{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 第七讲:求解$Ax=0$,主变量,特解\n", "\n", "举例:$3 \\times 4$矩阵\n", "$\n", "A=\n", "\\begin{bmatrix}\n", "1 & 2 & 2 & 2\\\\\n", "2 & 4 & 6 & 8\\\\\n", "3 & 6 & 8 & 10\\\\\n", "\\end{bmatrix}\n", "$,求$Ax=0$的特解:\n", "\n", "找出主变量(pivot variable):\n", "$$\n", "A=\n", "\\begin{bmatrix}\n", "1 & 2 & 2 & 2\\\\\n", "2 & 4 & 6 & 8\\\\\n", "3 & 6 & 8 & 10\\\\\n", "\\end{bmatrix}\n", "\\underrightarrow{消元}\n", "\\begin{bmatrix}\n", "\\underline{1} & 2 & 2 & 2\\\\\n", "0 & 0 & \\underline{2} & 4\\\\\n", "0 & 0 & 0 & 0\\\\\n", "\\end{bmatrix}\n", "=U\n", "$$\n", "\n", "主变量(pivot variable,下划线元素)的个数为2,即矩阵$A$的秩(rank)为2,即$r=2$。\n", "\n", "主变量所在的列为主列(pivot column),其余列为自由列(free column)。\n", "\n", "自由列中的变量为自由变量(free variable),自由变量的个数为$n-r=4-2=2$。\n", "\n", "通常,给自由列变量赋值,去求主列变量的值。如,令$x_2=1, x_4=0$求得特解\n", "$x=c_1\\begin{bmatrix}-2\\\\1\\\\0\\\\0\\\\\\end{bmatrix}$;\n", "再令$x_2=0, x_4=1$求得特解\n", "$x=c_2\\begin{bmatrix}2\\\\0\\\\-2\\\\1\\\\\\end{bmatrix}$。\n", "\n", "该例还能进一步简化,即将$U$矩阵化简为$R$矩阵(Reduced row echelon form),即简化行阶梯形式。\n", "\n", "在简化行阶梯形式中,主元上下的元素都是$0$:\n", "$$\n", "U=\n", "\\begin{bmatrix}\n", "\\underline{1} & 2 & 2 & 2\\\\\n", "0 & 0 & \\underline{2} & 4\\\\\n", "0 & 0 & 0 & 0\\\\\n", "\\end{bmatrix}\n", "\\underrightarrow{化简}\n", "\\begin{bmatrix}\n", "\\underline{1} & 2 & 0 & -2\\\\\n", "0 & 0 & \\underline{1} & 2\\\\\n", "0 & 0 & 0 & 0\\\\\n", "\\end{bmatrix}\n", "=R\n", "$$\n", "\n", "将$R$矩阵中的主变量放在一起,自由变量放在一起(列交换),得到\n", "\n", "$$\n", "R=\n", "\\begin{bmatrix}\n", "\\underline{1} & 2 & 0 & -2\\\\\n", "0 & 0 & \\underline{1} & 2\\\\\n", "0 & 0 & 0 & 0\\\\\n", "\\end{bmatrix}\n", "\\underrightarrow{列交换}\n", "\\left[\n", "\\begin{array}{c c | c c}\n", "1 & 0 & 2 & -2\\\\\n", "0 & 1 & 0 & 2\\\\\n", "\\hline\n", "0 & 0 & 0 & 0\\\\\n", "\\end{array}\n", "\\right]\n", "=\n", "\\begin{bmatrix}\n", "I & F \\\\\n", "0 & 0 \\\\\n", "\\end{bmatrix}\n", "\\textrm{,其中}I\\textrm{为单位矩阵,}F\\textrm{为自由变量组成的矩阵}\n", "$$\n", "\n", "计算零空间矩阵$N$(nullspace matrix),其列为特解,有$RN=0$。\n", "\n", "$$\n", "x_{pivot}=-Fx_{free} \\\\\n", "\\begin{bmatrix}\n", "I & F \\\\\n", "\\end{bmatrix}\n", "\\begin{bmatrix}\n", "x_{pivot} \\\\\n", "x_{free} \\\\\n", "\\end{bmatrix}=0 \\\\\n", "N=\\begin{bmatrix}\n", "-F \\\\\n", "I \\\\\n", "\\end{bmatrix}\n", "$$\n", "\n", "在本例中\n", "$\n", "N=\n", "\\begin{bmatrix}\n", "-2 & 2 \\\\\n", "0 & -2 \\\\\n", "1 & 0 \\\\\n", "0 & 1 \\\\\n", "\\end{bmatrix}\n", "$,与上面求得的两个$x$特解一致。\n", "\n", "另一个例子,矩阵\n", "$\n", "A=\n", "\\begin{bmatrix}\n", "1 & 2 & 3 \\\\\n", "2 & 4 & 6 \\\\\n", "2 & 6 & 8 \\\\\n", "2 & 8 & 10 \\\\\n", "\\end{bmatrix}\n", "\\underrightarrow{消元}\n", "\\begin{bmatrix}\n", "1 & 2 & 3 \\\\\n", "0 & 2 & 2 \\\\\n", "0 & 0 & 0 \\\\\n", "0 & 0 & 0 \\\\\n", "\\end{bmatrix}\n", "\\underrightarrow{化简}\n", "\\begin{bmatrix}\n", "1 & 0 & 1 \\\\\n", "0 & 1 & 1 \\\\\n", "0 & 0 & 0 \\\\\n", "0 & 0 & 0 \\\\\n", "\\end{bmatrix}\n", "=R\n", "$\n", "\n", "矩阵的秩仍为$r=2$,有$2$个主变量,$1$个自由变量。\n", "\n", "同上一例,取自由变量为$x_3=1$,求得特解\n", "$\n", "x=c\n", "\\begin{bmatrix}\n", "-1 \\\\\n", "-1 \\\\\n", "1 \\\\\n", "\\end{bmatrix}\n", "$" ] } ], "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 }