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docs: 修正部分 repo，添加 24-25 线代 I 历年卷答案
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‎讲义/LALU-answers.tex‎

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@@ -120,6 +120,7 @@ \chapter{线性代数I（H）期末历年卷试题集}
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% 线代I期末答案
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\chapter{线性代数I（H）期末历年卷答案}
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\input{./历年卷/2023-2024-1final-answer.tex}
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\input{./历年卷/2024-2025-1final-answer.tex}
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% 线代II期中/练习
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\chapter{线性代数II（H）期中/小测历年卷试题集}

‎讲义/专题/3 有限维线性空间.tex‎

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@@ -530,7 +530,7 @@ \subsection{极大线性无关组的求法}
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已知 $\mathbf{R}^4$ 的一个子集 $S = \{a_1, a_2, a_3, a_4\}$, 其中
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\[
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a_1 = (1,1,0,1), \quad a_2 = (0,1,2,4), \quad
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a_3 = (2,1,-2,2), \quad a_4 = (0,1,1,1).
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a_3 = (2,1,-2,-2), \quad a_4 = (0,1,1,1).
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\]
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试求 $\spa(S)$ 的维数及其一组基$B$。
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\end{example}

‎讲义/专题/8 相抵标准形.tex‎

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\sigma(\alpha_i)=0 & i=r+1,\ldots,n
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\end{cases}\]
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其中$e_i$表示第$i$个位置为1，其余位置全为0的列向量. 因此我们根据线性映射矩阵表示的定义得到
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\[\sigma(\alpha_1,\ldots,\alpha_r,\alpha_{r+1},\alpha_n)=(\sigma(\alpha_1),\ldots,\sigma(\alpha_r),\beta_{r+1},\ldots,\beta_m)\begin{pmatrix}
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\[\sigma(\alpha_1,\ldots,\alpha_r,\alpha_{r+1},\ldots,\alpha_n)=(\sigma(\alpha_1),\ldots,\sigma(\alpha_r),\beta_{r+1},\ldots,\beta_m)\begin{pmatrix}
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E_r & O \\ O & O
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\end{pmatrix}.\]
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根据\autoref{thm:换基公式}，设$B_1$到$B_1'$的过渡矩阵为$Q$，$B_2'$到$B_2$的过渡矩阵为$P$，则有$PAQ=U_r$，其中$P$和$Q$因是过渡矩阵所以可逆，由此得证.

‎讲义/历年卷/2024-2025-1final-answer.tex‎

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\section{2024-2025学年线性代数I（H）期末答案}
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\begin{enumerate}
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\item 增广矩阵为
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\begin{align*}
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\begin{pmatrix}[ccc|c]
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1 & a & 1 & 3 \\
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1 & 1 & 1 & 4 \\
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1 & 2a & 1 & 4
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\end{pmatrix} & \to
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\begin{pmatrix}[ccc|c]
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1 & a & 1 & 3 \\
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0 & 1 - a & 0 & 1 \\
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0 & a & 0 & 1
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\end{pmatrix}
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\end{align*}
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故有 \((1 - a)x_{2} = ax_{2} = 1 \implies a = \dfrac{1}{2}, x_{2} = 2, x_{1} + x_{3} = 2\).
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即一般解为 \((x_{1}, x_{2}, x_{3})^\mathrm{T}=(0,2,2)^\mathrm{T}+k(1,0,-1)^\mathrm{T}\).
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\item 对 \(X=(x_1,x_2,x_3)^\mathrm{T}\)，
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\begin{align*}
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X^\mathrm{T}AX &= 2x_{1}^{2}+x_{2}^{2}+3x_{3}^{2}+2tx_{1}x_{2}+4x_{1}x_{3} \\
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&= 2\left(x_{1}+\dfrac{t}{2}x_{2}+x_{3}\right)^{2}+\left(x_{3}-tx_{2}\right)^{2}+\left(1-\dfrac{3}{2}t^{2}\right)x_{2}^{2}
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\end{align*}
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因此：
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\begin{enumerate}
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\item \(A\)正定\(\iff 1-\dfrac{3}{2}t^{2}>0\)，即\(\vert t\vert<\sqrt{\dfrac{2}{3}}\).
