# 数学代写|数值分析代写numerical analysis代考|LU Factorization

## 数学代写|数值分析代写numerical analysis代考|LU Factorization

The process of Gaussian elimination for the entries of the matrix $A$ does not depend on $\boldsymbol{b}$, although the changes to $\boldsymbol{b}$ do depend on the entries in $A$. We can separate the two parts of this process. This leads us to the LU factorization, the most common method for solving small-to-moderate systems of linear equations. The main difference between Gaussian elimination and LU factorization is just saving the multipliers $m_{i k}$. We can put the multipliers into a matrix
$$L=\left[\begin{array}{ccccc} 1 & & & & \ m_{21} & 1 & & & \ m_{31} & m_{32} & 1 & & \ \vdots & \vdots & \vdots & \ddots & \ m_{n 1} & m_{n 2} & m_{n 3} & \cdots & 1 \end{array}\right] .$$
The remarkable property of this matrix is that $L U=A$ where $U$ is the upper triangular matrix remaining after $A$ is overwritten in Gaussian elimination, that is, storing the multipliers enables us to reconstruct the matrix $A$. While it can be somewhat difficult to see this directly, there is a recursive version of the LU factorization that makes this easier to see.
Write
$$L=\left[\begin{array}{cc} 1 & \tilde{L} \ m & \tilde{L} \end{array}\right], \quad A=\left[\begin{array}{ll} \alpha & \boldsymbol{r}^T \ \boldsymbol{c} & \tilde{A} \end{array}\right] \quad(n \times n) .$$
Note that $\tilde{L}$ is also lower triangular (that is, all non-zeros occur on or below the main diagonal). Note that the first row of $A$ is not changed by Gaussian elimination. The remaining upper triangular matrix after elimination is
$$U=\left[\begin{array}{r} \alpha \boldsymbol{r}^T \ \widetilde{U} \end{array}\right]$$
where $\tilde{U}$ is also upper triangular. Note that the multipliers in Gaussian elimination are obtained by setting $\boldsymbol{m}$ to $\boldsymbol{c} / \alpha$ (since $\alpha=a_{11}$ and $m_{i 1} \leftarrow a_{i 1} / a_{11}$ ). The effect of the first stage of Gaussian elimination (for $k=1$ ) is to replace $a_{i j} \leftarrow a_{i j}-m_{i 1} a_{1, j}$ for $i, j>1$. This corresponds to replacing $\tilde{A}$ with $\widetilde{A}^{\prime}=\widetilde{A}-m r^T$. Under our implicit induction hypothesis, Gaussian elimination to $\widetilde{A}^{\prime}$ is equivalent to factoring $\widetilde{A}^{\prime}=\widetilde{L} \widetilde{U}$.

## 数学代写|数值分析代写numerical analysis代考|Permutation Matrices

A permutation matrix is the identity matrix with the rows shuffled. As a result, every entry is either zero or one; every column has exactly one entry that is one; and every row has exactly one entry that is one. Examples include the following:
$$\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{llll} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \end{array}\right] .$$
Permutation matrices can be represented efficiently in memory using simple arrays of integers. The above examples would be represented by
$$[1,2], \quad[2,1,3], \quad[3,1,2,4]$$
An array $[\pi(1), \pi(2), \ldots, \pi(n)]$ represents the permutation matrix $P$ where $P \boldsymbol{e}j=$ $\boldsymbol{e}{\pi(j)}, j=1,2, \ldots, n$. Note that every integer from one to $n$ is listed in the array exactly once, so that $\pi$ is a permutation of ${1,2, \ldots, n}$.

Permutation matrices $P$ shuffle the entries of a vector, so that $P x$ has the same entries as $\boldsymbol{x}$, but in a different order. This means that $|P \boldsymbol{x}|_p=|x|_p$ for any $1 \leq$ $p \leq \infty$. In particular, we can take $p=2$, so that $|P \boldsymbol{x}|_2^2=|\boldsymbol{x}|_2^2$ and $\boldsymbol{x}^T P^T P \boldsymbol{x}=$ $\boldsymbol{x}^T \boldsymbol{I} \boldsymbol{x}$ for all $\boldsymbol{x}$. Thus $P^T P=I$ and $P^T=P^{-1}$. That is, permutation matrices are orthogonal.

The vector array subscripting features of MATLAB, Julia, and R enable us to apply permutation matrices without forming an actual permutation matrix. Array comprehensions in Python can achieve the same effect.

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|LU Factorization

$$L=\left[\begin{array}{lll} 1 & \tilde{L} m & \tilde{L} \end{array}\right], \quad A=\left[\begin{array}{llll} \alpha & \boldsymbol{r}^T & \boldsymbol{c} & \tilde{A} \end{array}\right] \quad(n \times n) .$$

$$U=\left[\alpha \boldsymbol{r}^T \widetilde{U}\right]$$

## 数学代写|数值分析代写numerical analysis代考|Permutation Matrices

$$[1,2], \quad[2,1,3], \quad[3,1,2,4]$$

$|P \boldsymbol{x}|_2^2=|\boldsymbol{x}|_2^2$ 和 $\boldsymbol{x}^T P^T P \boldsymbol{x}=\boldsymbol{x}^T \boldsymbol{I} \boldsymbol{x}$ 对全部 $\boldsymbol{x}$. 因此 $P^T P=I$ 和 $P^T=P^{-1}$. 也就是 说，置换矩阵是正交的。

MATLAB、Julia 和 R 的向量数组下标特性使我们能够在不形成实际置换矩阵的情况下 应用置换矩阵。Python 中的数组理解可以达到相同的效果。

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