张祖锦2023年数学专业真题分类70天之第61天
1381、 7、 (15 分) 设 $\displaystyle V$ 是 $\displaystyle n$ 维线性空间, $\displaystyle \varphi$ 为 $\displaystyle V$ 上的线性变换, 且 $\displaystyle \varphi$ 的特征多项式为
$$\begin{aligned} f(x)=(x-\lambda_1)^{m_1}(x-\lambda_2)^{m_2}\left(\lambda_1\neq \lambda_2\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
其中 $\displaystyle m_1+m_n=n$. (1)、 (5 分) 证明: $\displaystyle \ker\left((\varphi-\lambda_1\mathscr{E})^{m_1}\right)$ 是 $\displaystyle \varphi$ 的不变子空间, 其中 $\displaystyle \mathscr{E}$ 是恒等变换; (2)、 (10 分) 证明:
$$\begin{aligned} V=\ker\left((\varphi-\lambda_1\mathscr{E})^{m_1}\right)+\ker\left((\varphi-\lambda_2\mathscr{E})^{m_2}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
[张祖锦注: 参考解答其实证明了第 2 问是直和.] (东北师范大学2023年高等代数与解析几何考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 对 $\displaystyle \alpha\in \ker\left((\varphi-\lambda_1\mathscr{E})^{m_1}\right)$,
$$\begin{aligned} (\varphi-\lambda_1\mathscr{E})^{m_1}\varphi(\alpha) =\varphi(\varphi-\lambda_1\mathscr{E})^{m_1}\alpha =\varphi 0=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \varphi(\alpha)\in \ker\left((\varphi-\lambda_1\mathscr{E})^{m_1}\right)$. 这就证明了 $\displaystyle \ker\left((\varphi-\lambda_1\mathscr{E})^{m_1}\right)$ 是 $\displaystyle \varphi$ 的不变子空间. (2)、 (2-1)、 我们先给出一般结论. 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb{P}$ 上的 $\displaystyle n$ 维线性空间, $\displaystyle \sigma$ 是 $\displaystyle V$ 上的线性变换, $\displaystyle f(x),g(x)\in\mathbb{P}[x]$, $\displaystyle \left(f(x),g(x)\right)=1$, $\displaystyle h(x)=f(x)g(x)$. 记线性变换 $\displaystyle h(\sigma), f(\sigma)$ 和 $\displaystyle g(\sigma)$ 的核为 $\displaystyle \ker h(\sigma), \ker f(\sigma)$ 和 $\displaystyle \ker g(\sigma)$. 则
$$\begin{aligned} \ker h(\sigma)=\ker f(\sigma)\oplus \ker g(\sigma). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
事实上, 由 $\displaystyle (f,g)=1$ 知 $\displaystyle \exists\ u,v,\mathrm{ s.t.} uf+vg=1$. 于是
$$\begin{aligned} \alpha\in \ker h(\sigma)&\Rightarrow \alpha=u(\sigma)f(\sigma)\alpha+v(\sigma)g(\alpha)\alpha \in \ker g(\sigma)+\ker f(\sigma). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这表明 $\displaystyle \ker h(\sigma)=\ker f(\sigma)+\ker g(\sigma)$. 又由
$$\begin{aligned} \alpha\in \ker f(\sigma)\cap \ker g(\sigma) &\Rightarrow f(\sigma)\alpha=0=g(\sigma)\alpha\\\\ &\Rightarrow \alpha=u(\sigma)f(\sigma)\alpha+v(\sigma)g(\alpha)\alpha=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \ker h(\sigma)=\ker f(\sigma)\oplus\ker g(\sigma)$. (2-2)、 由 Hamilton-Cayley 定理知 $\displaystyle \ker f(\varphi)=V$. 再由
$$\begin{aligned} \left((x-\lambda_1)^{m_1},(x-\lambda_2)^{m_2}\right)=1 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
及第 i 步知
$$\begin{aligned} V=\ker\left((\varphi-\lambda_1\mathscr{E})^{m_1}\right)\oplus\ker\left((\varphi-\lambda_2\mathscr{E})^{m_2}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1382、 5、 设 $\displaystyle \mathbb{F}$ 是一个数域, $\displaystyle V=M_2(\mathbb{F})$ 是 $\displaystyle \mathbb{F}$ 上的所有 $\displaystyle 2$ 阶矩阵构成的线性空间,
$$\begin{aligned} \mathscr{A}(X)=X^\star,\forall\ X\in V, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
