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

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• 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|Triangular Matrices

Lemma 2.4 (Inverse of a Block Triangular Matrix) Suppose
$$\boldsymbol{A}=\left[\begin{array}{cc} \boldsymbol{A}{11} & \boldsymbol{A}{12} \ \mathbf{0} & \boldsymbol{A}{22} \end{array}\right]$$ where $\boldsymbol{A}, \boldsymbol{A}{11}$ and $\boldsymbol{A}{22}$ are square matrices. Then $\boldsymbol{A}$ is nonsingular if and only if both $A{11}$ and $A_{22}$ are nonsingular. In that case
$$\boldsymbol{A}^{-1}=\left[\begin{array}{cc} \boldsymbol{A}{11}^{-1} & \boldsymbol{C} \ \mathbf{0} & \boldsymbol{A}{22}^{-1} \end{array}\right]$$
for some matrix $\boldsymbol{C}$.
Proof Suppose $A$ is nonsingular. We partition $B:=A^{-1}$ conformally with $A$ and have
$$\boldsymbol{B A}-\left[\begin{array}{ll} \boldsymbol{B}{11} & \boldsymbol{B}{12} \ \boldsymbol{B}{21} & \boldsymbol{B}{22} \end{array}\right]\left[\begin{array}{cc} \boldsymbol{A}{11} & \boldsymbol{A}{12} \ \mathbf{0} & \boldsymbol{A}{22} \end{array}\right]-\left[\begin{array}{ll} \boldsymbol{I} & \mathbf{0} \ \mathbf{0} & \boldsymbol{I} \end{array}\right]-\boldsymbol{I}$$ Using block-multiplication we find $$B{11} A_{11}=I, B_{21} A_{11}=\mathbf{0}, B_{21} A_{12}+B_{22} A_{22}=I, B_{11} A_{12}+B_{12} A_{22}=0 .$$

## 线性代数作业代写linear algebra代考|Algorithms for Triangular Systems

A nonsingular triangular linear system $\boldsymbol{A x}=\boldsymbol{b}$ is easy to solve. By Lemma $2.5 \boldsymbol{A}$ has nonzero diagonal elements. Consider first the lower triangular case. For $n=3$ the system is
$$\left[\begin{array}{ccc} a_{11} & 0 & 0 \ a_{21} & a_{22} & 0 \ a_{31} & a_{32} & a_{33} \end{array}\right]\left[\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right]=\left[\begin{array}{l} b_{1} \ b_{2} \ b_{3} \end{array}\right] .$$
From the first equation we find $x_{1}=b_{1} / a_{11}$. Solving the second equation for $x_{2}$ we obtain $x_{2}=\left(b_{2}-a_{21} x_{1}\right) / a_{22}$. Finally the third equation gives $x_{3}=\left(b_{3}-a_{31} x_{1}-\right.$ $\left.a_{32} x_{2}\right) / a_{33}$. This process is known as forward substitution. In general
$$x_{k}=\left(b_{k}-\sum_{j=1}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n .$$
When $\boldsymbol{A}$ is a lower triangular band matrix the number of arithmetic operations necessary to find $\boldsymbol{x}$ can be reduced. Suppose $\boldsymbol{A}$ is a lower triangular $d$-banded, so that $a_{k, j}=0$ for $j \notin\left{l_{k}, l_{k}+1, \ldots, k\right.$ for $k=1,2, \ldots, n$, and where $l_{k}:=\max (1, k-d)$, see Fig. 3.2. For a lower triangular $d$-band matrix the calculation in (3.7) can be simplified as follows
$$x_{k}=\left(b_{k}-\sum_{j=l_{k}}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n .$$
Note that (3.8) reduces to (3.7) if $d=n$. Letting $A\left(k, l_{k}:(k-1)\right) * x\left(l_{k}:(k-1)\right)$ denote the sum $\sum_{j=l_{k}}^{k-1} a_{k j} x_{j}$ we arrive at the following algorithm, where the initial ” $\mathrm{r}$ ” in the name signals that this algorithm is row oriented. The algorithm takes a nonsingular lower triangular $d$-banded matrix $\boldsymbol{A} \in \mathbb{C}^{n \times n}$, and $\boldsymbol{b} \in \mathbb{C}^{n}$, as input, and returns an $\boldsymbol{x} \in \mathbb{C}^{n}$ so that $\boldsymbol{A x}=\boldsymbol{b}$. For each $k$ we take the inner product of a part of a row with the already computed unknowns.

## 线性代数作业代写linear algebra代考|Triangular Matrices

$$\boldsymbol{A}=\left[\begin{array}{llll} \boldsymbol{A 1 1} & \boldsymbol{A 1 2} \mathbf{0} & \boldsymbol{A 2 2} \end{array}\right]$$

$$\boldsymbol{A}^{-1}=\left[\begin{array}{lll} \boldsymbol{A} 11^{-1} & \boldsymbol{C} 0 & \boldsymbol{A} 22^{-1} \end{array}\right]$$

$$\boldsymbol{B A}-\left[\begin{array}{llll} \boldsymbol{B} 11 & \boldsymbol{B} 12 & \boldsymbol{B} 21 & \boldsymbol{B} 22 \end{array}\right]\left[\begin{array}{lll} \boldsymbol{A} 11 & \boldsymbol{A} 12 \mathbf{0} & \boldsymbol{A} 22 \end{array}\right]-\left[\begin{array}{llll} \boldsymbol{I} & \mathbf{0} & \mathbf{0} & \boldsymbol{I} \end{array}\right]-\boldsymbol{I}$$

$$B 11 A_{11}=I, B_{21} A_{11}=\mathbf{0}, B_{21} A_{12}+B_{22} A_{22}=I, B_{11} A_{12}+B_{12} A_{22}=0 .$$

## 线性代数作业代写linear algebra代考|Algorithms for Triangular Systems

$$\left[\begin{array}{lllllllll} a_{11} & 0 & 0 & a_{21} & a_{22} & 0 & a_{31} & a_{32} & a_{33} \end{array}\right]\left[\begin{array}{lll} x_{1} & x_{2} & x_{3} \end{array}\right]=\left[\begin{array}{lll} b_{1} & b_{2} & b_{3} \end{array}\right] .$$

$$x_{k}=\left(b_{k}-\sum_{j=1}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n$$

# 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