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|Pivot Strategies

The choice of pivot element in (3.13) is not unique. In partial pivoting we select the largest element
$$\left|a_{r_{k}, k}^{(k)}\right|:=\max \left{\left|a_{i, k}^{(k)}\right|: k \leq i \leq n\right}$$
with $r_{k}$ the smallest such index in case of a tie. The following example illustrating that small pivots should be avoided.

Example 3.4 Applying Gaussian elimination without row interchanges to the linear system
$$\begin{array}{r} 10^{-4} x_{1}+2 x_{2}=4 \ x_{1}+x_{2}=3 \end{array}$$
we obtain the upper triangular system
\begin{aligned} 10^{-4} x_{1}+2 x_{2} &=4 \ \left(1-2 \times 10^{4}\right) x_{2} &=3-4 \times 10^{4} \end{aligned}
The exact solution is
$$x_{2}=\frac{-39997}{-19999} \approx 2, \quad x_{1}=\frac{4-2 x_{2}}{10^{-4}}=\frac{20000}{19999} \approx 1$$
Suppose we round the result of each arithmetic operation to three digits. The solutions $\mathrm{fl}\left(x_{1}\right)$ and $\mathrm{fl}\left(x_{2}\right)$ computed in this way is
$$\mathrm{fl}\left(x_{2}\right)=2, \quad \mathrm{fl}\left(x_{1}\right)=0$$

线性代数作业代写linear algebra代考|The LU and LDU Factorizations

Gaussian elimination without row interchanges is one way of computing an LU factorization of a matrix. There are other ways that can be advantageous for certain kind of problems. Here we consider the general theory of LU factorizations. Recall that $\boldsymbol{A}=\boldsymbol{L} \boldsymbol{U}$ is an $\mathbf{L} \mathbf{U}$ factorization of $\boldsymbol{A} \in \mathbb{C}^{-n \times n}$ if $L \in \mathbb{C}^{-n \times n}$ is lower triangular and $U \in \mathbb{C}^{n \times n}$ is upper triangular, i.e.,
$$\boldsymbol{L}=\left[\begin{array}{ccc} l_{1,1} & \cdots & 0 \ \vdots & \ddots & \vdots \ l_{n, 1} & \cdots & l_{n, n} \end{array}\right], \quad \boldsymbol{U}=\left[\begin{array}{ccc} u_{1,1} & \cdots & u_{1, n} \ \vdots & \ddots & \vdots \ 0 & \cdots & u_{n, n} \end{array}\right]$$
To find an LU factorization there is one equation for each of the $n^{2}$ elements in $\boldsymbol{A}$, and $\boldsymbol{L}$ and $\boldsymbol{U}$ contain a total of $n^{2}+n$ unknown elements. There are several ways to restrict the number of unknowns to $n^{2}$.

L1U: $\quad l_{i i}=1$ all $i$,
LU1: $\quad u_{i i}=1$ all $i$
LDU: $\quad \boldsymbol{A}=\boldsymbol{L} \boldsymbol{D} \boldsymbol{U}, l_{i i}=u_{i i}=1$ all $i, \boldsymbol{D}=\operatorname{diag}\left(d_{11}, \ldots, d_{n n}\right)$.

线性代数作业代写linear algebra代考|Pivot Strategies

(3.13) 中枢轴元素的选择不是唯一的。在部分旋转中，我们选择最大的元素
$\backslash$ left $\mid a_{-}\left{r_{-}{k}, k\right} \wedge{(k)} \backslash$ right $\mid:=\backslash$ max $\backslash$ left ${\backslash$ left|a_{i, $k} \wedge{(k)} \backslash$ right $\mid: k \backslash l e q$ 我 $\backslash$ leq $\left.n \backslash r i g h t\right}$

$$10^{-4} x_{1}+2 x_{2}=4 x_{1}+x_{2}=3$$

$$10^{-4} x_{1}+2 x_{2}=4\left(1-2 \times 10^{4}\right) x_{2}=3-4 \times 10^{4}$$

$$x_{2}=\frac{-39997}{-19999} \approx 2, \quad x_{1}=\frac{4-2 x_{2}}{10^{-4}}=\frac{20000}{19999} \approx 1$$

$$\mathrm{fl}\left(x_{2}\right)=2, \quad \text { fl }\left(x_{1}\right)=0$$

线性代数作业代写linear algebra代考|The LU and LDU Factorizations

$$\boldsymbol{L}=\left[\begin{array}{lllllll} l_{1,1} & \cdots & 0 & \ddots & \vdots l_{n, 1} & \cdots & l_{n, n} \end{array}\right], \quad \boldsymbol{U}=\left[\begin{array}{llllllll} u_{1,1} & \cdots & u_{1, n} & \ddots & \vdots & 0 & \cdots & u_{n, n} \end{array}\right]$$

L1U: $\quad l_{i i}=1$ 全部i

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