Book4_Power-of-Matrix/Book4_Ch04_Python_Codes/Bk4_Ch04_13.ipynb

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    "Chapter 04\n",
    "\n",
    "# 迹\n",
    "Book_4《矩阵力量》 | 鸢尾花书：从加减乘除到机器学习 (第二版)"
   ]
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    "该代码定义了一个 $3 \\times 3$ 矩阵 $A$，并计算其迹（trace）。矩阵 $A$ 的定义为：\n",
    "\n",
    "$$\n",
    "A = \\begin{bmatrix} 1 & -1 & 0 \\\\ 3 & 2 & 4 \\\\ -2 & 0 & 3 \\end{bmatrix}\n",
    "$$\n",
    "\n",
    "矩阵的迹是其主对角线元素之和，公式为：\n",
    "\n",
    "$$\n",
    "\\text{tr}(A) = A_{11} + A_{22} + A_{33}\n",
    "$$\n",
    "\n",
    "因此，矩阵 $A$ 的迹为：\n",
    "\n",
    "$$\n",
    "\\text{tr}(A) = 1 + 2 + 3 = 6\n",
    "$$\n",
    "\n",
    "此代码展示了如何使用 `np.trace` 函数计算矩阵的迹。"
   ]
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  {
   "cell_type": "markdown",
   "id": "ba8dee01-3887-4a76-93a9-274b079171a1",
   "metadata": {},
   "source": [
    "## 导入所需库"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "650779c1-a190-44cc-a778-127b2c189962",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np  # 导入NumPy库，用于数值计算"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "5ff0a2c5-dcea-4891-84d4-7a72436e6772",
   "metadata": {},
   "source": [
    "## 定义矩阵A"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "a1a28546-6d6c-4f65-a94f-c8cb27e5ead5",
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    {
     "data": {
      "text/plain": [
       "array([[ 1, -1,  0],\n",
       "       [ 3,  2,  4],\n",
       "       [-2,  0,  3]])"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = np.array([[1, -1, 0],  # 定义矩阵A\n",
    "              [3, 2, 4],\n",
    "              [-2, 0, 3]])\n",
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7b8f3355-005a-4b5f-be18-41ada700fc7b",
   "metadata": {},
   "source": [
    "## 计算矩阵A的迹"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "9e5b87ef-8278-4900-b60c-6e96a60e5eef",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "6"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "tr_A = np.trace(A)  # 计算矩阵A的迹\n",
    "tr_A"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "85a80909-2aac-49ed-bb7a-f8cc6b80ee7d",
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "ecd322f4-f919-4be2-adc3-69d28ef25e69",
   "metadata": {},
   "outputs": [],
   "source": []
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