{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 第四讲:$A$ 的 $LU$ 分解\n", "\n", "$AB$的逆矩阵:\n", "$$\n", "\\begin{aligned}\n", "A \\cdot A^{-1} = I & = A^{-1} \\cdot A\\\\\n", "(AB) \\cdot (B^{-1}A^{-1}) & = I\\\\\n", "\\textrm{则} AB \\textrm{的逆矩阵为} & B^{-1}A^{-1}\n", "\\end{aligned}\n", "$$\n", "\n", "$A^{T}$的逆矩阵:\n", "$$\n", "\\begin{aligned}\n", "(A \\cdot A^{-1})^{T} & = I^{T}\\\\\n", "(A^{-1})^{T} \\cdot A^{T} & = I\\\\\n", "\\textrm{则} A^{T} \\textrm{的逆矩阵为} & (A^{-1})^{T}\n", "\\end{aligned}\n", "$$\n", "\n", "## 将一个 $n$ 阶方阵 $A$ 变换为 $LU$ 需要的计算量估计:\n", "\n", "1. 第一步,将$a_{11}$作为主元,需要的运算量约为$n^2$\n", "$$\n", "\\begin{bmatrix}\n", "a_{11} & a_{12} & \\cdots & a_{1n} \\\\\n", "a_{21} & a_{22} & \\cdots & a_{2n} \\\\\n", "\\vdots & \\vdots & \\ddots & \\vdots \\\\\n", "a_{n1} & a_{n2} & \\cdots & a_{nn} \\\\\n", "\\end{bmatrix}\n", "\\underrightarrow{消元}\n", "\\begin{bmatrix}\n", "a_{11} & a_{12} & \\cdots & a_{1n} \\\\\n", "0 & a_{22} & \\cdots & a_{2n} \\\\\n", "0 & \\vdots & \\ddots & \\vdots \\\\\n", "0 & a_{n2} & \\cdots & a_{nn} \\\\\n", "\\end{bmatrix}\n", "$$\n", "\n", "2. 以此类推,接下来每一步计算量约为$(n-1)^2、(n-2)^2、\\cdots、2^2、1^2$。\n", "\n", "3. 则将 $A$ 变换为 $LU$ 的总运算量应为$O(n^2+(n-1)^2+\\cdots+2^2+1^2)$,即$O(\\frac{n^3}{3})$。\n", "\n", "置换矩阵(Permutation Matrix):\n", "\n", "3阶方阵的置换矩阵有6个:\n", "$$\n", "\\begin{bmatrix}\n", "1 & 0 & 0 \\\\\n", "0 & 1 & 0 \\\\\n", "0 & 0 & 1 \\\\\n", "\\end{bmatrix}\n", "\\begin{bmatrix}\n", "0 & 1 & 0 \\\\\n", "1 & 0 & 0 \\\\\n", "0 & 0 & 1 \\\\\n", "\\end{bmatrix}\n", "\\begin{bmatrix}\n", "0 & 0 & 1 \\\\\n", "0 & 1 & 0 \\\\\n", "1 & 0 & 0 \\\\\n", "\\end{bmatrix}\n", "\\begin{bmatrix}\n", "1 & 0 & 0 \\\\\n", "0 & 0 & 1 \\\\\n", "0 & 1 & 0 \\\\\n", "\\end{bmatrix}\n", "\\begin{bmatrix}\n", "0 & 1 & 0 \\\\\n", "0 & 0 & 1 \\\\\n", "1 & 0 & 0 \\\\\n", "\\end{bmatrix}\n", "\\begin{bmatrix}\n", "0 & 0 & 1 \\\\\n", "1 & 0 & 0 \\\\\n", "0 & 1 & 0 \\\\\n", "\\end{bmatrix}\n", "$$\n", "\n", "$n$阶方阵的置换矩阵有$\\binom{n}{1}=n!$个。" ] } ], "metadata": { "kernelspec": { "display_name": "Python [default]", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.2" } }, "nbformat": 4, "nbformat_minor": 0 }