### 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Linear Systems

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等楖率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Simultaneous Linear Equations

Here we consider the solution of simultaneous linear equations of the form
$$A \mathbf{x}=\mathbf{b}$$
where $A$ is a matrix of coefficients, $\mathbf{b}$ is a given vector, and $\mathbf{x}$ is the vector of unknowns to be determined. In the first instance, we assume $A$ is square with $n$ rows and $n$ columns, and $\mathbf{x}, \mathbf{b} \in \mathbb{R}^{n}$. At least one element of $\mathbf{b}$ is non-zero. The equations have a unique solution if and only if $A$ is a non-singular matrix, i.e., the inverse $A^{-1}$ exists. The solution is then $\mathbf{x}=A^{-1} \mathbf{b}$. There is no need to calculate $A^{-1}$ explicitly, since the vector $A^{-1} \mathbf{b}$ needs to be calculated and the calculation of $A^{-1}$ would be an intermediate step. The calculation of a matrix inverse is usually avoided unless the elements of the inverse itself are required for other purposes, since this can lead to unnecessary loss of accuracy.
If $A$ is singular, there exist non-zero vectors $v$ such that
$$A \mathbf{v}=\mathbf{0}$$
These vectors lie in the null space of $A$. That is the space of all vectors mapped to zero when multiplied by $A$. If $\mathbf{x}$ is a solution of (2.1) then so is $\mathbf{x}+\mathbf{v}$, since
$$A(\mathbf{x}+\mathbf{v})=A \mathbf{x}+A \mathbf{v}=\mathbf{b}+\mathbf{0}=\mathbf{b}$$
In this case there are infinitely many solutions.
The result of $A$ applied to all vectors in $\mathbb{R}^{n}$ is called the image of $A$. If $\mathbf{b}$ does not lie in the image of $A$, then no vector $\mathbf{x}$ satisfies (2.1) and there is no solution. In this case the equations are inconsistent. This situation can also occur when $A$ is singular.

The solution of Equation $(2.1)$ is trivial if the matrix $A$ is either lower

triangular or upper triangular, i.e.,
$$\left(\begin{array}{cccc} a_{1,1} & 0 & \cdots & 0 \ \vdots & \ddots & \ddots & \vdots \ a_{n-1,1} & \cdots & a_{n-1, n-1} & 0 \ a_{n, 1} & \cdots & \cdots & a_{n, n} \end{array}\right) \text { or }\left(\begin{array}{cccc} a_{1,1} & \cdots & \cdots & a_{1, n} \ 0 & a_{2,2} & \cdots & a_{2, n} \ \vdots & \ddots & \ddots & \vdots \ 0 & \cdots & 0 & a_{n, n} \end{array}\right) \text {. }$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Gaussian Elimination and Pivoting

Given a set of simultaneous equations $A \mathbf{x}=\mathbf{b}$, the solution $\mathbf{x}$ is unchanged if any of the following operations is performed:

1. Multiplication of an equation by a non-zero constant.
2. Addition of the multiple of one equation to another.
3. Interchange of two equations.
The same operations have to be performed on both sides of the equal sign. These operations are used to convert the system of equations to the trivial case, i.e., upper or lower triangular form. This is called Gaussian elimination. By its nature there are a many different ways to go about this. The usual strategy is called pivotal strategy and we see below that this in general avoids the accumulation of errors and in some situations is crucially important.
A pivot entry is usually required to be at least distinct from zero and often well away from it. Finding this element is called pivoting. Once the pivot element is found, interchange of rows (and possibly columns) may follow to bring the pivot element into a certain position. Pivoting can be viewed as sorting rows (and possibly columns) in a matrix. The swapping of rows is equivalent to multiplying $A$ by a permutation matrix. In practice the matrix elements are, however, rarely moved, since this would cost too much time. Instead the algorithms keep track of the permutations. Pivoting increases the overall computational cost of an algorithm. However, sometimes pivoting is necessary for the algorithm to work at all, at other times it increases the numerical stability. We illustrate this with two examples.

