Example 4.1 Consider the system of two simultaneous nonlinear equations in two unknowns, $x_1$ and $x_2$, defined by
\begin{aligned} x_1^2+x_2^2-1 & =0, \ 5 x_1^2+21 x_2^2-9 & =0 . \end{aligned}
Here $\boldsymbol{x}=\left(x_1, x_2\right)^{\mathrm{T}}$ and $\boldsymbol{f}=\left(f_1, f_2\right)^{\mathrm{T}}$ with
\begin{aligned} & f_1\left(x_1, x_2\right)=x_1^2+x_2^2-1, \ & f_2\left(x_1, x_2\right)=5 x_1^2+21 x_2^2-9 . \end{aligned}
The equation $\boldsymbol{f}(\boldsymbol{x})=\mathbf{0}$ has four solutions:
$$\begin{array}{ll} \boldsymbol{\xi}_1=(-\sqrt{3} / 2,1 / 2)^{\mathrm{T}}, & \boldsymbol{\xi}_2=(\sqrt{3} / 2,1 / 2)^{\mathrm{T}}, \ \boldsymbol{\xi}_3=(-\sqrt{3} / 2,-1 / 2)^{\mathrm{T}}, & \boldsymbol{\xi}_4=(\sqrt{3} / 2,-1 / 2)^{\mathrm{T}} . \end{array}$$

## 数学代写数值分析代写Numerical analysis代考|Monotone matrices

$$A=\left(\begin{array}{lll} a & -b-c & d \end{array}\right)$$

$$A^{-1}=\frac{1}{a d-b c}\left(\begin{array}{lll} d & b c & a \end{array}\right),$$

## 数学代写|数值分析代写Numerical analysis代考|Simultaneous iteration

$$f_i\left(x_1, \ldots, x_n\right)=0, \quad i=1, \ldots, n,$$

$$x_1^2+x_2^2-1=0,5 x_1^2+21 x_2^2-9=0 .$$

$$f_1\left(x_1, x_2\right)=x_1^2+x_2^2-1, \quad f_2\left(x_1, x_2\right)=5 x_1^2+21 x_2^2-9 .$$

$$\boldsymbol{\xi}_1=(-\sqrt{3} / 2,1 / 2)^{\mathrm{T}}, \quad \boldsymbol{\xi}_2=(\sqrt{3} / 2,1 / 2)^{\mathrm{T}}, \boldsymbol{\xi}_3=(-\sqrt{3} / 2,-1 / 2)^{\mathrm{T}}, \quad \boldsymbol{\xi}_4=(\sqrt{3} / 2,-1 / 2)^{\mathrm{T}} .$$

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