其中 $\displaystyle X^\star$ 为 $\displaystyle X$ 的伴随矩阵. (1)、 证明: $\displaystyle \mathscr{A}$ 为 $\displaystyle V$ 上的线性变换; (2)、 求 $\displaystyle \mathscr{A}$ 在 $\displaystyle E_{11}, E_{12}, E_{21}, E_{22}$ 下的矩阵 $\displaystyle A$, 并求 $\displaystyle V$ 的一组基, 使得 $\displaystyle \mathscr{A}$ 在该基下的矩阵为对角阵 $\displaystyle \varLambda$, 并求出对角阵. (东南大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 设 $\displaystyle X=\left(\begin{array}{cccccccccccccccccccc}x_{11}&x_{12}\\\\ x_{21}&x_{22}\end{array}\right)$, 则 $\displaystyle X^\star=\left(\begin{array}{cccccccccccccccccccc}x_{22}&-x_{12}\\\\ -x_{21}&x_{11}\end{array}\right)$. 对 $\displaystyle \forall\ k,l\in\mathbb{F}$,
$$\begin{aligned} X=\left(\begin{array}{cccccccccccccccccccc}x_{11}&x_{12}\\\\ x_{21}&x_{22}\end{array}\right), Y=\left(\begin{array}{cccccccccccccccccccc}y_{11}&yx_{12}\\\\ y_{21}&y_{22}\end{array}\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
我们有
$$\begin{aligned} &\mathscr{A}(kX+lY)=\mathscr{A}\left(\begin{array}{cccccccccccccccccccc}kx_{11}+ly_{11}&kx_{12}+ly_{12}\\\\ kx_{21}+ly_{21}&kx_{22}+ly_{22}\end{array}\right)\\\\ =&\left(\begin{array}{cccccccccccccccccccc}kx_{22}+ly_{22}&-kx_{12}-ly_{12}\\\\ -kx_{21}-ly_{21}&kx_{11}+ly_{11}\end{array}\right)\\\\ =&k\left(\begin{array}{cccccccccccccccccccc}x_{22}&-x_{12}\\\\ -x_{21}&x_{11}\end{array}\right)+l\left(\begin{array}{cccccccccccccccccccc}y_{22}&-y_{12}\\\\ -y_{21}&y_{11}\end{array}\right) =k\mathscr{A} X+l\mathscr{A} Y. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \mathscr{A}$ 为 $\displaystyle V$ 上的线性变换. (2)、 由
$$\begin{aligned} \mathscr{A} E_{11}=E_{22}, \mathscr{A} E_{12}=-E_{12}, \mathscr{A} E_{21}=-E_{21}, \mathscr{A} E_{22}=E_{11} \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} \mathscr{A}(E_{11},E_{12},E_{21},E_{22})=&(E_{11},E_{12},E_{21},E_{22})A,\\\\ A=&\left(\begin{array}{cccccccccccccccccccc}0&0&0&1\\\\ 0&-1&0&0\\\\ 0&0&-1&0\\\\ 1&0&0&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,-1,-1,-1$. 由
$$\begin{aligned} E-A\to\left(\begin{array}{cccccccccccccccccccc}1&0&0&-1\\\\ 0&1&0&0\\\\ 0&0&1&0\\\\ 0&0&0&0 \end{array}\right), -E-A\to\left(\begin{array}{cccccccccccccccccccc}1&0&0&1\\\\ 0&0&0&0\\\\ 0&0&0&0\\\\ 0&0&0&0 \end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,-1$ 的特征向量分别为
$$\begin{aligned} \xi_1=\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0\\\\1 \end{array}\right), \xi_2=\left(\begin{array}{cccccccccccccccccccc}0\\\\1\\\\0\\\\0 \end{array}\right), \xi_3=\left(\begin{array}{cccccccccccccccccccc}0\\\\0\\\\1\\\\0 \end{array}\right), \xi_4=\left(\begin{array}{cccccccccccccccccccc}-1\\\\0\\\\0\\\\1 \end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
于是
$$\begin{aligned} &P=(\xi_1,\xi_2,\xi_3,\xi_4)=\left(\begin{array}{cccccccccccccccccccc} 1&0&0&-1\\\\ 0&1&0&0\\\\ 0&0&1&0\\\\ 1&0&0&1 \end{array}\right)\\\\ \Rightarrow& P^{-1}AP=\mathrm{diag}\left(1,-1,-1,-1\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这表明 $\displaystyle \mathscr{A}$ 在 $\displaystyle V$ 的基
$$\begin{aligned} (E_{11},E_{12},E_{21},E_{22})\xi_1=E_2, (E_{11},E_{12},E_{21},E_{22})\xi_2=E_{12},\\\\ (E_{11},E_{12},E_{21},E_{22})\xi_3=E_{21}, (E_{11},E_{12},E_{21},E_{22})\xi_4=\left(\begin{array}{cccccccccccccccccccc}-1&0\\\\ 0&1\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
