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# 数学代写|数值分析代写Numerical analysis代考|MATH2722 Iterative Methods for Linear Systems

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## 数学代写|数值分析代写Numerical analysis代考|Linear Iterative Systems

We begin with the basic definition of an iterative system of linear equations.
Definition 7.1. A linear iterative system takes the form
$$\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}, \quad \mathbf{u}^{(0)}=\mathbf{a} .$$
The coefficient matrix $T$ has size $n \times n$. We will consider both real and complex systems, and so the iterates ${ }^{\dagger} \mathbf{u}^{(k)}$ are vectors either in $\mathbb{R}^n$ (which assumes that the coefficient matrix $T$ is also real) or in $\mathbb{C}^n$. For $k=1,2,3, \ldots$, the solution $\mathbf{u}^{(k)}$ is uniquely determined by the initial conditions $\mathbf{u}^{(0)}=\mathbf{a}$.
Powers of Matrices
The solution to the general linear iterative system (7.1) is, at least at first glance, immediate. Clearly,
$$\mathbf{u}^{(1)}=T \mathbf{u}^{(0)}=T \mathbf{a}, \quad \mathbf{u}^{(2)}=T \mathbf{u}^{(1)}=T^2 \mathbf{a}, \quad \mathbf{u}^{(3)}=T \mathbf{u}^{(2)}=T^3 \mathbf{a}$$

and, in general,
$$\mathbf{u}^{(k)}=T^k \mathbf{a} .$$
Thus, the iterates are simply determined by multiplying the initial vector a by the successive powers of the coefficient matrix $T$. And so, unlike differential equations, proving the existence and uniqueness of solutions to an iterative system is completely trivial.

However, unlike real or complex scalars, the general formulae and qualitative behavior of the powers of a square matrix are not nearly so immediately apparent. (Before continuing, the reader is urged to experiment with simple $2 \times 2$ matrices, trying to detect patterns.) To make progress, recall how we managed to solve linear systems of differential equations by suitably adapting the known exponential solution from the scalar version. In the iterative case, the scalar solution formula (2.8) is written in terms of powers, not exponentials. This motivates us to try the power ansatz
$$\mathbf{u}^{(k)}=\lambda^k \mathbf{v},$$
in which $\lambda$ is a scalar and $\mathbf{v}$ is a fixed vector, as a possible solution to the system. We find
$$\mathbf{u}^{(k+1)}=\lambda^{k+1} \mathbf{v}, \quad \text { while } \quad T \mathbf{u}^{(k)}=T\left(\lambda^k \mathbf{v}\right)=\lambda^k T \mathbf{v}$$

## 数学代写|数值分析代写Numerical analysis代考|Gaussian Elimination — Regular Case

With the basic matrix arithmetic operations in hand, let us now return to our primary task. The goal is to develop a systematic method for solving linear systems of equations. While we could continue to work directly with the equations, matrices provide a convenient alternative that begins by merely shortening the amount of writing, but ultimately leads to profound insight into the structure of linear systems and their solutions.

We begin by replacing the system (3.2) by its matrix constituents. It is convenient to ignore the vector of unknowns, and form the augmented matrix
$$M=(A \mid \mathbf{b})=\left(\begin{array}{cccc|c} a_{11} & a_{12} & \ldots & a_{1 n} & b_1 \ a_{21} & a_{22} & \ldots & a_{2 n} & b_2 \ \vdots & \vdots & \ddots & \vdots & \vdots \ a_{m 1} & a_{m 2} & \ldots & a_{m n} & b_m \end{array}\right)$$
which is an $m \times(n+1)$ matrix obtained by tacking the right hand side vector onto the original coefficient matrix. The extra vertical line is included just to remind us that the last column of this matrix plays a special role. For example, the augmented matrix for the system (4.1), i.e.,
\begin{aligned} x+2 y+z & =2, \ 2 x+6 y+z & =7, \ x+y+4 z & =3, \end{aligned} \quad \text { is } \quad M=\left(\begin{array}{lll|l} 1 & 2 & 1 & 2 \ 2 & 6 & 1 & 7 \ 1 & 1 & 4 & 3 \end{array}\right)

## 数学代写|数值分析代写NUMERICAL ANALYSIS代考|LINEAR ITERATIVE SYSTEMS

$$\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}, \quad \mathbf{u}^{(0)}=\mathbf{a} .$$

$$而且，总的来说，$$
\mathbf{u}^{(k)}=T^k \mathbf{a} .
$$因此，通过将初始向量 a 乘以系数矩阵的连续帛来简单地确定迭代 T. 因此，与微分方程不同，证明迭代系统解的存在性和唯一性是完全微不足道的。 然而，与实标量或复标量不同的是，方阵的幕的一般公式和定性行为并不那么明显。 Beforecontinuing, thereaderisurgedtoexperimentwithsimple \ 2 \times 2 \$$ matrices, tryingtodetectpatterns. 为了取得进步，回想一下我们如何通过适当地 调整标量版本的已知指数解来设法求解微分方程的线性系统。在迭代情况下，标量解公式 $2.8$ 是用掦而不是指数来写的。这促使我们尝试使用 power ansatz
$$\mathbf{u}^{(k)}=\lambda^k \mathbf{v}$$

$$\mathbf{u}^{(k+1)}=\lambda^{k+1} \mathbf{v}, \quad \text { while } \quad T \mathbf{u}^{(k)}=T\left(\lambda^k \mathbf{v}\right)=\lambda^k T \mathbf{v}$$

## 数学代写|数值分析代写NUMERICAL ANALYSIS代考|GAUSSIAN ELIMINATION – REGULAR CASE

$$x+2 y+z=2,2 x+6 y+z \quad=7, x+y+4 z=3, \quad \text { is } \quad M=\left(\begin{array}{lll|l|l|l|l|l|l|l|l} 1 & 2 & 1 & 22 & 6 & 1 & 7 & 1 & 1 & 4 & 3 \end{array}\right)$$

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