Homework 05¶
约 274 个字 4 张图片 预计阅读时间 1 分钟
7.2.3¶
（a）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda-2 & 1\\1 & \lambda-2\end{pmatrix}|=(\lambda-2)^2-1=0\Rightarrow\lambda_1=3,\lambda_2=1\)
\(\therefore\rho(\mathbf{A})=3>1\)，不收敛
（b）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda & -1\\-1 & \lambda-1\end{pmatrix}|=\lambda(\lambda-1)-1=0\Rightarrow\lambda_1=\frac{1+\sqrt{5}}{2},\lambda_2=\frac{1-\sqrt{5}}{2}\)
\(\therefore\rho(\mathbf{A})=\frac{1+\sqrt{5}}{2}>1\)，不收敛
（c）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda & -\frac{1}{2}\\-\frac{1}{2} & \lambda\end{pmatrix}|=\lambda^2-\frac{1}{4}=0\Rightarrow\lambda_1=\frac{1}{2},\lambda_2=-\frac{1}{2}\)
\(\therefore\rho(\mathbf{A})=\frac{1}{2}<1\)，收敛
（d）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda-1 & -1\\2 & \lambda+2\end{pmatrix}|=\lambda^2+\lambda=0\Rightarrow\lambda_1=0,\lambda_2=-1\)
\(\therefore\rho(\mathbf{A})=1\)，不收敛
（e）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda-2 & -1 & 0\\-1 & \lambda-2 & 0\\0 & 0 & \lambda-3\end{pmatrix}|=0\Rightarrow\lambda_1=\lambda_2=3,\lambda_3=1\)
\(\therefore\rho(\mathbf{A})=3>1\)，不收敛
（f）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda+1 & -2 & 0\\0 & \lambda-3 & -4\\0 & 0 & \lambda-7\end{pmatrix}|=0\Rightarrow\lambda_1=7,\lambda_2=3,\lambda_3=-1\)
\(\therefore\rho(\mathbf{A})=7>1\)，不收敛
（g）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda-1 & -1 & -1\\-2 & \lambda-3 & -2\\-1 & -1 & \lambda-2\end{pmatrix}|=0\Rightarrow\lambda_1=\lambda_2=1,\lambda_3=5\)
\(\therefore\rho(\mathbf{A})=5>1\)，不收敛
（h）\(|\lambda\mathbf{E}-\mathbf{A}|=|\begin{pmatrix}\lambda-3 & -2 & 1\\-1 & \lambda+2 & -3\\-2 & 0 & \lambda-4\end{pmatrix}|=0\Rightarrow\lambda_1=3,\lambda_2=4,\lambda_3=-2\)
\(\therefore\rho(\mathbf{A})=4>1\)，不收敛
7.3.13¶