下的矩阵为对角阵 $\displaystyle \mathrm{diag}(1,-1,-1,-1)$.跟锦数学微信公众号. 在线资料/公众号/
1383、 9、 设 $\displaystyle \mathscr{A}$ 是 $\displaystyle n$ 维线性空间 $\displaystyle V$ 上的线性变换, $\displaystyle \mathscr{A}$ 在 $\displaystyle V$ 的某组基下的矩阵为 $\displaystyle A$,
$$\begin{aligned} V_1=&\left\{\alpha\in V; \mathscr{A}\alpha=0\right\},\\\\ V_2=&\left\{\alpha=\mathscr{A}(\beta); \beta\in V\right\}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
证明以下是命题等价: (1)、 $\displaystyle V=V_1+V_2$; (2)、 $\displaystyle V_1+V_2$ 是直和; (3)、 $\displaystyle \mathrm{rank} A=\mathrm{rank}(A^2)$. (东南大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 $\displaystyle (1)\Rightarrow (2)$: 由
$$\begin{aligned} \dim(V_1\cap V_2)=&\dim V_1+\dim V_2-\dim(V_1+V_2)\\\\ =&\dim \ker \mathscr{A}+\dim \mathrm{im} \mathscr{A}-\dim V=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle V_1\cap V_2=\left\{0\right\}$. 联合 $\displaystyle V_1+V_2=V$ 知 $\displaystyle V_1\oplus V_2=V$. (2)、 $\displaystyle (2)\Rightarrow (1)$: 显然成立. (3)、 $\displaystyle (2)\Rightarrow (3)$: 由 $\displaystyle V_1\cap V_2=\left\{0\right\}$ 知 $\displaystyle 0=V_1\cap V_2=\ker \mathscr{A}\cap \mathrm{im} \mathscr{A}$, 而
$$\begin{aligned} \mathscr{A}^2\alpha=0\Rightarrow \mathscr{A}\alpha\in \mathrm{im} \mathscr{A}\cap \ker \mathscr{A}\Rightarrow \mathscr{A}\alpha=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
又显然 $\displaystyle \mathscr{A}\alpha=0\Rightarrow \mathscr{A}^2\alpha=0$, 我们有
$$\begin{aligned} &\left\{\alpha\in V; \mathscr{A}^2\alpha=0\right\}=\left\{\alpha\in V; \mathscr{A}\alpha=0\right\}\\\\ \Rightarrow&\dim\left\{\alpha\in V; \mathscr{A}^2\alpha=0\right\}=\dim\left\{\alpha\in V; \mathscr{A}\alpha=0\right\}\\\\ \Leftrightarrow&n-\mathrm{rank}(A^2)=n-\mathrm{rank} A\Leftrightarrow \mathrm{rank}(A^2)=\mathrm{rank} A.\qquad(I) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(4)、 $\displaystyle (3)\Rightarrow (2)$: 由 $\displaystyle \left\{\alpha\in V; \mathscr{A}\alpha=0\right\}\subset \left\{\alpha\in V; \mathscr{A}^2\alpha=0\right\}$ 知 $\displaystyle (I)$ 可往回推, 得 $\displaystyle \mathscr{A}^2\alpha=0\Rightarrow \mathscr{A}\alpha=0$. 而
$$\begin{aligned} \alpha\in V_1\cap V_2\Rightarrow& \mathscr{A}\alpha=0; \exists\ \beta\in V,\mathrm{ s.t.} \alpha=\mathscr{A}\beta\\\\ \Rightarrow&0=\mathscr{A}\alpha=\mathscr{A}^2\beta\Rightarrow 0=\mathscr{A}\beta=\alpha. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle V_1\cap V_2=\left\{0\right\}$. 这表明
$$\begin{aligned} \dim(V_1+V_2)=&\dim V_1+\dim V_2-\dim(V_1\cap V_2)\\\\ =&\dim \ker \mathscr{A}+\dim \mathrm{im}\mathscr{A}-0=n=\dim V. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
从而 $\displaystyle V_1+V_2=V$. 联合 $\displaystyle V_1\cap V_2=\left\{0\right\}$ 知 $\displaystyle V_1\oplus V_2=V$.跟锦数学微信公众号. 在线资料/公众号/
1384、 (3)、 设 $\displaystyle V_1,V_2$ 为 $\displaystyle n$ 维线性空间的子空间, 且
$$\begin{aligned} \dim(V_1+V_2)=\dim V_1+1, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则 $\displaystyle \dim V_2-\dim(V_1\cap V_2)=\underline{\ \ \ \ \ \ \ \ \ \ }$. (福州大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / $\displaystyle \dim V_2-\dim(V_1\cap V_2)=\dim(V_1+V_2)-\dim V_1=1$.跟锦数学微信公众号. 在线资料/公众号/
