Book4_Power-of-Matrix/Book4_Ch04_Python_Codes/Bk4_Ch04_09.ipynb

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    "Chapter 04\n",
    "\n",
    "# 矩阵乘幂比较\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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    "\n",
    "该代码展示了 `np.matrix` 和 `np.array` 在平方操作中的不同行为。\n",
    "\n",
    "1. **矩阵平方（`np.matrix` 类型）**：当 $A$ 是 `np.matrix` 类型时，`A**2` 计算的是矩阵乘法，即 $A @ A$。对于矩阵：\n",
    "\n",
    "   $$\n",
    "   A = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}\n",
    "   $$\n",
    "\n",
    "   其平方 $A^2$ 计算为：\n",
    "\n",
    "   $$\n",
    "   A^2 = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix} @ \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix} = \\begin{bmatrix} 7 & 15 \\\\ 10 & 22 \\end{bmatrix}\n",
    "   $$\n",
    "\n",
    "2. **逐元素平方（`np.array` 类型）**：当 $B$ 是 `np.array` 类型时，`B**2` 计算的是逐元素平方，即每个元素单独平方。对于数组：\n",
    "\n",
    "   $$\n",
    "   B = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}\n",
    "   $$\n",
    "\n",
    "   逐元素平方的结果为：\n",
    "\n",
    "   $$\n",
    "   B\\odot B = \\begin{bmatrix} 1^2 & 3^2 \\\\ 2^2 & 4^2 \\end{bmatrix} = \\begin{bmatrix} 1 & 9 \\\\ 4 & 16 \\end{bmatrix}\n",
    "   $$\n",
    "\n",
    "此代码展示了 `np.matrix` 和 `np.array` 在平方操作上的关键差异：一个执行矩阵乘法，另一个进行逐元素运算。"
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  {
   "cell_type": "markdown",
   "id": "9734eeec-2b37-40f9-add5-c4ccd409546a",
   "metadata": {},
   "source": [
    "## 导入所需库"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "4b42a3b3-504c-45eb-8c6b-2830e6a11be8",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np  # 导入NumPy库，用于数值计算"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "79f50411-80c3-487e-b20d-e42867cea09a",
   "metadata": {},
   "source": [
    "## 定义矩阵A并计算其矩阵平方"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "5879a711-cd82-4352-a41f-c9b950865d71",
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   "source": [
    "A = np.matrix([[1, 3],  # 定义为np.matrix类型的矩阵A\n",
    "               [2, 4]])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "02a0e4c0-e50d-4955-8e72-351468e22d57",
   "metadata": {},
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     "text": [
      "[[ 7 15]\n",
      " [10 22]]\n"
     ]
    }
   ],
   "source": [
    "print(A**2)  # 打印矩阵A的矩阵平方结果"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8dee9c28-b297-4f4d-9f0b-b0c8880d792e",
   "metadata": {},
   "source": [
    "## 定义数组B并计算其逐元素平方"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "7ab6ed46-fca3-46de-aa69-8d1a16b67d24",
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   "outputs": [],
   "source": [
    "B = np.array([[1, 3],  # 定义为np.array类型的数组B\n",
    "              [2, 4]])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "8e1194bb-59fa-40a2-b0a2-e2c85b7cc667",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[[ 1  9]\n",
      " [ 4 16]]\n"
     ]
    }
   ],
   "source": [
    "print(B**2)  # 打印数组B的逐元素平方结果"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "85a80909-2aac-49ed-bb7a-f8cc6b80ee7d",
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   "source": []
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