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\item \(A\)半正定\(\iff 1-\dfrac{3}{2}t^{2} \geq 0\)，即\(\vert t\vert \leq \sqrt{\dfrac{2}{3}}\).
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\item \(X^\mathrm{T}AX\)的负惯性指数为\(1\)，即\(\vert t\vert > \sqrt{\dfrac{2}{3}}\).
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\end{enumerate}
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\item \(A=\begin{pmatrix}
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1 & 0 & 25 & 8 \\
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1 & 1 & 2 & 2 \\
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0 & 0 & 4 & 3 \\
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0 & 0 & 6 & 5
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\end{pmatrix}\)，计算得
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\begin{gather*}
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A_{11}=2, A_{12}=-2, A_{13}=0, A_{14}=0, \\
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A_{21}=0, A_{22}=2, A_{23}=0, A_{24}=0, \\
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A_{31}=-77, A_{32}=79, A_{33}=5, A_{34}=-6, \\
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A_{41}=43, A_{42}=-47, A_{43}=-3, A_{44}=4.
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\end{gather*}
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故
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\(\vert A\vert = 4\times5 - 3\times6 = 2\)，
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\(A^{-1}=\begin{pmatrix}
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1 & 0 & -\dfrac{77}{2} & \dfrac{43}{2} \\[0.5em]
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-1 & 1 & \dfrac{79}{2} & -\dfrac{47}{2} \\[0.5em]
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0 & 0 & \dfrac{5}{2} & -\dfrac{3}{2} \\
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0 & 0 & -3 & 2
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\end{pmatrix}\).
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\item
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\begin{enumerate}
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\item 分别验证 \(W\) 关于加法和数乘封闭性即可.
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\item 显然 \(x^{2}-1\)，\(x - 1\) 为一组基.
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\item 令 \(\sigma(x^{i})=x^{i}-1\)，\(i = 0,1,2\)，我们证明此即为所求.
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一方面我们验证其满足题设。显然 \(\sigma(1)=0\)，且对 \(\forall f \in \mathbf{R}[x]_{3}\)，设 \(f(x)=a_{0}+a_{1}x+a_{2}x^{2}\)，则 \(\sigma(f(x))=a_{1}(x - 1)+a_{2}(x^{2}-1)\)，且 \(x - 1\)，\(x^{2}-1\) 为 \(W\) 的一组基，故 \(\mathrm{Im}(\sigma)=W\).
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另一方面，由一组基上的像可唯一确定一个线性映射，故 \(\sigma\) 唯一，即其为所求.
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\end{enumerate}
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\item
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\begin{enumerate}
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\item 设 \(k_1\beta+k_2 A\beta+\cdots k_n A^{n-1}\beta=0\). 同时左乘\(A\)得 \[ k_1 A\beta+\cdots+k_n A^n\beta=k_1 A\beta+\cdots+k_{n-1} A^{n-1}\beta=0. \]
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一直如此可得到 \(k_1 A^{n-1}\beta=0\). 由条件 \(A^i\beta\neq 0, \enspace\forall 0\leq i \leq n-1 \implies k_1=0\)，将其回代可得到 $k_2 = \cdots = k_n = 0$. 即这些向量线性无关，且长度为n，因此是一组基.
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\item 由(1)知对 \(\forall \alpha \in \mathbf{R}^{n}\)，存在 \(k_{1}, k_{2}, \cdots, k_{n}\) 使得 \(\alpha=k_{1}\beta + k_{2}A\beta+\cdots + k_{n}A^{n - 1}\beta\).
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从而对任意$\alpha$，\(A^n\alpha=k_1 A^n\beta+\cdots+k_n A^{2n-1}\beta=0 \implies A^n=0\).
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\item 即证\(\dim N(A)=1\).