1385、 (4)、 设 $\displaystyle \varphi$ 是线性空间 $\displaystyle V$ 到 $\displaystyle W$ 的线性映射, $\displaystyle \xi_1,\cdots,\xi_n$ 及 $\displaystyle \eta_1,\cdots,\eta_n$ 分别是 $\displaystyle V$ 和 $\displaystyle W$ 的基, $\displaystyle A$ 是 $\displaystyle \varphi$ 在这两个基下的矩阵, 则 $\displaystyle \varphi$ 为单射的充分必要条件是 $\displaystyle \mathrm{rank} A=\underline{\ \ \ \ \ \ \ \ \ \ }$. (福州大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / $\displaystyle \varphi$ 是单射 $\displaystyle \Leftrightarrow\ker \varphi=\left\{0\right\}$
$$\begin{aligned} \Leftrightarrow 0=\dim \ker \varphi=\dim V-\dim \mathrm{im} \varphi =n-\mathrm{rank} A\Leftrightarrow \mathrm{rank} A=n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1386、 (3)、 设向量 $\displaystyle \alpha_1,\cdots,\alpha_n\ (n > 1)$ 线性无关,
$$\begin{aligned} \beta=\alpha_1+\cdots+\alpha_n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
证明: $\displaystyle \beta-\alpha_1,\cdots,\beta-\alpha_n$ 线性无关. (福州大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / 由
$$\begin{aligned} &0=\sum_{i=1}^n k_i(\beta-\alpha_i) \stackrel{k=\sum_{i=1}^n k_i}{\Rightarrow} \sum_{i=1}^n k_i\alpha_i=k\beta=k\sum_{i=1}^n \alpha_i\\\\ \Rightarrow& \sum_{i=1}^n (k_i-k)\alpha_i=0\Rightarrow k_1=\cdots=k_n=k\\\\ \Rightarrow& k=\sum_{i=1}^n k_i=nk\stackrel{n > 1}{\Rightarrow}k=0\Rightarrow k_1=\cdots=k_n=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \beta-\alpha_1,\cdots,\beta-\alpha_n$ 线性无关.跟锦数学微信公众号. 在线资料/公众号/
1387、 (4)、 (12 分) 已知 $\displaystyle \varPhi: \mathbb{R}^n\to\mathbb{R}^s, \varPsi: \mathbb{R}^m\to\mathbb{R}^n$ 都为线性映射. 证明:
$$\begin{aligned} \dim \ker (\varPhi\varPsi)\leq \dim\ker \varPhi+\dim\ker\varPsi, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
其中 $\displaystyle \ker$ 为核空间. (福州大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / 分别取定 $\displaystyle \mathbb{R}^m,\mathbb{R}^n,\mathbb{R}^s$ 的一组基后, 设 $\displaystyle \varPhi,\varPsi$ 在相应基下的矩阵分别为 $\displaystyle A_{s\times n},B_{n\times m}$, 则
$$\begin{aligned} &\dim \ker (\varPhi\varPsi)\leq \dim\ker \varPhi+\dim\ker\varPsi\\\\ \Leftrightarrow&m-\mathrm{im}\mathrm{im}(\varPhi\varPsi)\leq \left(n-\dim \mathrm{im}\varPhi\right)+\left(m-\dim\mathrm{im}\varPsi\right)\\\\ \Leftrightarrow&\dim \mathrm{im} \varPhi+\dim\mathrm{im}\varPsi\leq n+\dim \mathrm{im}(\varPhi\varPsi)\\\\ \Leftrightarrow&\mathrm{rank} A+\mathrm{rank} B\leq n+\mathrm{rank}(AB). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
事实上, 这可由
$$\begin{aligned} \left(\begin{array}{cccccccccccccccccccc}AB&\\\\ &E_n\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}AB&-A\\\\ 0&E_n\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}0&-A\\\\ B&E_n\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}0&A\\\\ B&E_n\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
得到.跟锦数学微信公众号. 在线资料/公众号/