Consider the three simultaneous equations where the diagonal of the matrix consists entirely of zeros,
$$\left(\begin{array}{lll} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{array}\right)\left(\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right)=\left(\begin{array}{l} 1 \ 2 \ 4 \end{array}\right)$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|LU Factorization

Another possibility to solve a linear system is to factorize $A$ into a lower triangular matrix $L$ (i.e., $L_{i, j}=0$ for $ij$ ), that is, $A=L U$. This is called $L U$ factorization. The linear system then becomes $L(U \mathbf{x})=\mathbf{b}$, which we decompose into $L \mathbf{y}=\mathbf{b}$ and $U \mathbf{x}=\mathbf{y}$. Both these systems can be solved easily by back substitution.
Other applications of the LU factorization are

1. Calculation of determinant:
$$\operatorname{det} A=(\operatorname{det} L)(\operatorname{det} U)=\left(\prod_{k=1}^{n} L_{k, k}\right)\left(\prod_{k=1}^{n} U_{k, k}\right) .$$
2. Non-singularity testing: $A=L U$ is non-singular if and only if all the diagonal elements of $L$ and $U$ are nonzero.
3. Calculating the inverse: The inverse of triangular matrices can be easily calculated directly. Subsequently $A^{-1}=U^{-1} L^{-1}$.

In the following we derive how to obtain the LU factorization. We denote the columns of $L$ by $\mathbf{l}{1}, \ldots, \mathbf{l}{n}$ and the rows of $U$ by $\mathbf{u}{1}^{T}, \ldots, \mathbf{u}{n}^{T}$. Thus
$$A=L U=\left(\mathbf{1}{1} \ldots \mathbf{l}{n}\right)\left(\begin{array}{c} \mathbf{u}{1}^{T} \ \vdots \ \mathbf{u}{n}^{T} \end{array}\right)=\sum_{k=1}^{n} \mathbf{l}{k} \mathbf{u}{k}^{T}$$
Assume that $A$ is nonsingular and that the factorization exists. Hence the diagonal elements of $L$ are non-zero. Since $\mathbf{l}{k} \mathbf{u}{k}^{T}$ stays the same if $\mathbf{l}{k}$ is replaced by $\alpha \mathbf{l}{k}$ and $\mathbf{u}{k}$ by $\alpha^{-1} \mathbf{u}{k}$, where $\alpha \neq 0$, we can assume that all diagonal elements of $L$ are equal to 1 .

Since the first $k-1$ components of $\mathbf{l}{k}$ and $\mathbf{u}{k}$ are zero, each matrix $\mathbf{l}{k} \mathbf{u}{k}^{T}$ has zeros in the first $k-1$ rows and columns. It follows that $\mathbf{u}{1}^{T}$ is the first row of $A$ and $l{1}$ is the first column of $A$ divided by $A_{1,1}$ so that $L_{1,1}=1$.
Having found $\mathbf{l}{1}$ and $\mathbf{u}{1}$, we form the matrix $A_{1}=A-\mathbf{l}{1} \mathbf{u}{1}^{T}=\sum_{k=2}^{n} \mathbf{l}{k} \mathbf{u}{k}^{T}$.

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Simultaneous Linear Equations

(一种1,10⋯0 ⋮⋱⋱⋮ 一种n−1,1⋯一种n−1,n−10 一种n,1⋯⋯一种n,n) 或者 (一种1,1⋯⋯一种1,n 0一种2,2⋯一种2,n ⋮⋱⋱⋮ 0⋯0一种n,n).

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Gaussian Elimination and Pivoting

1. 一个方程乘以一个非零常数。
2. 将一个方程的倍数加到另一个方程。
3. 两个方程的互换。
必须在等号的两边执行相同的操作。这些操作用于将方程组转换为平凡的情况，即上三角或下三角形式。这称为高斯消元法。就其性质而言，有许多不同的方法可以解决这个问题。通常的策略称为关键策略，我们在下面看到，这通常可以避免错误的累积，并且在某些情况下至关重要。
通常要求枢轴条目至少与零不同，并且通常远离零。找到这个元素称为旋转。一旦找到枢轴元素，就可以交换行（可能还有列），以将枢轴元素带到某个位置。透视可以被视为对矩阵中的行（可能还有列）进行排序。行的交换相当于相乘一种通过置换矩阵。然而，在实践中，矩阵元素很少移动，因为这会花费太多时间。相反，算法会跟踪排列。旋转增加了算法的总体计算成本。然而，有时旋转对于算法的工作来说是必要的，有时它会增加数值稳定性。我们用两个例子来说明这一点。

(011 101 110)(X1 X2 X3)=(1 2 4)

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|LU Factorization

LU 分解的其他应用是

1. 行列式的计算：
这⁡一种=(这⁡大号)(这⁡在)=(∏ķ=1n大号ķ,ķ)(∏ķ=1n在ķ,ķ).
2. 非奇点测试：一种=大号在是非奇异的当且仅当大号和在是非零的。
3. 计算逆矩阵：可以很容易地直接计算三角矩阵的逆矩阵。随后一种−1=在−1大号−1.

## 广义线性模型代考

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## MATLAB代写

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