【柯】代数学引论 第2章 §3.线性映射. 矩阵的运算
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\(Page.81\\1.\ 设X=[x_{1},x_{2},···,x_{n}],Y=[y_{1},y_{2},···,y_{n}]\\\quad a)\quad \varphi (X+Y)=[x_{1}+y_{1},x_{2}+y_{2},···,x_{n}+y_{n}]=\varphi (X)+\varphi (Y)\\\qquad \quad \varphi (kX)=[kx_{1},kx_{2},···,kx_{n}]=k\varphi (X)\\\quad b)\quad \varphi (X+Y)=[x_{1}+y_{1},x^{2}_{2}+y^{2}_{2},···,x^{n}_{n}+y^{n}_{n}]\neq \varphi (X)+\varphi (Y)\\\quad c)\quad \varphi (X+Y)=[x_{1}+y_{1},x_{1}+y_{1}+x_{2}+y_{2},···,\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}y_{i}]=\varphi (X)+\varphi (Y)\\\qquad \quad \varphi (kX)=[kx_{1},kx_{1}+kx_{2},···,k\sum_{i=1}^{n}x_{i}]=k\varphi(X)\\\quad 故a,c为线性映射\)
\(2.\ 归纳证明,略\)
\(3.\ \begin{pmatrix}0&-1\\ 1&-1\end{pmatrix}\begin{pmatrix}0&-1\\ 1&-1\end{pmatrix}\begin{pmatrix}0&-1\\ 1&-1\end{pmatrix}\\=\begin{pmatrix}-1&1\\ -1&0\end{pmatrix}\begin{pmatrix}0&-1\\ 1&-1\end{pmatrix}\\=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\)
\(4.\ \begin{pmatrix}p_{11}&p_{12}&···&p_{1n}\\p_{21}&p_{22}&···&p_{2n}\\···&···&···&···\\p_{n1}&p_{n2}&···&p_{nn}\end{pmatrix}\begin{pmatrix}x_{1}\\x_{2}\\···\\x_{n}\end{pmatrix}=\begin{pmatrix}\sum_{i=1}^{n}p_{1i}*x_{i}\\\sum_{i=1}^{n}p_{2i}*x_{i}\\···\\\sum_{i=1}^{n}p_{ni}*x_{i}\end{pmatrix}\\~\\“\Rightarrow ”:\sum_{j=1}^{n}\sum_{i=1}^{n}p_{ji}*x_{i}\\\quad \ \ =\sum_{i=1}^{n}x_{i}(\sum_{j=1}^{n}p_{ji})\\\quad \ \ =\sum_{i=1}^{n}x_{i}=1\\~\\“\Leftarrow ”:\sum_{j=1}^{n}\sum_{i=1}^{n}p_{ji}*x_{i}\\\quad \ \ =\sum_{i=1}^{n}x_{i}(\sum_{j=1}^{n}p_{ji})\\\quad \ \ 当且仅当\sum_{j=1}^{n}p_{ji}=1时.成立\\\quad \ \ 即P为随机矩阵\)
\(5.\ 略\)
\(6.\ 归纳证明,略\)
\(7.\ 设C=A+B\\\quad \ r_{A}=rankA=dim<A_{(1)},A_{(2)},···,A_{(r_{A})}> \\\quad r_{B}=rankA=dim<B_{(1)},B_{(2)},···,B_{(r_{B})}> \\\quad C_{(k)}=A_{(k)}+B_{(k)}=\sum_{i=1}^{k}\alpha _{i}A_{(i)}+\sum_{i=1}^{k}\beta _{i}B_{(i)} \\\quad <C_{(1)},···,C_{k}>=<\alpha _{1}A_{(1)}+\beta _{1}B_{(1)},···,\alpha _{k}A_{(k)}+\beta _{k}B_{(k)}> \\\quad rankC=rank(A+B)=dim<C_{(1)},···,C_{k}>\ \leqslant r_{A}+r_{B}\)
\(8.\ \begin{pmatrix}A&O\\E_{s}&B\end{pmatrix}=\begin{pmatrix}E_{m}&A\\O&E_{s}\end{pmatrix}\begin{pmatrix}O&AB\\E_{s}&O\end{pmatrix}\begin{pmatrix}E_{s}&B\\O&-E_{n}\end{pmatrix}\\\quad rankA+rankB\leqslant rank(\begin{pmatrix}O&AB\\E_{s}&O\end{pmatrix})=rank(AB)+s\\\quad 即rankA+rankB-s\leqslant rank(AB)\)
\(9.\ (AB)C=O\Rightarrow (AB)C^{(k)}=O\\\quad 故C^{(k)}为(AB)X=O的解，又(AB)X=0的基础解系有n-rank(AB)个线性无关的解\\\quad 故rankC+rank(AB)\leqslant n\\\quad rank(AB)\leqslant min{rankA,rankB}=n\\\quad 综上rankA+rankB+rankC\leqslant 2n\)
\(10.\ \left\{\begin{array}{l}0\quad A=O\\1 \quad else\end{array}\right.\)
\(11.\ A^{T}*(A^{T})^{-1}=E\Rightarrow [(A^{T})^{-1}]^{T}*A=E\Rightarrow [(A^{T})^{-1}]^{T}=A^{-1}=[(A^{-1})^{-T}]^{T}\Rightarrow (A^{T})^{-1}=(A^{-1})^{T}\Rightarrow A^{-1}=(A^{-1})^{T}\)
\(12.\ 略\)
\(13.\ 验证略\\\quad ad=bc=0\Rightarrow \frac{a}{c}=\frac{b}{d}=k\Rightarrow rankA\neq n\ 故不存在A^{-1}\)
\(14.\ 略\)
\(15.\ 略\)
\(16.\ A^{m}=O\Rightarrow \left | A\right |=0\Rightarrow ad-bc=0\\\quad A^{m}=A^{m-2}((a+d)A-(ad-bc)E)=(a+d)A^{m-1}=···=(a+d)^{m-1}A=0\\\quad 当A=O时,显然成立\\\quad 当a+d=0即a=-d时,A^{2}=\begin{pmatrix}a^{2}+bc&b(a+d)\\c(a+d)&d^{2}+bc\end{pmatrix}=O\\\quad Q.E.D.\)
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