1388、 3、 设 $\displaystyle \varphi$ 是有限维复线性空间 $\displaystyle V$ 上的线性变换, 对任意的 $\displaystyle t\in \mathbb{C}$, $\displaystyle \ker(\varphi^2-tI_V)$ 最多是 $\displaystyle 1$ 维的. 证明: (1)、 若 $\displaystyle \lambda$ 是 $\displaystyle \varphi$ 的特征值, 且 $\displaystyle \lambda\neq 0$, 则 $\displaystyle -\lambda$ 不是 $\displaystyle \varphi$ 的特征值. (2)、 若 $\displaystyle \lambda$ 是 $\displaystyle \varphi$ 的特征值, 则 $\displaystyle \ker(\varphi-\lambda I_V)$ 是 $\displaystyle 1$ 维的. (3)、 存在一个复多项式 $\displaystyle p(x)$, 使得 $\displaystyle p(\varphi^2)=\varphi$. (复旦大学2023年代数(第4,6,7,8题没做)考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 设 $\displaystyle 0\neq \alpha\in V$ 使得 $\displaystyle \varphi(\alpha)=\lambda \alpha$. 用反证法证明题目. 若 $\displaystyle -\lambda$ 也是 $\displaystyle \varphi$ 的特征值, 对应的特征向量为 $\displaystyle \beta$, 则 $\displaystyle \varphi(\beta)=-\lambda \beta$. 于是
$$\begin{aligned} \varphi^2(\alpha)=\lambda^2\alpha, \varphi^2(\beta)=\lambda^2\beta. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由 $\displaystyle \ker(\varphi^2-\lambda^2I_V)$ 至多是一维的知
$$\begin{aligned} &\exists\ \lambda\in\mathbb{C},\mathrm{ s.t.} \beta=\lambda\alpha \stackrel{\varphi(\beta)=\lambda \beta}{\Rightarrow}\varphi(\alpha)=-\lambda \alpha\\\\ \stackrel{\varphi(\alpha)=\lambda \alpha}{\Rightarrow}&2\lambda \alpha=0\stackrel{\lambda\neq 0}{\Rightarrow}\alpha=0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这是一个矛盾. 故有结论. (2)、 设 $\displaystyle \alpha$ 是 $\displaystyle \varphi$ 的属于特征值 $\displaystyle \lambda$ 的特征向量, 则 $\displaystyle \varphi(\alpha)=\lambda \alpha$. 故 $\displaystyle m=\dim \ker(\varphi-\lambda I_V)\geq 1$. 设 $\displaystyle \varepsilon_1,\cdots,\varepsilon_m$ 是 $\displaystyle \ker(\varphi-\lambda I_V)$ 的一组基, 则
$$\begin{aligned} \varphi(\varepsilon_i)=\lambda \varepsilon_i\Rightarrow \varphi^2(\varepsilon_i)=\lambda^2\varepsilon_i \Rightarrow \varepsilon_i\in \ker (\varphi^2-\lambda^2I_V). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由题设’$\dim\ker(\varphi^2-\lambda^2I_V)\leq 1$‘知 $\displaystyle m\leq 1$. 故确有 $\displaystyle m=1$. (3)、 由第 2 步知 $\displaystyle \varphi$ 在 $\displaystyle V$ 的某组基下的矩阵为 Jordan 标准形
$$\begin{aligned} J=\mathrm{diag}\left(J_{n_1}(\lambda_1),\cdots,J_{n_s}(\lambda_s)\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
其中 $\displaystyle \lambda_i$ 互异 [若不然, 某两个 $\displaystyle \lambda_i$ 相同, 则 $\displaystyle \ker(\varphi-\lambda_iI_V)\geq 2$ 了, 矛盾]. 由
$$\begin{aligned} J_{n_i}^2(\lambda_i)=\lambda_i^2+n_i\lambda_iJ_{n_i}(0)+J_{n_i}^2(0) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知若 $\displaystyle \lambda_i=0$, 则 $\displaystyle \dim \ker (\varphi^2-\lambda_i^2I_V)=\dim \ker \varphi^2\geq 2$. 这是一个矛盾. 故个 $\displaystyle \lambda_i\neq 0$. 又由第 1 步知
$$\begin{aligned} i\neq j\Rightarrow \lambda_i\neq -\lambda_j\stackrel{\lambda_i\neq\lambda_j}{\Rightarrow}\lambda_i^2\neq \lambda_j^2. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
为证 $\displaystyle \exists\ p(x)\in \mathbb{C}[x],\mathrm{ s.t.} p(\varphi^2)=\varphi$, 仅需验证存在多项式 $\displaystyle p(x)$ 使得
$$\begin{aligned} a\in \mathbb{C}\backslash\left\{0\right\}, p\left(J_k^2(a)\right)=J_k(a)s=aE_k+J_k(0). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
设 $\displaystyle J=J_k^2(a)-a^2E_k=aJ_k(0)+J_k^2(0)$, 则仅需证明存在 $\displaystyle q(x)$, 使得 $\displaystyle q(J)=J_k(0)$. 事实上, $\displaystyle \frac{1}{a^{k-1}}J^{k-1}=J_k^{k-1}(0)$, $\displaystyle \frac{1}{a^{k-2}}J^{k-2}$ 减去某 $\displaystyle bJ_k^{k-1}(0)$ 等于 $\displaystyle J_k^{k-2}(0)$, $\displaystyle \frac{1}{a^{k-2}}J^{k-3}$ 减去某 $\displaystyle cJ_k^{k-2}(0)+dJ_k^{k-1}(0)$ 等于 $\displaystyle J_k^{k-3}(0)$, 等等. 一直做下去知 $\displaystyle J_k^{k-1}(0), J_k^{k-2}(0), J_k^{k-3}(0),\cdots, J_k(0)$ 是 $\displaystyle J$ 的多项式. 证毕.跟锦数学微信公众号. 在线资料/公众号/
