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    "# 第十讲 四个基本子空间\n",
    "\n",
    "对于$m \\times n$矩阵$A$，$rank(A)=r$有：\n",
    "\n",
    "* 行空间$C(A^T) \\in \\mathbb{R}^n, dim C(A^T)=r$，基见例1。\n",
    "\n",
    "* 零空间$N(A) \\in \\mathbb{R}^n, dim N(A)=n-r$，自由元所在的列即可组成零空间的一组基。\n",
    "\n",
    "* 列空间$C(A) \\in \\mathbb{R}^m, dim C(A)=r$，主元所在的列即可组成列空间的一组基。\n",
    "\n",
    "* 左零空间$N(A^T) \\in \\mathbb{R}^m, dim N(A^T)=m-r$，基见例2。\n",
    "\n",
    "例1，对于行空间\n",
    "$\n",
    "A=\n",
    "\\begin{bmatrix}\n",
    "1 & 2 & 3 & 1 \\\\\n",
    "1 & 1 & 2 & 1 \\\\\n",
    "1 & 2 & 3 & 1 \\\\\n",
    "\\end{bmatrix}\n",
    "\\underrightarrow{消元、化简}\n",
    "\\begin{bmatrix}\n",
    "1 & 0 & 1 & 1 \\\\\n",
    "0 & 1 & 1 & 0 \\\\\n",
    "0 & 0 & 0 & 0 \\\\\n",
    "\\end{bmatrix}\n",
    "=R\n",
    "$\n",
    "\n",
    "由于我们做了行变换，所以A的列空间受到影响，$C(R) \\neq C(A)$，而行变换并不影响行空间，所以可以在$R$中看出前两行就是行空间的一组基。\n",
    "\n",
    "所以，可以得出无论对于矩阵$A$还是$R$，其行空间的一组基，可以由$R$矩阵的前$r$行向量组成（这里的$R$就是第七讲提到的简化行阶梯形式）。\n",
    "\n",
    "例2，对于左零空间，有$A^Ty=0 \\rightarrow (A^Ty)^T=0^T\\rightarrow y^TA=0^T$，因此得名。\n",
    "\n",
    "采用Gauss-Jordan消元，将增广矩阵$\\left[\\begin{array}{c|c}A_{m \\times n} & I_{m \\times m}\\end{array}\\right]$中$A$的部分划为简化行阶梯形式$\\left[\\begin{array}{c|c}R_{m \\times n} & E_{m \\times m}\\end{array}\\right]$，此时矩阵$E$会将所有的行变换记录下来。\n",
    "\n",
    "则$EA=R$，而在前几讲中，有当$A'$是$m$阶可逆方阵时，$R'$即是$I$，所以$E$就是$A^{-1}$。\n",
    "\n",
    "本例中\n",
    "\n",
    "$$\n",
    "\\left[\\begin{array}{c|c}A_{m \\times n} & I_{m \\times m}\\end{array}\\right]=\n",
    "\\left[\n",
    "\\begin{array}\n",
    "{c c c c|c c c}\n",
    "1 & 2 & 3 & 1 & 1 & 0 & 0 \\\\\n",
    "1 & 1 & 2 & 1 & 0 & 1 & 0 \\\\\n",
    "1 & 2 & 3 & 1 & 0 & 0 & 1 \\\\\n",
    "\\end{array}\n",
    "\\right]\n",
    "\\underrightarrow{消元、化简}\n",
    "\\left[\n",
    "\\begin{array}\n",
    "{c c c c|c c c}\n",
    "1 & 0 & 1 & 1 & -1 & 2 & 0 \\\\\n",
    "0 & 1 & 1 & 0 & 1 & -1 & 0 \\\\\n",
    "0 & 0 & 0 & 0 & -1 & 0 & 1 \\\\\n",
    "\\end{array}\n",
    "\\right]\n",
    "=\\left[\\begin{array}{c|c}R_{m \\times n} & E_{m \\times m}\\end{array}\\right]\n",
    "$$\n",
    "\n",
    "则\n",
    "\n",
    "$$\n",
    "EA=\n",
    "\\begin{bmatrix}\n",
    "-1 & 2  & 0 \\\\\n",
    "1  & -1 & 0 \\\\\n",
    "-1 & 0  & 1 \\\\\n",
    "\\end{bmatrix}\n",
    "\\cdot\n",
    "\\begin{bmatrix}\n",
    "1 & 2 & 3 & 1 \\\\\n",
    "1 & 1 & 2 & 1 \\\\\n",
    "1 & 2 & 3 & 1 \\\\\n",
    "\\end{bmatrix}\n",
    "=\n",
    "\\begin{bmatrix}\n",
    "1 & 0 & 1 & 1 \\\\\n",
    "0 & 1 & 1 & 0 \\\\\n",
    "0 & 0 & 0 & 0 \\\\\n",
    "\\end{bmatrix}\n",
    "=R\n",
    "$$\n",
    "\n",
    "\n",
    "很明显，式中$E$的最后一行对$A$的行做线性组合后，得到$R$的最后一行，即$0$向量，也就是$y^TA=0^T$。\n",
    "\n",
    "最后，引入矩阵空间的概念，矩阵可以同向量一样，做求和、数乘。\n",
    "\n",
    "举例，设所有$3 \\times 3$矩阵组成的矩阵空间为$M$。则上三角矩阵、对称矩阵、对角矩阵（前两者的交集）。\n",
    "\n",
    "观察一下对角矩阵，如果取\n",
    "$\n",
    "\\begin{bmatrix}\n",
    "1 & 0 & 0 \\\\\n",
    "0 & 0 & 0 \\\\\n",
    "0 & 0 & 0 \\\\\n",
    "\\end{bmatrix} \\quad\n",
    "\\begin{bmatrix}\n",
    "1 & 0 & 0 \\\\\n",
    "0 & 3 & 0 \\\\\n",
    "0 & 0 & 0 \\\\\n",
    "\\end{bmatrix} \\quad\n",
    "\\begin{bmatrix}\n",
    "0 & 0 & 0 \\\\\n",
    "0 & 0 & 0 \\\\\n",
    "0 & 0 & 7 \\\\\n",
    "\\end{bmatrix}\n",
    "$\n",
    "，可以发现，任何三阶对角矩阵均可用这三个矩阵的线性组合生成，因此，他们生成了三阶对角矩阵空间，即这三个矩阵是三阶对角矩阵空间的一组基。"
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