AnsForHw4.ipynb

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    "手写拍照，上传照片在附件。非手写者0分。\n",
    "\n",
    "***exercise : Show that the vectors $\\begin{bmatrix}1 \\\\ i \\end{bmatrix}$ and $\\begin{bmatrix}1 \\\\ 1 \\end{bmatrix}$ are capable of spanning $\\mathbb{C}^2$ as a complex linear space.***\n",
    "\n",
    "练习：证明向量 $\\begin{bmatrix}1 \\\\i \\end{bmatrix}$ 和 $\\begin{bmatrix}1 \\\\1 \\end{bmatrix}$ 能够作为复数线性空间 $\\mathbb{C}^2$ 的一组基。\n"
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    "To solve this problem, we need to show that any vector in $\\mathbb{C}^2$ can be expressed as a linear combination of the vectors $\\begin{bmatrix}1 \\\\i \\end{bmatrix}$ and $\\begin{bmatrix}1 \\\\1 \\end{bmatrix}$.\n",
    "\n",
    "Let $\\begin{bmatrix}a \\\\b \\end{bmatrix}$ be an arbitrary vector in $\\mathbb{C}^2$, where $a$ and $b$ are complex numbers. We want to find complex numbers $x$ and $y$ such that $\\begin{bmatrix}a \\\\b \\end{bmatrix} = x\\begin{bmatrix}1 \\\\i \\end{bmatrix} + y\\begin{bmatrix}1 \\\\1 \\end{bmatrix}$.\n",
    "\n",
    "Setting up the equation, we have:\n",
    "$a = x + y$\n",
    "$b = ix + y$\n",
    "\n",
    "We can solve this system of equations to find the values of $x$ and $y$. Subtracting the second equation from the first equation, we get:\n",
    "$a - b = (1 - i)x$\n",
    "\n",
    "Simplifying, we have:\n",
    "$x = \\frac{a - b}{1 - i}$\n",
    "\n",
    "Substituting this value of $x$ into either of the original equations, we can solve for $y$:\n",
    "$y = a - x = a - \\frac{a - b}{1 - i} = \\frac{b - ia}{1 - i}$\n",
    "\n",
    "Therefore, we have found that for any vector $\\begin{bmatrix}a \\\\b \\end{bmatrix}$ in $\\mathbb{C}^2$, we can express it as a linear combination of $\\begin{bmatrix}1 \\\\i \\end{bmatrix}$ and $\\begin{bmatrix}1 \\\\1 \\end{bmatrix}$ by choosing appropriate values of $x$ and $y$. This shows that the vectors $\\begin{bmatrix}1 \\\\i \\end{bmatrix}$ and $\\begin{bmatrix}1 \\\\1 \\end{bmatrix}$ span $\\mathbb{C}^2$ as a complex linear space."
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