1389、 4、 记 $\displaystyle M_n(\mathbb{R})$ 为全体 $\displaystyle n$ 阶实矩阵组成的, 作为 $\displaystyle \mathbb{R}$ 上的线性空间. 求满足如下条件的所有 $\displaystyle \mathbb{R}$ 上的线性子空间 $\displaystyle V\subset M_n(\mathbb{R})$:
$$\begin{aligned} \mbox{对任意的 $\displaystyle A\in V$, 若 $\displaystyle A\neq 0$, 则 $\displaystyle A$ 可逆}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(复旦大学2023年代数(第4,6,7,8题没做)考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 /跟锦数学微信公众号. 在线资料/公众号/
1390、 (3)、 设
$$\begin{aligned} \alpha_1=(2,0,0)^\mathrm{T}, \alpha_2=(-1,3,0)^\mathrm{T}, \alpha_3=(7,-4,0)^\mathrm{T}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则向量组 $\displaystyle \alpha_1,\alpha_2,\alpha_3$ 的秩为 $\displaystyle \underline{\ \ \ \ \ \ \ \ \ \ }$. (广西大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / 由
$$\begin{aligned} (\alpha_1,\alpha_2,\alpha_3)\to\left(\begin{array}{cccccccccccccccccccc}1&0&\frac{17}{6}\\\\ 0&1&-\frac{4}{3}\\\\ 0&0&1\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知应填 $\displaystyle 2$.跟锦数学微信公众号. 在线资料/公众号/
1391、 (6)、 设
$$\begin{aligned} (\alpha_1,\alpha_2,\alpha_3,\alpha_4)=\left(\begin{array}{cccccccccccccccccccc}1&1&1&1\\\\ 1&1&-1&-1\\\\ 1&-1&1&-1\\\\ 1&-1&-1&1\end{array}\right), (\beta_1,\beta_2,\beta_3,\beta_4)=\left(\begin{array}{cccccccccccccccccccc}2&1&-1&1\\\\ 0&3&1&0\\\\ 5&3&2&1\\\\ 6&6&1&3\end{array}\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则基 $\displaystyle \alpha_1,\alpha_2,\alpha_3,\alpha_4$ 到 $\displaystyle \beta_1,\beta_2,\beta_3,\beta_4$ 的过渡矩阵为 $\displaystyle \underline{\ \ \ \ \ \ \ \ \ \ }$. (广西大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / 设
$$\begin{aligned} A=\left(\begin{array}{cccccccccccccccccccc}1&1&1&1\\\\ 1&1&-1&-1\\\\ 1&-1&1&-1\\\\ 1&-1&-1&1\end{array}\right), B=\left(\begin{array}{cccccccccccccccccccc}2&1&-1&1\\\\ 0&3&1&0\\\\ 5&3&2&1\\\\ 6&6&1&3\end{array}\right), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则
$$\begin{aligned} (\beta_1,\beta_2,\beta_3,\beta_4)=(\alpha_1,\alpha_2,\alpha_3,\alpha_4)A^{-1}B. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故过渡矩阵为 $\displaystyle A^{-1}B=\frac{1}{4}\left(\begin{array}{cccccccccccccccccccc}13&13&3&5\\\\ -9&-5&-3&-3\\\\ 1&-5&-1&-1\\\\ 3&1&-3&3\end{array}\right)$.跟锦数学微信公众号. 在线资料/公众号/
1392、 3、 设 $\displaystyle \mathscr{A}$ 是数域 $\displaystyle \mathbb{F}$ 上的线性空间 $\displaystyle V$ 上的线性变换, $\displaystyle W$ 是 $\displaystyle \mathscr{A}$ 的非平凡不变子空间, 在 $\displaystyle W$ 中取一个基 $\displaystyle \alpha_1,\cdots,\alpha_r$, 把它扩称成 $\displaystyle V$ 的一个基
$$\begin{aligned} \alpha_1,\cdots,\alpha_r,\alpha_{r+1},\cdots,\alpha_n, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
$\displaystyle \mathscr{A}$ 在 $\displaystyle \alpha_1,\cdots,\alpha_n$ 下的矩阵为 $\displaystyle \left(\begin{array}{cccccccccccccccccccc}A_1&A_3\\\\ 0&A_2\end{array}\right)$, 其中 $\displaystyle A_1$ 为 $\displaystyle r$ 阶方阵. 定义
$$\begin{aligned} \overline{\mathscr{A}}: V/W\to V/W, \alpha+W\mapsto \mathscr{A}\alpha+W. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
