高等代数教学辅导

\begin{equation*} \begin{array}{lcl} A(\lambda) & \xrightarrow{E(1,2(1))} & \begin{pmatrix} 1&\lambda^2+\lambda&0\\ \lambda&\lambda&-\lambda\\ 1+\lambda^2&\lambda^2&-\lambda^2 \end{pmatrix}\\ & \xrightarrow[E(1,2(-\lambda^2- \lambda))]{ } & \begin{pmatrix} 1&0&0\\ \lambda&-\lambda^3- \lambda^2+\lambda&-\lambda\\ 1+\lambda^2&-\lambda^4- \lambda^3- \lambda&-\lambda^2 \end{pmatrix}\\ & \xrightarrow{{\begin{array}{c} E(2,1(-\lambda))\\E(3,1(- \lambda^2-1)) \end{array}}} & \begin{pmatrix} 1&0&0\\ 0&-\lambda^3- \lambda^2+\lambda&-\lambda\\ 0&-\lambda^4- \lambda^3- \lambda&-\lambda^2 \end{pmatrix}\\ & \xrightarrow[E(2,3)]{} & \begin{pmatrix} 1&0&0\\ 0&-\lambda&-\lambda^3- \lambda^2+\lambda\\ 0&-\lambda^2&-\lambda^4- \lambda^3- \lambda \end{pmatrix}\\ & \xrightarrow{E(3,2(- \lambda))} & \begin{pmatrix} 1&0&0\\ 0&-\lambda&-\lambda^3- \lambda^2+\lambda\\ 0&0&-\lambda^2- \lambda \end{pmatrix}\\ & \xrightarrow[E(2,3(- \lambda^2- \lambda+1))]{}& \begin{pmatrix} 1&0&0\\ 0&-\lambda&0\\ 0&0&-\lambda^2- \lambda \end{pmatrix}\\ & \xrightarrow{E(2(-1)),E(3(-1))} & \begin{pmatrix} 1&0&0\\ 0&\lambda&0\\ 0&0&\lambda^2+\lambda \end{pmatrix} \end{array} \end{equation*}

\begin{equation*} \begin{array}{ccl}A(\lambda) & \xrightarrow[{\begin{array}{c} E(1,4)\\E(2,3) \end{array}}]{}&\begin{pmatrix} \lambda^2&0&0&0\\ 0&\lambda^2-\lambda&0&0\\ 0&0&(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ &\xrightarrow{E(1,2(1)) }&\begin{pmatrix} \lambda^2&\lambda^2-\lambda&0&0\\ 0&\lambda^2-\lambda&0&0\\ 0&0&(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ &\xrightarrow[E(1,2(-1))]{}&\begin{pmatrix} \lambda^2&-\lambda&0&0\\ 0&\lambda^2-\lambda&0&0\\ 0&0&(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ & \xrightarrow[E(1,2)]{}&\begin{pmatrix} -\lambda&\lambda^2&0&0\\ \lambda^2-\lambda&0&0&0\\ 0&0&(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ & \xrightarrow{E(1,3(1))}&\begin{pmatrix} -\lambda&\lambda^2&(\lambda-1)^2&0\\ \lambda^2-\lambda&0&0&0\\ 0&0&(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ & \xrightarrow[E(3,1(\lambda-2))]{}&\begin{pmatrix} -\lambda&\lambda^2&1&0\\ \lambda^2-\lambda&0&(\lambda^2- \lambda)(\lambda-2)&0\\ 0&0&(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ & \xrightarrow[E(1,3)]{}&\begin{pmatrix} 1&\lambda^2&-\lambda&0\\ (\lambda^2- \lambda)(\lambda-2)&0&\lambda^2-\lambda&0\\ (\lambda-1)^2& 0&0&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ & \xrightarrow[{\begin{array}{c} E(1,2(-\lambda^2))\\E(1,3(\lambda)) \end{array}}]{{\begin{array}{c} E(2,1(-(\lambda^2- \lambda)(\lambda-2)))\\E(3,1(-(\lambda-1)^2)) \end{array}}}&\begin{pmatrix} 1&0&0&0\\ 0&-\lambda^3(\lambda- 1)(\lambda-2)&\lambda (\lambda-1)^3&0\\ 0& -\lambda^2(\lambda-1)^2&\lambda(\lambda-1)^2&0\\ 0&0&0&\lambda^2-\lambda \end{pmatrix}\\ & \xrightarrow[E(2,4)]{E(2,4)}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)^2&-\lambda^2(\lambda-1)^2\\ 0&0&\lambda(\lambda-1)^3&-\lambda^3(\lambda-1)(\lambda-2) \end{pmatrix}\\ & \xrightarrow{E(4,3(1- \lambda))}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)^2&-\lambda^2(\lambda-1)^2\\ 0&0&0&\lambda^2(\lambda-1) \end{pmatrix}\\ & \xrightarrow{E(3,4(1))}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)^2&-\lambda^2(\lambda-1)(\lambda-2)\\ 0&0&0&\lambda^2(\lambda-1) \end{pmatrix}\\ &\xrightarrow[E(3,4(\lambda-1))]{}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)^2&\lambda(\lambda-1)\\ 0&0&0&\lambda^2(\lambda-1) \end{pmatrix}\\ & \xrightarrow[E(3,4)]{}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)&\lambda(\lambda-1)^2\\ 0&0&\lambda^2(\lambda-1) &0 \end{pmatrix}\\ & \xrightarrow[E(3,4(1- \lambda))]{E(4,3(-\lambda))}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)&0\\ 0&0&0&-\lambda^2(\lambda-1)^2 \end{pmatrix}\\ & \xrightarrow{E(4(-1))}&\begin{pmatrix} 1&0&0&0\\ 0&\lambda(\lambda- 1)&0&0\\ 0& 0&\lambda(\lambda-1)&0\\ 0&0&0&\lambda^2(\lambda-1)^2 \end{pmatrix}. \end{array} \end{equation*}

