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149 changes: 69 additions & 80 deletions 习题参考答案/专题/1 预备知识.tex
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@@ -1,3 +1,4 @@
\phantomsection
\section*{1 预备知识}
\addcontentsline{toc}{section}{1 预备知识}

Expand All @@ -6,87 +7,75 @@ \section*{1 预备知识}
\centerline{\heiti A组}
\begin{enumerate}
\item

\item
\item
\begin{align*}
A=&
\begin{pmatrix}
1 & 1 & 1 & 4 & -3 \\
2 & 1 & 3 & 5 & -5 \\
1 & -1 & 3 & -2 & -1 \\
3 & 1 & 5 & 6 & -7
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 1 & 1 & 4 & -3 \\
0 & -1 & 1 & -3 & 1 \\
0 & -2 & 2 & -6 & 2 \\
0 & -2 & 2 & -6 & 2
\end{pmatrix}
\rightarrow \\
&
\begin{pmatrix}
1 & 1 & 1 & 4 & -3 \\
0 & 1 & -1 & 3 & -1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 0 & 2 & 1 & -2 \\
0 & 1 & -1 & 3 & -1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{align*}
令 $ x_3 = k_1, x_4 = k_2, x_5 = k_3$，有 $x_1 = -2k_1 - k_2 + 2k_3$，$x_2 = k_1 + -3k_2 + k_3 $ ，
则
\begin{equation*}
X = (x_1, x_2, x_3, x_4, x_5)^\mathrm{T} = k_1 \begin{pmatrix} -2 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}
+ k_2 \begin{pmatrix} -1 \\ -3 \\ 0 \\ 1 \\ 0 \end{pmatrix}
+ k_3 \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\quad(k_1, k_2, k_3 \in \mathbf{R})
\end{equation*}
\item
\begin{align*}
\bar{A}=&
\begin{pmatrix}
1 & -1 & 2 & -2 & 3 & 1 \\
2 & -1 & 5 & -9 & 8 & -1 \\
3 & -2 & 7 & -11 & 11 & 0 \\
1 & -1 & 1 & -1 & 3 & 3
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & -1 & 2 & -2 & 3 & 1 \\
0 & 1 & 1 & -5 & 2 & -3 \\
0 & 1 & 1 & -5 & 2 & -3 \\
0 & 0 & -1 & 1 & 0 & 2
\end{pmatrix}
\rightarrow \\
&
\begin{pmatrix}
1 & -1 & 2 & -2 & 3 & 1 \\
0 & 1 & 1 & -5 & 2 & -3 \\
0 & 0 & -1 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 0 & 0 & -4 & 5 & 4 \\
0 & 1 & 0 & -4 & 2 & 1 \\
0 & 0 & 1 & -1 & 0 & -2 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{align*}
令 $ x_4 = k_1, x_5 = k_2, x_6 = k_3$，有 $x_1 = 4k_1 - 5k_2 - 4k_3$，$x_2 = 4k_1 - 2k_2 + k_3, x_3 = k_1 + 2k_3 $ ，
则
\begin{equation*}
X = (x_1, x_2, x_3, x_4, x_5, x_6)^\mathrm{T} = k_1 \begin{pmatrix} 4 \\ 4 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}
+ k_2 \begin{pmatrix} -5 \\ -2 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}
+ k_3 \begin{pmatrix} -4 \\ 1 \\ 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\quad(k_1, k_2, k_3 \in \mathbf{R})
\end{equation*}

\item \begin{align*}
A ={} &
\begin{pmatrix}
1 & 1 & 1 & 4 & -3 \\
2 & 1 & 3 & 5 & -5 \\
1 & -1 & 3 & -2 & -1 \\
3 & 1 & 5 & 6 & -7
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 1 & 1 & 4 & -3 \\
0 & -1 & 1 & -3 & 1 \\
0 & -2 & 2 & -6 & 2 \\
0 & -2 & 2 & -6 & 2
\end{pmatrix} \rightarrow \\
&
\begin{pmatrix}
1 & 1 & 1 & 4 & -3 \\
0 & 1 & -1 & 3 & -1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 0 & 2 & 1 & -2 \\
0 & 1 & -1 & 3 & -1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{align*}
令 $ x_3 = k_1, x_4 = k_2, x_5 = k_3$，有 $x_1 = -2k_1 - k_2 + 2k_3,\enspace\allowbreak x_2 = k_1 + -3k_2 + k_3 $，则
\[ X = (x_1, x_2, x_3, x_4, x_5)^\mathrm{T} = k_1 \begin{pmatrix} -2 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + k_2 \begin{pmatrix} -1 \\ -3 \\ 0 \\ 1 \\ 0 \end{pmatrix} + k_3 \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \qquad k_1, k_2, k_3 \in \mathbf{R} \]

\item \begin{align*}
\bar{A} ={} &
\begin{pmatrix}
1 & -1 & 2 & -2 & 3 & 1 \\
2 & -1 & 5 & -9 & 8 & -1 \\
3 & -2 & 7 & -11 & 11 & 0 \\
1 & -1 & 1 & -1 & 3 & 3
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & -1 & 2 & -2 & 3 & 1 \\
0 & 1 & 1 & -5 & 2 & -3 \\
0 & 1 & 1 & -5 & 2 & -3 \\
0 & 0 & -1 & 1 & 0 & 2
\end{pmatrix} \rightarrow \\
&
\begin{pmatrix}
1 & -1 & 2 & -2 & 3 & 1 \\
0 & 1 & 1 & -5 & 2 & -3 \\
0 & 0 & -1 & 1 & 0 & 2 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 0 & 0 & -4 & 5 & 4 \\
0 & 1 & 0 & -4 & 2 & 1 \\
0 & 0 & 1 & -1 & 0 & -2 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{align*}
令 $ x_4 = k_1, x_5 = k_2, x_6 = k_3$，有 $x_1 = 4k_1 - 5k_2 - 4k_3,\enspace\allowbreak x_2 = 4k_1 - 2k_2 + k_3,\enspace\allowbreak x_3 = k_1 + 2k_3 $，则
\[ X = (x_1, x_2, x_3, x_4, x_5, x_6)^\mathrm{T} = k_1 \begin{pmatrix} 4 \\ 4 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + k_2 \begin{pmatrix} -5 \\ -2 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + k_3 \begin{pmatrix} -4 \\ 1 \\ 2 \\ 0 \\ 0 \\ 1 \end{pmatrix} \qquad k_1, k_2, k_3 \in \mathbf{R} \]

\item 见教材P33例3. 无解.
\end{enumerate}

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78 changes: 49 additions & 29 deletions 习题参考答案/专题/10 矩阵运算进阶（I）.tex
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\phantomsection
\section*{10 矩阵运算进阶（I）}
\addcontentsline{toc}{section}{10 矩阵运算进阶（I）}

Expand All @@ -6,49 +7,68 @@ \section*{10 矩阵运算进阶（I）}
\centerline{\heiti A组}
\begin{enumerate}
\item 由题意有 $P_2AP_1 = E$，从而有 $A=P_2^{-1}P_1^{-1}=P_2P_1^{-1}$.

\item 由题意知 $B = E_{ij}A$，所以 $BA^{-1}=E_{ij}$ 从而 $B$ 可逆，同时可得 $AB^{-1}=A(A^{-1}E_{ij}^{-1})=E_{ij}$.

\item $Q = (\alpha_1+\alpha_2,\alpha_2,\alpha_3)=(\alpha_1,\alpha_2,\alpha_3)\begin{pmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}=P\begin{pmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.

记 $E_{21}=\begin{pmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$，有 $Q^{-1}AQ=E_{21}^{-1}P^{-1}APE_{21}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix}$.

记 $E_{21}=\begin{pmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$，有 $Q^{-1}AQ=E_{21}^{-1}P^{-1}APE_{21}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix}$.

\item 见本章例 10.5，此处不赘述.
\end{enumerate}

\centerline{\heiti B组}
\begin{enumerate}
\item 此处仅给出答案，具体过程略.
\begin{enumerate}
\item $\begin{pmatrix}a_{21} & a_{22} & a_{23} \\ a_{11} & a_{12} & a_{13} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}$.
\item $\begin{pmatrix}-a_{11} & -a_{12} & -a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}$.
\item $\begin{pmatrix}-a_{13} & -a_{12} & a_{11} \\ a_{23} & a_{22} & -a_{21} \\ a_{33} & a_{32} & -a_{31}\end{pmatrix}$.
\item $\begin{pmatrix}-a_{11}-a_{12} & -a_{12}-a_{13} & -a_{13}-a_{11} \\ a_{21}+a_{22} & a_{22}+a_{23} & a_{23}+a_{21} \\ a_{31}+a_{32} & a_{32}+a_{33} & a_{33}+a_{31}\end{pmatrix}$.
\end{enumerate}
\begin{enumerate}
\item $\begin{pmatrix}a_{21} & a_{22} & a_{23} \\ a_{11} & a_{12} & a_{13} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}$.

\item $\begin{pmatrix}-a_{11} & -a_{12} & -a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}$.

\item $\begin{pmatrix}-a_{13} & -a_{12} & a_{11} \\ a_{23} & a_{22} & -a_{21} \\ a_{33} & a_{32} & -a_{31}\end{pmatrix}$.

\item $\begin{pmatrix}-a_{11}-a_{12} & -a_{12}-a_{13} & -a_{13}-a_{11} \\ a_{21}+a_{22} & a_{22}+a_{23} & a_{23}+a_{21} \\ a_{31}+a_{32} & a_{32}+a_{33} & a_{33}+a_{31}\end{pmatrix}$.
\end{enumerate}

\item \begin{enumerate}
\item 略.
\item 设 $A=(a_{ij})_{3\times 2}$，$e_{ij}$ 为 $\mathbf{R}^{3\times 2}$ 的自然基。因为 $PAQ = \begin{pmatrix}a_{12}+a_{22} & 0 \\ a_{22} & 0 \\ 0 & 0\end{pmatrix}$，
所以 $\sigma(e_{12}) = e_{11},\sigma(e_{22}) = e_{11}+e_{21},\sigma(e_{11})=\sigma(e_{21})=\sigma(e_{31})=\sigma(e_{32})=0$.

于是 $\mathrm{Ker}\sigma = \mathrm{span}(e_{11},e_{21},e_{31},e_{32}),\mathrm{Im}\sigma = \mathrm{span}(e_{11},e_{11}+e_{21})$.
\item 令 $B_1=\{e_{12},e_{22},e_{11},e_{21},e_{31},e_{32}\},B_2=\{e_{11},e_{11}+e_{21},e_{12},e_{22},e_{31},e_{32}\}$，则均为 $\mathbf{R^{3\times 2}}$ 的基，且 $\sigma(\epsilon)=(\eta)\begin{pmatrix}E_2 & 0 \\ 0 & 0\end{pmatrix}$.
\end{enumerate}
\item 略.

\item 设 $A=(a_{ij})_{3\times 2}$，$e_{ij}$ 为 $\mathbf{R}^{3\times 2}$ 的自然基. 因为 $PAQ = \begin{pmatrix}a_{12}+a_{22} & 0 \\ a_{22} & 0 \\ 0 & 0\end{pmatrix}$，所以 $\sigma(e_{12}) = e_{11},\sigma(e_{22}) = e_{11}+e_{21},\sigma(e_{11})=\sigma(e_{21})=\sigma(e_{31})=\sigma(e_{32})=0$.

于是 $\ker\sigma = \spa (e_{11},e_{21},e_{31},e_{32}),\im\sigma = \spa (e_{11},e_{11}+e_{21})$.

\item 令 $B_1=\{e_{12},e_{22},e_{11},e_{21},e_{31},e_{32}\},B_2=\{e_{11},e_{11}+e_{21},e_{12},e_{22},e_{31},e_{32}\}$，则均为 $\mathbf{R^{3\times 2}}$ 的基，且 $\sigma(\varepsilon)=(\eta)\begin{pmatrix}E_2 & 0 \\ 0 & 0\end{pmatrix}$.
\end{enumerate}

\item 见教材 P147/例 5
\end{enumerate}

\centerline{\heiti C组}
\begin{enumerate}
\item 使用数学归纳法。当 $n=1$ 时，$A=\begin{pmatrix}a\end{pmatrix}(a\neq 0)$，$B$ 取任意一阶矩阵均成立；
假设 $n-1$ 阶成立，$A = \begin{pmatrix}A_1 & \alpha \\ \beta & a_{nn}\end{pmatrix}$，其中 $A_1$ 为 $n-1$ 阶矩阵且存在 $n-1$ 阶下三角矩阵 $B_1$ 使得 $B_1A_1$ 为上三角矩阵，则有
\[\begin{pmatrix}B_1 & O \\ O & 1\end{pmatrix}\begin{pmatrix}A_1 & \alpha \\ O & a_{nn}-\beta A^{-1}\alpha\end{pmatrix} = \begin{pmatrix}B_1A_1 & B_1\alpha \\ O & a_{nn}-\beta A_1^{-1}\alpha\end{pmatrix}\]为上三角矩阵.
而 \[\begin{pmatrix}E_{n-1} & O \\ -\beta A_1^{-1} & 1\end{pmatrix}\begin{pmatrix}A_1 & \alpha \\ \beta & a_{nn}\end{pmatrix}=\begin{pmatrix}A_1 & \alpha \\ O & a_{nn}-\beta A_1^{-1} \alpha\end{pmatrix}\]
故 $B=\begin{pmatrix}B_1 & O \\ O & 1\end{pmatrix}\begin{pmatrix}E_{n-1} & O \\ -\beta A_1^{-1} & 1\end{pmatrix}$ 符合条件（$B_1$ 为下三角矩阵，故 $B$ 也是）.
\item 任取 $\alpha\in W$，有 $A_{12}\alpha=0$.
\[\begin{pmatrix}O_{k\times 1} \\ \alpha\end{pmatrix}=A^{-1}A\begin{pmatrix}O_{k\times 1} \\ \alpha\end{pmatrix}=A^{-1}\begin{pmatrix}A_{11} & A_{12} \\ A_{21} & A_{22}\end{pmatrix}\begin{pmatrix}O_{k\times 1} \\ \alpha\end{pmatrix}=A^{-1}\begin{pmatrix}O_{l\times 1} \\ A_{22}\alpha\end{pmatrix}\]
\[=\begin{pmatrix}B_{11} & B_{12} \\ B_{21} & B_{22}\end{pmatrix}\begin{pmatrix}O_{l\times 1} \\ A_{22}\alpha\end{pmatrix}=\begin{pmatrix}B_{12}A_{22}\alpha \\ B_{22}A_{22}\alpha\end{pmatrix}\]
故 $B_{12}A_{22}\alpha=0$，故我们可以推测以下定义：$\sigma\in L(W,U),\sigma(\alpha)=A_{22}\alpha$.
证明单射、满射即可.

单射：$\sigma(\alpha)=\sigma(\beta)\Rightarrow A_{22}(\alpha-\beta)=0$. 又有 $A_{12}\alpha=A_{12}\beta=0\Rightarrow A_{12}(\alpha-\beta)=0$，可得 $\begin{pmatrix}A_{12} \\ A_{22}\end{pmatrix}(\alpha-\beta) = 0$. 由于 $\begin{pmatrix}A_{12} \\ A_{22}\end{pmatrix}$ 列满秩，故有 $\alpha=\beta$. 成立.

满射：$\forall \gamma\in U,B_{12}\gamma = 0.$ 类似我们一开始的步骤，$\begin{pmatrix}O_{l\times 1} \\ \gamma\end{pmatrix}=AA^{-1}\begin{pmatrix}O_{l\times 1} \\ \gamma\end{pmatrix}= \cdots = \begin{pmatrix}A_{12}B_{22}\gamma \\ A_{22}B_{22}\gamma\end{pmatrix}$，故 $\exists B_{22}\gamma \in W,A_{22}B_{22}\gamma=\gamma\in U$. 故满射成立，证毕.
\item 使用数学归纳法. 当 $n=1$ 时，$A=\begin{pmatrix}a\end{pmatrix}(a\neq 0)$，$B$ 取任意一阶矩阵均成立；假设 $n-1$ 阶成立，$A = \begin{pmatrix}A_1 & \alpha \\ \beta & a_{nn}\end{pmatrix}$，其中 $A_1$ 为 $n-1$ 阶矩阵且存在 $n-1$ 阶下三角矩阵 $B_1$ 使得 $B_1A_1$ 为上三角矩阵，则有
\[\begin{pmatrix}B_1 & O \\ O & 1\end{pmatrix}\begin{pmatrix}A_1 & \alpha \\ O & a_{nn}-\beta A^{-1}\alpha\end{pmatrix} = \begin{pmatrix}B_1A_1 & B_1\alpha \\ O & a_{nn}-\beta A_1^{-1}\alpha\end{pmatrix}\]为上三角矩阵.
而 \[\begin{pmatrix}E_{n-1} & O \\ -\beta A_1^{-1} & 1\end{pmatrix}\begin{pmatrix}A_1 & \alpha \\ \beta & a_{nn}\end{pmatrix}=\begin{pmatrix}A_1 & \alpha \\ O & a_{nn}-\beta A_1^{-1} \alpha\end{pmatrix}\]
故 $B=\begin{pmatrix}B_1 & O \\ O & 1\end{pmatrix}\begin{pmatrix}E_{n-1} & O \\ -\beta A_1^{-1} & 1\end{pmatrix}$ 符合条件（$B_1$ 为下三角矩阵，故 $B$ 也是）.

\item 证明：$ \forall \alpha \in W,\enspace A_{12} \alpha = \vec{0} $.
\begin{align*}
\begin{pmatrix} O_{k \times 1} \\ \alpha \end{pmatrix}
& = A^{-1}A \begin{pmatrix} O_{k \times 1} \\ \alpha \end{pmatrix} = A^{-1} \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} \begin{pmatrix} O_{k \times 1} \\ \alpha \end{pmatrix} = A^{-1} \begin{pmatrix} O_{l \times 1} \\ A_{22} \alpha \end{pmatrix} \\
& = \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} \begin{pmatrix} O_{l \times 1} \\ A_{22} \alpha \end{pmatrix} = \begin{pmatrix} B_{12} A_{22} \alpha \\ B_{22} A_{22} \alpha \end{pmatrix}
\end{align*}
故 $ B_{12} A_{22} \alpha = \vec{0} $，故我们可以推测如下定义：$ \sigma \in \mathcal{L}(W,U),\enspace \sigma(\alpha) = A_{22} \alpha $.
只需证明 $ \sigma $ 是单射且满射即可.

单射：$ \sigma(\alpha) = \sigma(\beta) \implies A_{22}(\alpha - \beta) = \vec{0} $. 又有 $ A_{12} \alpha = A_{12} \beta = \vec{0} \implies A_{12}( \alpha - \beta) = \vec{0} $. 故 $ \begin{pmatrix} A_{12} \\ A_{22} \end{pmatrix} (\alpha - \beta) = \vec{0} $. 由于 $ \begin{pmatrix} A_{12} \\ A_{22} \end{pmatrix} $ 列满秩（$ A $ 可逆），故 $ \alpha = \beta $.

满射：$ \forall \gamma \in U,\enspace B_{12} \gamma = \vec{0} $.
\begin{align*}
\begin{pmatrix} O_{l \times 1} \\ \gamma \end{pmatrix}
& = A A^{-1} \begin{pmatrix} O_{l \times 1} \\ \gamma \end{pmatrix} = A \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} \begin{pmatrix} O_{l \times 1} \\ \gamma \end{pmatrix} \\
& = A \begin{pmatrix} O_{k \times 1} \\ B_{22} \gamma \end{pmatrix} = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} \begin{pmatrix} O_{k \times 1} \\ B_{22} \gamma \end{pmatrix} \\
& = \begin{pmatrix} A_{12} B_{22} \gamma \\ A_{22} B_{22} \gamma \end{pmatrix}
\end{align*}
故 $ \exists B_{22} \gamma \in W,\enspace A_{22} B_{22} \gamma = \gamma \in U $.
\end{enumerate}

\clearpage
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