证明: (1)、 $\displaystyle \overline{\mathscr{A}}$ 是 $\displaystyle V/W$ 上的线性变换; (2)、 $\displaystyle A_2$ 是 $\displaystyle \overline{\mathscr{A}}$ 在基 $\displaystyle \alpha_{r+1}+W,\cdots,\alpha_n+W$ 下的矩阵. (广西大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 由题设, $\displaystyle \mathscr{A}(\alpha_1,\cdots,\alpha_r)=(\alpha_1,\cdots,\alpha_r)A_1$, 而 $\displaystyle W$ 是 $\displaystyle \mathscr{A}$ 的不变子空间. 由
$$\begin{aligned} &\alpha+W=\beta+W\Leftrightarrow \alpha-\beta\in W\\\\ \Rightarrow& \mathscr{A}(\alpha-\beta)\in W\Rightarrow \mathscr{A}\alpha+W=\mathscr{A}\beta+W \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \overline{\mathscr{A}}$ 是良定义的, 即不依赖于 $\displaystyle \alpha$ 的选取. 又由
$$\begin{aligned} &\overline{\mathscr{A}}\left[k(\alpha+W)+l(\beta+W)\right] =\overline{\mathscr{A}}\left[(k\alpha+l\beta)+W\right]\\\\ =&\mathscr{A}(k\alpha+l\beta)+W =k\mathscr{A}\alpha+l\mathscr{A}\beta+W\\\\ =&k(\mathscr{A} \alpha+W)+l(\mathscr{A}\beta+W) =k\overline{\mathscr{A}}(\alpha+W)+l\overline{\mathscr{A}}(\beta+W) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \overline{\mathscr{A}}$ 是 $\displaystyle V/W$ 上的线性变换. (2)、 对 $\displaystyle r+1\leq i\leq n$,
$$\begin{aligned} &\overline{\mathscr{A}}(\alpha_i+W)=\mathscr{A} \alpha_i+W =\left[\sum_{j=1}^r (A_3)_{ji}\alpha_j +\sum_{j=r+1}^n (A_2)_{ji}\alpha_j\right]+W\\\\ &\stackrel{1\leq j\leq r\Rightarrow \alpha_j\in W}{=} \sum_{j=r+1}^n (A_2)_{ji}\alpha_j+W =\sum_{j=r+1}^n (A_2)_{ji}(\alpha_j+W). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \overline{\mathscr{A}}$ 在基 $\displaystyle \alpha_{r+1}+W,\cdots,\alpha_n+W$ 下的矩阵为 $\displaystyle A_2$.跟锦数学微信公众号. 在线资料/公众号/
1393、 7、 设线性空间 $\displaystyle \mathbb{R}^3$ 的一组基
$$\begin{aligned} \alpha_1=(1,1,1)^\mathrm{T}, \alpha_2=(0,1,1)^\mathrm{T}, \alpha_3=(0,0,1)^\mathrm{T} \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
在线性变换 $\displaystyle \mathscr{T}$ 下的象分别为
$$\begin{aligned} \mathscr{T}(\alpha_1)=&\beta_1=(1,0,-1)^\mathrm{T},\\\\ \mathscr{T}(\alpha_2)=&\beta_2=(1,2,0)^\mathrm{T},\\\\ \mathscr{T}(\alpha_3)=&\beta_3=(-1,0,-1)^\mathrm{T}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 求线性变换 $\displaystyle \mathscr{T}$ 在基 $\displaystyle \alpha_1,\alpha_2,\alpha_3$ 下的矩阵; (2)、 求 $\displaystyle \mathscr{T}(\beta_1), \mathscr{T}(\beta_2), \mathscr{T}(\beta_3)$. (哈尔滨工程大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 由
$$\begin{aligned} (\alpha_1,\alpha_2,\alpha_3)=&(e_1,e_2,e_3)A, A=\left(\begin{array}{cccccccccccccccccccc}1&0&0\\\\ 1&1&0\\\\ 1&1&1\end{array}\right),\\\\ (\beta_1,\beta_2,\beta_3)=&(e_1,e_2,e_3)B, B=\left(\begin{array}{cccccccccccccccccccc}1&1&-1\\\\ 0&2&0\\\\ -1&0&-1\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} &\mathscr{T}(\alpha_1,\alpha_2,\alpha_3)=(\beta_1,\beta_2,\beta_3) =(e_1,e_2,e_3)B=(\alpha_1,\alpha_2,\alpha_3)A^{-1}B. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故线性变换 $\displaystyle \mathscr{T}$ 在基 $\displaystyle \alpha_1,\alpha_2,\alpha_3$ 下的矩阵为
$$\begin{aligned} A^{-1}B=\left(\begin{array}{cccccccccccccccccccc}1&1&-1\\\\-1&1&1\\\\-1&-2&-1\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 由