\begin{equation*} \begin{array}{rcl} \lambda E-A & \xrightarrow[E(1,3)]{}& \begin{pmatrix} -1&0&\lambda+1\\2&\lambda-2&-3\\\lambda-4&-1& 5 \end{pmatrix}\\ & \xrightarrow[E(1,3(\lambda+1))]{{\begin{array}{c} E(2,1(2))\\E(3,1( \lambda -4)) \end{array}}}&\begin{pmatrix} -1&0&0\\0&\lambda-2&2 \lambda-1\\0&-1&\lambda^2-3 \lambda+1 \end{pmatrix}\\ & \xrightarrow{E(2,3)} & \begin{pmatrix} -1&0&0\\0&-1&\lambda^2-3 \lambda+1\\0&\lambda-2&2 \lambda-1 \end{pmatrix}\\ & \xrightarrow[E(2,3(\lambda^2-3 \lambda+1))]{{\begin{array}{c}E(3,2(\lambda-2))\\E(1(-1))\\E(2(-1))\end{array}}}&\begin{pmatrix} 1&0&0\\0&1&0\\0&0&\lambda^3-5 \lambda^2+9 \lambda-3 \end{pmatrix}, \end{array} \end{equation*}

\begin{equation*} \begin{array}{rcl} \lambda E-A & \xrightarrow[E(1,3)]{} & \begin{pmatrix} -1&-1&\lambda-3\\0&\lambda-4&0\\\lambda-5&-1&1 \end{pmatrix}\\ & \xrightarrow[{\begin{array}{c} E(1,2(-1))\\E(1,3( \lambda -3)) \end{array}}]{E(3,1(\lambda-5))}&\begin{pmatrix} -1&0&0\\0&\lambda-4&0\\0&4-\lambda&(\lambda-4)^2 \end{pmatrix}\\ & \xrightarrow{E(3,2(1))} & \begin{pmatrix} -1&0&0\\0&\lambda-4&0\\0&0&(\lambda-4)^2 \end{pmatrix}\\ & \xrightarrow{E(1(-1))}&\begin{pmatrix} 1&0&0\\0&\lambda-4&0\\0&0&(\lambda-4)^2 \end{pmatrix}, \end{array} \end{equation*}

\begin{equation*} \lambda E_n-A=P(\lambda){\rm diag} (1,\dots ,1,d_1(\lambda),d_2(\lambda),\dots ,d_k(\lambda))Q(\lambda), \end{equation*}