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任取 \(X_{1}, X_{2} \in N(A)\)，设 \(X_{1}=\displaystyle\sum\limits_{i = 1}^{n}a_{i}A^{i - 1}\beta\)，\(X_{2}=\displaystyle\sum\limits_{i = 1}^{n}b_{i}A^{i - 1}\beta\). 则对 \(\forall k_{1}, k_{2} \in \mathbf{R}\)，由 \(k_{1}X_{1}+k_{2}X_{2} \in N(A)\)，有
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\[
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A(k_{1}X_{1}+k_{2}X_{2})=\sum_{i = 1}^{n}(k_{1}a_{i}+k_{2}b_{i})A^{i}\beta=\sum_{i = 1}^{n - 1}(k_{1}a_{i}+k_{2}b_{i})A^{i}\beta = 0.
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\]
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由 \(A\beta, \cdots, A^{n - 1}\beta\) 线性无关知 \(k_{1}a_{i}+k_{2}b_{i}=0, \enspace\forall i = 1,2,\cdots, n - 1\).
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由 \(k_{1}, k_{2}\) 任意性知 \(a_{i}=b_{i}=0, \enspace\forall i = 1,2,\cdots, n - 1\).
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故 \(X_{1}=a_{n}A^{n - 1}\beta\)，\(X_{2}=b_{n}A^{n - 1}\beta\)，从而 \(X_{1}\) 与 \(X_{2}\) 线性相关，故 \(\dim N(A)=1\).
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\end{enumerate}
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\item
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\begin{enumerate}
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\item
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\begin{align*}
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T(x_{1}, x_{2}, x_{3}, x_{4})^\mathrm{T} &= (x_{1}-x_{2}+x_{4}, x_{1}+2x_{3}+2x_{4}, x_{1}+x_{2}+4x_{3}+3x_{4})^\mathrm{T} \\
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&=\begin{pmatrix}
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1 & -2 & 0 & 1 \\
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1 & 0 & 2 & 2 \\
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1 & 1 & 4 & 3
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\end{pmatrix}\begin{pmatrix}
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x_{1} \\
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x_{2} \\
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x_{3} \\
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x_{4}
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\end{pmatrix}
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\end{align*}
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故所求即为 \(A=\begin{pmatrix}
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1 & -2 & 0 & 1 \\
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1 & 0 & 2 & 2 \\
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1 & 1 & 4 & 3
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\end{pmatrix}\).
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\item
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求 $AX=0$ 可得 $x_2 = 0, x_4 = -2x_3, x_1 = 2x_3$，即$X=(x_1,x_2,x_3,x_4)^\mathrm{T} = k(2,0,1,-2)^\mathrm{T}$，核空间为$\operatorname{span}\{(2,0,1,-2)^\mathrm{T}\}$.
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取出发空间的一组自然基$e_1,\ldots,e_4$，则$T(e_1)=(1,1,1)^\mathrm{T}$，$T(e_2)=(-2,0,1)^\mathrm{T}$，$T(e_3)=(0,2,4)^\mathrm{T}$，$T(e_4)=(1,2,3)^\mathrm{T}$. 从而可以对 $A$ 作初等列变换：
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\[
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\begin{pmatrix}
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1 & -2 & 0 & 1 \\
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1 & 0 & 2 & 2 \\
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1 & 1 & 4 & 3
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\end{pmatrix}
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\to
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\begin{pmatrix}
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1 & 0 & 0 & 0 \\
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1 & 2 & 2 & 1 \\
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1 & 3 & 4 & 2
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\end{pmatrix}
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\to
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\begin{pmatrix}
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1 & 0 & 0 & 0 \\
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1 & 2 & 1 & 0 \\
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1 & 3 & 2 & 0
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\end{pmatrix}
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\to
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\begin{pmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 1 & 0 \\
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0 & 1 & 2 & 0
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\end{pmatrix}
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\]
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可得像空间的一组基为 $T(e_1), T(e_2), T(e_3)$.
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故像空间为$\operatorname{span}\{(1,1,1)^\mathrm{T}, (-2,0,1)^\mathrm{T}, (0,2,4)^\mathrm{T}\}$.
