# 线性代数网课代修|数值线性代数代写Numerical linear algebra代考|MTH3320

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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|Counting Operations

It is useful to have a number which indicates the amount of work an algorithm requires. In this book we measure this by estimating the total number of (complex) arithmetic operations. We count both additions, subtractions, multiplications and divisions, but not work on indices. As an example we show that the LU factorization of a full matrix of order $n$ using Gaussian elimination requires exactly
$$N_{L U}:=\frac{2}{3} n^{3}-\frac{1}{2} n^{2}-\frac{1}{6} n$$
operations. Let $M, D, A, S$ be the number of (complex) multiplications, divisions, additions, and subtractions. In (3.2) the multiplications and subtractions occur in the calculation of $a_{i j}^{k+1}=a_{i j}^{(k)}-l_{i k}^{(k)} a_{k j}^{(k)}$ which is carried out $(n-k)^{2}$ times. Moreover,each calculation involves one subtraction and one multiplication. Thus we find $M+$ $S=2 \sum_{k=1}^{n-1}(n-k)^{2}=2 \sum_{m=1}^{n-1} m^{2}=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)$. For each $k$ there are $n-k$ divisions giving a sum of $\sum_{k=1}^{n-1}(n-k)=\frac{1}{2} n(n-1)$. Since there are no additions we obtain the total
$$M+D+A+S=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)+\frac{1}{2} n(n-1)=N_{L U}$$
given by (3.9).
We are only interested in $N_{L U}$ when $n$ is large and for such $n$ the term $\frac{2}{3} n^{3}$ dominates. We therefore regularly ignore lower order terms and use number of operations both for the exact count and for the highest order term. We also say more loosely that the number of operations is $O\left(n^{3}\right)$. We will use the number of operations counted in one of these ways as a measure of the complexity of an algorithm and say that the complexity of LU factorization of a full matrix is $O\left(n^{3}\right)$ or more precisely $\frac{2}{3} n^{3}$.

## 线性代数作业代写linear algebra代考|The PLU Factorization

Theorem 3.1 shows that Gaussian elimination without row interchanges can fail on a nonsingular system. A simple example is $\left[\begin{array}{ll}0 & 1 \ 1 & 1\end{array}\right]\left[\begin{array}{l}x_{1} \ x_{2}\end{array}\right]=\left[\begin{array}{l}1 \ 1\end{array}\right]$. We show here that any nonsingular linear system can be solved by Gaussian elimination if we incorporate row interchanges.

Interchanging two rows (and/or two columns) during Gaussian elimination is known as pivoting. The element which is moved to the diagonal position $(k, k)$ is called the pivot element or pivot for short, and the row containing the pivot is called the pivot row. Gaussian elimination with row pivoting can be described as follows.

1. Choose $r_{k} \geq k$ so that $a_{r_{k}, k}^{(k)} \neq 0$.
2. Interchange rows $r_{k}$ and $k$ of $\boldsymbol{A}^{(k)}$.
3. Eliminate by computing $l_{i k}^{(k)}$ and $a_{i j}^{(k+1)}$ using (3.2).
To show that Gaussian elimination can always be carried to completion by using suitable row interchanges suppose by induction on $k$ that $\boldsymbol{A}^{(k)}$ is nonsingular. Since $\boldsymbol{A}^{(1)}=\boldsymbol{A}$ this holds for $k=1$. By Lemma $2.4$ the lower right diagonal block in $\boldsymbol{A}^{(k)}$ is nonsingular. But then at least one element in the first column of that block must be nonzero and it follows that $r_{k}$ exists so that $a_{r_{k}, k}^{(k)} \neq 0$. But then $\boldsymbol{A}^{(k+1)}$ is nonsingular since it is computed from $A^{(k)}$ using row operations preserving the nonsingularity. We conclude that $A^{(k)}$ is nonsingular for $k=1, \ldots, n$.

## 线性代数作业代写linear algebra代考|Counting Operations

$$N_{L U}:=\frac{2}{3} n^{3}-\frac{1}{2} n^{2}-\frac{1}{6} n$$

$S=2 \sum_{k=1}^{n-1}(n-k)^{2}=2 \sum_{m=1}^{n-1} m^{2}=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)$. 对于每个 $k$ 有 $n-k$ 除法给 出总和 $\sum_{k=1}^{n-1}(n-k)=\frac{1}{2} n(n-1)$. 由于没有加法，我们得到总数
$$M+D+A+S=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)+\frac{1}{2} n(n-1)=N_{L U}$$

## 线性代数作业代写linear algebra代考|The PLU Factorization

1. 选择 $r_{k} \geq k$ 以便 $a_{r_{k}, k}^{(k)} \neq 0$.
2. 交换行 $r_{k}$ 和 $k$ 的 $\boldsymbol{A}^{(k)}$.
3. 通过计算消除 $l_{i k}^{(k)}$ 和 $a_{i j}^{(k+1)}$ 使用 (3.2)。
为了证明高斯消元总是可以通过使用适当的行交换来完成，假设通过归纳 $k$ 那 $\boldsymbol{A}^{(k)}$ 是非 奇异的。自从 $\boldsymbol{A}^{(1)}=\boldsymbol{A}$ 这适用于 $k=1$.引理 $2.4$ 右下角块 $\boldsymbol{A}^{(k)}$ 是非奇异的。但是，该 块的第一列中至少有一个元素必须是非零的，并且它遵循 $r_{k}$ 存在使得 $a_{r_{k}, k}^{(k)} \neq 0$. 但是之 后 $\boldsymbol{A}^{(k+1)}$ 是非奇异的，因为它是从 $A^{(k)}$ 使用保留非奇异性的行操作。我们得出结论 $A^{(k)}$ 是非奇异的 $k=1, \ldots, n$.

# 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