\begin{equation*} \det (\lambda E_n-A)=\det P(\lambda)\det Q(\lambda)d_1(\lambda)d_2(\lambda)\cdots d_k(\lambda), \end{equation*}

\begin{equation*} \chi_A(\lambda)=cd_1(\lambda)d_2(\lambda)\cdots d_k(\lambda),c\in\mathbb{F}. \end{equation*}

\begin{equation*} \chi_A(\lambda)=d_1(\lambda)d_2(\lambda)\cdots d_k(\lambda). \end{equation*}

\begin{equation*} B=\begin{pmatrix} 0&2\\1&-1 \end{pmatrix},C=\begin{pmatrix} 0&-2\\1&3 \end{pmatrix}, \end{equation*}

\begin{equation*} A=\begin{pmatrix} B&0\\0&C \end{pmatrix}. \end{equation*}

\begin{equation*} \lambda E-B\simeq \begin{pmatrix} 1&0\\0&(\lambda+2)(\lambda-1) \end{pmatrix}, \end{equation*}

\begin{equation*} \lambda E-C\simeq \begin{pmatrix} 1&0\\0&(\lambda-1)(\lambda-2) \end{pmatrix}. \end{equation*}

\begin{equation*} A(\lambda)=\begin{pmatrix} 1&0&0&0\\ 0&(\lambda+2)(\lambda-1)&0&0\\ 0&0&1&0\\ 0&0&0&(\lambda-1)(\lambda-2) \end{pmatrix}, \end{equation*}

\begin{equation*} D_1(\lambda)=D_2(\lambda)=1,D_3(\lambda)=\lambda-1,D_4(\lambda)=(\lambda+2)(\lambda-2)(\lambda-1)^2, \end{equation*}

\begin{equation*} D_1(\lambda)=D_2(\lambda)=1,D_3(\lambda)=\lambda-1,D_4(\lambda)=(\lambda+2)(\lambda-2)(\lambda-1)^2. \end{equation*}

\begin{equation*} g_1(\lambda)=g_2(\lambda)=1,g_3(\lambda)=\lambda-1,g_4(\lambda)=(\lambda+2)(\lambda-2)(\lambda-1). \end{equation*}

\begin{equation*} \begin{pmatrix} f(\lambda)&0\\0&g(\lambda) \end{pmatrix}=P\ \begin{pmatrix} g(\lambda)&0\\0&f(\lambda) \end{pmatrix}P, \end{equation*}

\begin{equation*} \begin{array}{cl}\begin{pmatrix} f(\lambda)&0\\0&g(\lambda) \end{pmatrix}&\xrightarrow[E(2,1(v(\lambda) ))]{E(2,1(u(\lambda)))}\begin{pmatrix} f(\lambda)&0\\1&g(\lambda) \end{pmatrix}\stackrel{E(1,2)}{\longrightarrow}\begin{pmatrix} 1&g(\lambda)\\f(\lambda)&0 \end{pmatrix}\\&\xrightarrow[{\begin{array}{c} E(1,2(-g(\lambda)))\\E(2(-1)) \end{array}}]{{\begin{array}{c} E(2,1(-f(\lambda))) \end{array}}}\begin{pmatrix} 1&0\\0&f(\lambda)g(\lambda) \end{pmatrix}.\end{array} \end{equation*}

\begin{equation*} (\lambda E-A)^*=\begin{pmatrix} A_{11}(\lambda)&A_{21}(\lambda)&\cdots&A_{n1}(\lambda)\\ A_{12}(\lambda)&A_{22}(\lambda)&\cdots&A_{n2}(\lambda)\\ \vdots&\vdots&&\vdots\\ A_{1n}(\lambda)&A_{2n}(\lambda)&\cdots&A_{nn}(\lambda) \end{pmatrix}, \end{equation*}

\begin{equation*} M_{ij}(\lambda)=D_{n-1}(\lambda)B_{ij}(\lambda), \end{equation*}

\begin{equation*} B(\lambda)=\left((-1)^{i+j} B_{ij}(\lambda)\right)_{n\times n}, \end{equation*}