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\item 即求$P,Q$使得
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\[
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Q^{-1}AP =
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\begin{pmatrix}
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E_r & O \\
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O & O
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\end{pmatrix}
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\]
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% 利用行列变换得：
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% \begin{align*}
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% \begin{pmatrix}[cccc|cccc]
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% 1 & -2 & 0 & 1 & 1 & 0 & 0 \\
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% 1 & 0 & 2 & 2 & 0 & 1 & 0 \\
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% 1 & 1 & 4 & 3 & 0 & 0 & 1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}[cccc|cccc]
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% 1 & -2 & 0 & 1 & 1 & 0 & 0 \\
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% 0 & 2 & 2 & 1 & -1 & 1 & 0 \\
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% 0 & 3 & 4 & 2 & -1 & 0 & 1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}[cccc|cccc]
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% 1 & -2 & 0 & 1 & 1 & 0 & 0 \\
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% 0 & 1 & 2 & 1 & 0 & -1 & 1 \\
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% 0 & 0 & 2 & 1 & 1 & -3 & 2
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% \end{pmatrix}
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% \end{align*}
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% \[
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% \begin{pmatrix}
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% 1 & -2 & 0 & 1 \\
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% 0 & 1 & 2 & 1 \\
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% 0 & 0 & 2 & 1 \\
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% \hline
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% 1 & 0 & 0 & 0 \\
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% 0 & 1 & 0 & 0 \\
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% 0 & 0 & 1 & 0 \\
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% 0 & 0 & 0 & 1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}
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% 1 & 0 & 0 & 0 \\
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% 0 & 1 & 2 & 1 \\
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% 0 & 0 & 2 & 1 \\
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% \hline
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% 1 & 2 & 0 & -1 \\
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% 0 & 1 & 0 & 0 \\
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% 0 & 0 & 1 & 0 \\
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% 0 & 0 & 0 & 1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}
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% 1 & 0 & 0 & 0 \\
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% 0 & 1 & 0 & 0 \\
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% 0 & 0 & 1 & 1 \\
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% \hline
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% 1 & 2 & -1 & -3 \\
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% 0 & 1 & -1 & -1 \\
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% 0 & 0 & 1 & 0 \\
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% 0 & 0 & 0 & 1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}
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% 1 & 0 & 0 & 0 \\
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% 0 & 1 & 0 & 0 \\
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% 0 & 0 & 1 & 0 \\
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% \hline
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% 1 & 2 & -1 & -2 \\
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% 0 & 1 & -1 & 0 \\
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% 0 & 0 & 1 & -1 \\
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% 0 & 0 & 0 & 1
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% \end{pmatrix}
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% \]
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% 利用初等行变换求$Q^{-1}$：
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% \[
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% \begin{pmatrix}[ccc|ccc]
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% 1 & 0 & 0 & 1 & 0 & 0 \\
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% 0 & -1 & 1 & 0 & 1 & 0 \\
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% 1 & -3 & 2 & 0 & 0 & 1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}[ccc|ccc]
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% 1 & 0 & 0 & 1 & 0 & 0 \\
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% 0 & -1 & 1 & 0 & 1 & 0 \\
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% 0 & 3 & -2 & 1 & 0 & -1
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% \end{pmatrix}
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% \to
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% \begin{pmatrix}[ccc|ccc]
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% 1 & 0 & 0 & 1 & 0 & 0 \\
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% 0 & 1 & 0 & 1 & 2 & -1 \\
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% 0 & 0 & 1 & 1 & 3 & -1
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% \end{pmatrix}
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% \]
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% 故
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参考 LALU 8.5 节例 8.6，具体过程略，结果为：
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\[
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P =
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\begin{pmatrix}
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1 & 2 & -1 & -2 \\
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0 & 1 & -1 & 0 \\
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0 & 0 & 1 & -1 \\
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0 & 0 & -1 & 2
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\end{pmatrix},
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Q =
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\begin{pmatrix}
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1 & 0 & 0 \\
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1 & 2 & -1 \\
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1 & 3 & -1
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\end{pmatrix}.
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\]
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(尽管笔者记得考场上的结果是\(r(A)=2\)，可能是回忆卷出现误差，但是重要的还是求\(P,Q\)的方法.)
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\end{enumerate}
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\item
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\begin{enumerate}
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\item 由相抵标准型知存在 \(P, Q\) 为可逆矩阵，且
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\[
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A = P \diag(1,1,0,\ldots, 0)Q = P\diag(1,0,\ldots, 0)Q + P\diag(0,1,0,\ldots, 0)Q.
