# 线性代数网课代修|数值线性代数代写Numerical linear algebra代考|MATH3371

linearalgebra.me 为您的留学生涯保驾护航 在线性代数linear algebra作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的线性代数linear algebra代写服务。我们的专家在线性代数linear algebra代写方面经验极为丰富，各种线性代数linear algebra相关的作业也就用不着 说。

• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

## 线性代数作业代写linear algebra代考|Existence and Uniqueness

Consider the L1U factorization. Three things can happen. An L1U factorization exists and is unique, it exists, but it is not unique, or it does not exist. The $2 \times 2$ case illustrates this.

Example $3.5$ (LIU of $2 \times 2$ Matrix) Let $a, b, c, d \in \mathbb{C}$. An L1U factorization of $\boldsymbol{A}=\left[\begin{array}{ll}a & b \ c & d\end{array}\right]$ must satisfy the equations
$$\left[\begin{array}{ll} a & b \ c & d \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \ l_{1} & 1 \end{array}\right]\left[\begin{array}{cc} u_{1} & u_{2} \ 0 & u_{3} \end{array}\right]=\left[\begin{array}{cc} u_{1} & u_{2} \ u_{1} l_{1} & u_{2} l_{1}+u_{3} \end{array}\right]$$
for the unknowns $l_{1}$ in $L$ and $u_{1}, u_{2}, u_{3}$ in $\boldsymbol{U}$. The equations are
$$u_{1}=a, \quad u_{2}=b, \quad a l_{1}=c, \quad b l_{1}+u_{3}=d .$$
These equations do not always have a solution. Indeed, the main problem is the equation $a l_{1}=c$. There are essentially three cases

1. $a \neq 0$ : The matrix has a unique $\mathrm{L} 1 \mathrm{U}$ factorization.
2. $a=c=0$ : The L1U factorization exists, but it is not unique. Any value for $l_{1}$ can be used.
3. $a=0, c \neq 0:$ No L1U factorization exists.
Consider the four matrices
$$\boldsymbol{A}{1}:=\left[\begin{array}{cc} 2 & -1 \ -1 & 2 \end{array}\right], \quad \boldsymbol{A}{2}:=\left[\begin{array}{ll} 0 & 1 \ 1 & 1 \end{array}\right], \quad \boldsymbol{A}{3}:=\left[\begin{array}{ll} 0 & 1 \ 0 & 2 \end{array}\right], \quad \boldsymbol{A}{4}:=\left[\begin{array}{ll} 1 & 1 \ 1 & 1 \end{array}\right] .$$
From the previous discussion it follows that $\boldsymbol{A}{1}$ has a unique L1U factorization, $\boldsymbol{A}{2}$ has no L1U factorization, $\boldsymbol{A}{3}$ has an L1U factorization but it is not unique, and $\boldsymbol{A}{4}$ has a unique $\mathrm{L} 1 \mathrm{U}$ factorization even if it is singular.

## 线性代数作业代写linear algebra代考|Block LU Factorization

Suppose $A \in \mathbb{C}^{n \times n}$ is a block matrix of the form
$$\boldsymbol{A}:=\left[\begin{array}{ccc} \boldsymbol{A}{11} & \cdots & \boldsymbol{A}{1 m} \ \vdots & & \vdots \ \boldsymbol{A}{m 1} & \cdots & \boldsymbol{A}{m m} \end{array}\right],$$
where each diagonal block $\boldsymbol{A}{i i}$ is square. We call the factorization a block $\mathbf{L} \boldsymbol{1} \mathbf{U}$ factorization of $\boldsymbol{A}$. Here the $i$ th diagonal blocks $\boldsymbol{I}$ and $\boldsymbol{U}{i i}$ in $\boldsymbol{L}$ and $\boldsymbol{U}$ have the same size as $\boldsymbol{A}{i i}$, the $i$ th diagonal block in $\boldsymbol{A}$. Moreover, the $\boldsymbol{U}{i i}$ are not necessarily upper triangular. Block LU1 and block LDU factorizations are defined similarly.

The results for element-wise LU factorization carry over to block LU factorization as follows.

Theorem 3.5 (Block LU Theorem) Suppose $\boldsymbol{A} \in \mathbb{C}^{n \times n}$ is a block matrix of the form (3.19). Then A has a unique block $L U$ factorization (3.20) if and only if the leading principal block submatrices
$$\boldsymbol{A}{{k}}:=\left[\begin{array}{ccc} \boldsymbol{A}{11} & \cdots & \boldsymbol{A}{1 k} \ \vdots & & \vdots \ \boldsymbol{A}{k 1} & \cdots & \boldsymbol{A}_{k k} \end{array}\right]$$
are nonsingular for $k=1, \ldots, m-1$.

## 线性代数作业代写linear algebra代考|Existence and Uniqueness

$$u_{1}=a, \quad u_{2}=b, \quad a l_{1}=c, \quad b l_{1}+u_{3}=d .$$

1. $a \neq 0:$ 矩阵具有唯一性L1U因式分解。
2. $a=c=0: \mathrm{L} 1 \mathrm{U}$ 分解存在，但不是唯一的。任何价值 $l_{1}$ 可以使用。
3. $a=0, c \neq 0$ :不存在 L1U分解。
考虑四个矩阵
$$\boldsymbol{A} 1:=\left[\begin{array}{llll} 2 & -1 & -1 & 2 \end{array}\right], \quad \boldsymbol{A} 2:=\left[\begin{array}{llll} 0 & 1 & 1 & 1 \end{array}\right], \quad \boldsymbol{A} 3:=\left[\begin{array}{llll} 0 & 1 & 0 & 2 \end{array}\right], \quad \boldsymbol{A} 4:=\left[\begin{array}{llll} 1 & 1 & 1 & 1 \end{array}\right] .$$
从前面的讨论可以看出 $\boldsymbol{A} 1$ 具有独特的 L1U 分解， $\boldsymbol{A}$ 没有 L1U 分解， $\boldsymbol{A} 3$ 有一个 L1U
分解，但它不是唯一的，并且 $\boldsymbol{A} 4$ 有一个独特的L1U因式分解，即使它是单数的。

## 线性代数作业代写linear algebra代考|Block LU Factorization

$$\boldsymbol{A} k:=\left[\begin{array}{llllll} \boldsymbol{A} 11 & \cdots & \boldsymbol{A} 1 k & \vdots \boldsymbol{A} k 1 & \cdots & \boldsymbol{A}_{k k} \end{array}\right]$$

# 计量经济学代写

## 在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions