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更新矩阵和向量
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Didnelpsun
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@@ -22,6 +22,8 @@
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% 数学公式
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\usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref}
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% 超链接
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\usepackage{rotating}
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% 用于旋转对象（旋转包）
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\author{Didnelpsun}
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\title{矩阵}
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\date{}
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@@ -142,6 +144,72 @@ $=\left(\begin{array}{ccc}
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0 & 0 & 0
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\end{array}\right)$
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\section{初等变换}
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\subsection{可逆矩阵}
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若$A$和$B$等价，求一个可逆矩阵$P$，使得$PA=B$。只用右乘$P=BA^{-1}$。
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需要根据逻辑上的计算还原出左乘的初等矩阵。\medskip
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\textbf{例题：}$A=\left(\begin{array}{ccc}
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1 & 0 & 1 \\
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-1 & -1 & 1 \\
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0 & 2 & -4
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\end{array}\right)$，$B=\left(\begin{array}{ccc}
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1 & 0 & 1 \\
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0 & -1 & 2 \\
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0 & 0 & 0
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\end{array}\right)$，当$A\thicksim B$时，求$P$使得$PA=B$。.
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解：目标是将$A$变为$B$，所以第一步将第一列的第二行的-1变为0。即将第一行加到第二行。
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左乘$E_{21}(1)A=\left(\begin{array}{ccc}
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1 & 0 & 0 \\
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1 & 1 & 0 \\
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0 & 0 & 1
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\end{array}\right)\left(\begin{array}{ccc}
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1 & 0 & 1 \\
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-1 & -1 & 1 \\
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0 & 2 & -4
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\end{array}\right)=\left(\begin{array}{ccc}
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1 & 0 & 1 \\
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0 & -1 & 2 \\
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0 & 2 & -4
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\end{array}\right)=C$。\medskip
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然后对第二列进行消，首先将第三行加上第二行的两倍。
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$E_{32}(2)C=\left(\begin{array}{ccc}
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1 & 0 & 0 \\
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1 & 1 & 0 \\
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0 & 2 & 1
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\end{array}\right)\left(\begin{array}{ccc}
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1 & 0 & 1 \\
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0 & -1 & 2 \\
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0 & 2 & -4
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\end{array}\right)=\left(\begin{array}{ccc}
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1 & 0 & 1 \\
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1 & -1 & 2 \\
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0 & 0 & 0
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\end{array}\right)=B$。\medskip
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$\therefore E_{32}(2)E_{21}(1)A=B$。
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$P=E_{32}(2)E_{21}(1)=\left(\begin{array}{ccc}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 2 & 1
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\end{array}\right)\left(\begin{array}{ccc}
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1 & 0 & 0 \\
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1 & 1 & 0 \\
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0 & 0 & 1
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\end{array}\right)=\left(\begin{array}{ccc}
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1 & 0 & 0 \\
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1 & 1 & 0 \\
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2 & 2 & 1
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\end{array}\right)$。
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\section{逆矩阵}
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\subsection{定义法}
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@@ -170,8 +238,20 @@ $\because A+B=A(E+A^{-1}B)=A(B^{-1}+A^{-1})B$。
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$\therefore (A+B)^{-1}=B^{-1}(B^{-1}+A^{-1})^{-1}A^{-1}$。
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\subsection{初等变换}
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$\left[A\vdots B\right]\overset{r}{\thicksim}\left[E\vdots A^{-1}\right]$，$\left[\begin{array}{c}
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A \\
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B
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\end{array}\right]\overset{c}{\thicksim}\left[\begin{array}{c}
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E \\
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A^{-1}
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\end{array}\right]$。
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\subsection{分块矩阵}
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基于拉普拉斯展开式。
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对于一些分块矩阵的逆，若$A$，$B$都可逆，则：$\left[\begin{array}{cc}
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A & O \\
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O & B
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@@ -184,6 +264,108 @@ $\therefore (A+B)^{-1}=B^{-1}(B^{-1}+A^{-1})^{-1}A^{-1}$。
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\end{array}\right]^{-1}=\left[\begin{array}{cc}
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O & B^{-1} \\
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A^{-1} & O
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\end{array}\right]$。
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\end{array}\right]$。\medskip
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\textbf{例题：}已知$A=\left(\begin{array}{cc}
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B & O \\
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D & C
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\end{array}\right)$，其中$B$为$r\times r$可逆矩阵，$C$为$s\times s$可逆矩阵，求$A^{-1}$。
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解：$\because\vert A\vert=\left|\begin{array}{cc}
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B & O \\
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D & C
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\end{array}\right|=\vert B\vert\vert C\vert\neq0$，所以$A$可逆，设$A^{-1}=\left(\begin{array}{cc}
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X & Y \\
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Z & W
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\end{array}\right)$。
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$AA^{-1}=\left(\begin{array}{cc}
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E_r & O \\
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O & E_s
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\end{array}\right)=E_{r+s}$。即$\left(\begin{array}{cc}
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BX & BY \\
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DX+CZ & DY+CW
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\end{array}\right)=E_{r+s}$。
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$\therefore\left\{\begin{array}{l}
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BX=E \\
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BY=O \\
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DX+CZ=O \\
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DY+CW=E
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\end{array}\right.$，$\left\{\begin{array}{ll}
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B^{-1}BX=B^{-1}, & X=B^{-1}\\
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B^{-1}BY=O, & Y=O \\
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CZ=-DX=-DB^{-1}, & Z=-C^{-1}DB^{-1} \\
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CW=E, & W=C^{-1}
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\end{array}\right.$。
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$\therefore A^{-1}=\left(\begin{array}{cc}
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B^{-1} & O \\
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-C^{-1}DB^{-1} & C^{-1}
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\end{array}\right)$。\medskip
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当分块矩阵为三角矩阵时，对角线为原方块矩阵的逆矩阵，非0的一角为原矩阵，再左乘同行的逆矩阵，右乘同列的逆矩阵。\medskip
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$\therefore A=\left(\begin{array}{cc}
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B & D \\
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O & C
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\end{array}\right)$，$A^{-1}=\left(\begin{array}{cc}
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B^{-1} & -B^{-1}DC^{-1} \\
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O & C^{-1}
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\end{array}\right)$。\medskip
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当分块矩阵为副对角矩阵时，对角线为对角方块矩阵的逆矩阵，非0的一角为原矩阵，再左乘同行的逆矩阵，右乘同列的逆矩阵。\medskip
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$\therefore A=\left(\begin{array}{cc}
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O & B \\
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C & D
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\end{array}\right)$，$A^{-1}=\left(\begin{array}{cc}
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-C^{-1}DB^{-1} & C^{-1} \\
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B^{-1} & O
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\end{array}\right)$。\medskip
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$\therefore A=\left(\begin{array}{cc}
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D & B \\
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C & O
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\end{array}\right)$，$A^{-1}=\left(\begin{array}{cc}
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O & C^{-1} \\
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B^{-1} & -C^{-1}DB^{-1}
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\end{array}\right)$。\medskip
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$A=\left(\begin{array}{ccc}
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A_1 \\
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& \ddots \\
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& & A_n
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\end{array}\right)$，$A^{-1}=\left(\begin{array}{ccc}
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A_1^{-1} \\
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& \ddots \\
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& & A_n^{-1}
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\end{array}\right)$。\medskip
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$A=\left(\begin{array}{ccc}
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& & A_1 \\
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& \ddots \\
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A_n
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\end{array}\right)$，$A^{-1}=\left(\begin{array}{ccc}
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& & A_n^{-1} \\
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& \begin{turn}{80}$\ddots$\end{turn} \\
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A_1^{-1}
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\end{array}\right)$。
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\section{矩阵方程}
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含有未知矩阵的方程就是矩阵方程，需要将方程进行恒等变形，化为$AX=B$、$XA=B$或$AXB=C$的形式。
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若$A$、$B$可逆，且可以分别得到$X=A^{-1}B$，$X=BA^{-1}$，$X=A^{-1}CB^{-1}$。
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\textbf{例题：}设3阶方阵$A$，$B$满足$A^{-1}BA=6A+BA$，且$A=\left(\begin{array}{ccc}
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\dfrac{1}{3} & 0 & 0 \\
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0 & \dfrac{1}{4} & 0 \\
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0 & 0 & \dfrac{1}{5}
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\end{array}\right)$，求$B$。
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解：$A^{-1}BA=(6E+B)A$，$A^{-1}B=6E+B$，$A^{-1}B-B=6E$，$(A^{-1}-E)B=6E$。
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$\therefore B=6(A^{-1}-E)^{-1}$。
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\end{document}
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BIN
linear-algebra/exercise/3-vector/vector.pdf
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linear-algebra/exercise/3-vector/vector.pdf
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44
linear-algebra/exercise/3-vector/vector.tex
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44
linear-algebra/exercise/3-vector/vector.tex
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\documentclass[UTF8, 12pt]{ctexart}
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% UTF8编码，ctexart现实中文
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\usepackage{color}
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% 使用颜色
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\usepackage{geometry}
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\setcounter{tocdepth}{4}
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\setcounter{secnumdepth}{4}
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% 设置四级目录与标题
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\geometry{papersize={21cm,29.7cm}}
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% 默认大小为A4
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\geometry{left=3.18cm,right=3.18cm,top=2.54cm,bottom=2.54cm}
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% 默认页边距为1英尺与1.25英尺
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\usepackage{indentfirst}
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\setlength{\parindent}{2.45em}
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% 首行缩进2个中文字符
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\usepackage{setspace}
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\renewcommand{\baselinestretch}{1.5}
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% 1.5倍行距
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\usepackage{amssymb}
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% 因为所以
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\usepackage{amsmath}
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% 数学公式
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\usepackage[colorlinks,linkcolor=black,urlcolor=blue]{hyperref}
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% 超链接
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\author{Didnelpsun}
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\title{向量}
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\date{}
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\begin{document}
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\maketitle
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\pagestyle{empty}
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\thispagestyle{empty}
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\tableofcontents
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\thispagestyle{empty}
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\newpage
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\pagestyle{plain}
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\setcounter{page}{1}
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\section{线性相关性}
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\textbf{例题：}设$A$是$n$阶矩阵，若存在正整数$k$，使得线性方程组$A^kx=0$有解向量$\alpha$，且$A^{k-1}\alpha\neq0$，证明向量组$\alpha,A\alpha,\cdots,A^{k-1}\alpha$线性无关。
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证明：
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\end{document}
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