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\]
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设 \(P = (\beta_{1}, \beta_{2}, \cdots, \beta_{n})\)，\(Q^\mathrm{T}=(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n})\)，则
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\[
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P\diag(1,0,\cdots, 0)Q=(\beta_{1}, \cdots, \beta_{n})\begin{pmatrix}
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1 \\
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0 \\
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\vdots \\
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0
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\end{pmatrix}\begin{pmatrix}
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1 & 0 & \cdots & 0
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\end{pmatrix}\begin{pmatrix}
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\alpha_{1}^\mathrm{T} \\
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\alpha_{2}^\mathrm{T} \\
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\vdots \\
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\alpha_{n}^\mathrm{T}
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\end{pmatrix}=\beta_{1}\alpha_{1}^\mathrm{T}.
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\]
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同理 \(P\diag(0,1,0,\cdots, 0)Q=\beta_{2}\alpha_{2}^\mathrm{T}\)，即 \(A=\beta_{1}\alpha_{1}^\mathrm{T}+\beta_{2}\alpha_{2}^\mathrm{T}\)，得证.
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\item \(\beta_{1}=\alpha_{1}=(1,0,\cdots, 0)^\mathrm{T}\)，\(\beta_{2}=\alpha_{2}=(0,1,0,\cdots, 0)^\mathrm{T}\)，自行验证 \(A\) 的确可对角化.
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\item 对于将 \(A\) 对角化处理的矩阵 \(C\)（即 \(C^{-1}AC\) 中的 \(C\)）的列向量 \(X\)，要么 \(X \in \mathrm{span}\{\beta_{1}, \beta_{2}\}\)，要么 \(X \perp \alpha_{1}\) 且 \(X \perp \alpha_{2}\).
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证明：设 \(C=(X_{1}, \cdots, X_{n})\)，则
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\[
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AC = C\mathrm{diag}\left(\lambda_{1}, \cdots, \lambda_{n}\right) \iff \beta_{1}\alpha_{1}^\mathrm{T}X_{i}+\beta_{2}\alpha_{2}^\mathrm{T}X_{i}=\lambda_{i}X_{i}
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\]
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若 \(\lambda_{i} \neq 0\)，则 \(X_{i}\) 为 \(\beta_{1}\)，\(\beta_{2}\) 的线性扩张（注意到 \(\alpha_{1}^\mathrm{T}X_{i}\) 与 \(\alpha_{2}^\mathrm{T}X_{i}\) 都是实数）.
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若不然，由于 \(\beta_{1}\) 与 \(\beta_{2}\) 线性无关，有 \(\alpha_{1}^\mathrm{T}X_{i}=0\) 和 \(\alpha_{2}^\mathrm{T}X_{i}=0\)，即 \(X_{i} \perp \alpha_{1}\) 且 \(X_{i} \perp \alpha_{2}\).
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\end{enumerate}
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\item
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\begin{enumerate}
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\item 错. \(\alpha_{1}=(1,0)\)，\(\alpha_{2}=(-1,0)\)，\(\beta_{1}=(0,1)\)，\(\beta_{2}=(0,-2)\) 即有矛盾.
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\item 对. 设 \(A = (\alpha_{1}, \cdots, \alpha_{n})\)，则
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\[
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A^{2}=\begin{pmatrix}
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\alpha_{1}^\mathrm{T} \\
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\vdots \\
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\alpha_{n}^\mathrm{T}
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\end{pmatrix}\left(\alpha_{1}, \cdots, \alpha_{n}\right)
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\]
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其第 \(i\) 个主对角元元素为 \(\alpha_{i}^\mathrm{T}\alpha_{i}=\vert\alpha_{i}\vert^{2}=0\)，故 \(\alpha_{i}=0\)，即 \(A = 0\).
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\item 错. \(A=\begin{pmatrix}
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i & 1 \\
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1 & -i
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\end{pmatrix}\) 即为反例。
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\item 错. \(\sigma(x, y)=(y, 0)\) 即为反例.
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\end{enumerate}
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\end{enumerate}
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\clearpage

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