$$\begin{aligned} &\mathscr{T}(\beta_1,\beta_2,\beta_3)=\mathscr{T}(e_1,e_2,e_3)B =\mathscr{T}(\alpha_1,\alpha_2,\alpha_3)A^{-1}B\\\\ =&(\alpha_1,\alpha_2,\alpha_3)A^{-1}BA^{-1}B =(e_1,e_2,e_3)AA^{-1}BA^{-1}B\\\\ =&(e_1,e_2,e_3)BA^{-1}B =(e_1,e_2,e_3)\left(\begin{array}{cccccccccccccccccccc}1&4&1\\\\ -2&2&2\\\\ 0&1&2\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} \mathscr{T}(\beta_1)=(1,-2,0)^\mathrm{T}, \mathscr{T}(\beta_2)=(4,2,1)^\mathrm{T}, \mathscr{T}(\beta_3)=(1,2,2)^\mathrm{T}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1394、 9、 解答如下问题: (1)、 设 $\displaystyle V$ 为实数域上的 $\displaystyle n$ 维线性空间, $\displaystyle \mathscr{T}$ 为 $\displaystyle V$ 上的非零线性变换. 求证: 若 $\displaystyle \mathscr{T}$ 的不变子空间只有 $\displaystyle V$ 的平凡子空间, 则 $\displaystyle \mathscr{T}$ 可逆. (2)、 设 $\displaystyle V$ 为实数域上 $\displaystyle 2n+1$ 维线性空间, $\displaystyle \mathscr{T}_1$ 和 $\displaystyle \mathscr{T}_2$ 均为 $\displaystyle V$ 上的非零线性变换, 且 $\displaystyle \mathscr{T}_1\mathscr{T}_2+\mathscr{T}_2\mathscr{T}_1=\mathscr{O}$. 求证: $\displaystyle V$ 既有非零的 $\displaystyle \mathscr{T}_1$-子空间, 也有非平凡的 $\displaystyle \mathscr{T}_2$-子空间. (哈尔滨工程大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 $\displaystyle \ker\mathscr{T}$ 是 $\displaystyle \mathscr{T}$ 的不变子空间. 由 $\displaystyle \mathscr{T}$ 的不变子空间只有 $\displaystyle V$ 的平凡子空间知 $\displaystyle \ker\mathscr{T}=\left\{0\right\}\mbox{或} V$. 但 $\displaystyle \ker\mathscr{T}=V\Rightarrow \mathscr{T}=\mathscr{O}$, 与题设矛盾. 故 $\displaystyle \ker\mathscr{T}=\left\{0\right\}$, 而 $\displaystyle \mathscr{T}$ 单射. 又由
$$\begin{aligned} \dim \mathrm{im} T=\dim V-\dim \ker \mathscr{T}=\dim V\Rightarrow \mathrm{im} \mathscr{T}=V \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \mathscr{T}$ 是满射, 而是可逆的. (2)、 由奇数次实系数多项式一定有实根知 $\displaystyle \mathscr{T}_1$ 有一个实特征值 $\displaystyle \lambda$, 而
$$\begin{aligned} V_\lambda=\left\{\alpha\in V; \mathscr{T}_1\alpha=\lambda \alpha\right\}\left(\Rightarrow \dim V_\lambda\geq 1\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
就是非零的 $\displaystyle \mathscr{T}_1$-子空间. (2-1)、 若 $\displaystyle V_\lambda=V$, 则 $\displaystyle \mathscr{T}_1=\lambda\mathscr{E}$. 用反证法. 若 $\displaystyle \mathscr{T}_2$ 只有平凡的不变子空间, 则由第 1 步知 $\displaystyle \mathscr{T}_2$ 可逆,
$$\begin{aligned} &\forall\ 0\neq\alpha\in V, \mathscr{T}_2\alpha\neq 0\\\\ \Rightarrow&0=(\mathscr{T}_1\mathscr{T}_2+\mathscr{T}_2\mathscr{T}_1)\alpha =\lambda \mathscr{T}_2\alpha+\mathscr{T}_2(\lambda\alpha)=2\lambda \mathscr{T}_2\alpha\\\\ \Rightarrow&\lambda=0\Rightarrow V=V_0\Rightarrow \mathscr{T}_1=\mathscr{O},\mbox{与题设矛盾}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \mathscr{T}_2$ 确有非平凡的不变子空间. (2-2)、 若 $\displaystyle V_\lambda\subsetneq V$, 当 $\displaystyle \lambda=0$ 时,
$$\begin{aligned} \alpha\in V_0\Rightarrow \mathscr{T}_1\mathscr{T}_2\alpha=-\mathscr{T}_2\mathscr{T}_1\alpha=-\mathscr{T}_20=0=0\mathscr{T}_2\alpha \Rightarrow \mathscr{T}_2\alpha\in V_0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
蕴含 $\displaystyle V_\lambda$ 是 $\displaystyle \mathscr{T}_2$ 的非平凡不变子空间. 当 $\displaystyle \lambda\neq 0$ 时, 还用反证法. 若 $\displaystyle \mathscr{T}_2$ 只有平凡的不变子空间, 则由第 1 步知 $\displaystyle \mathscr{T}_2$ 可逆,
$$\begin{aligned} &\mathscr{T}_1\mathscr{T}_2+\mathscr{T}_2\mathscr{T}_1\Rightarrow \mathscr{T}_2^{-1}\mathscr{T}_1\mathscr{T}_2=-\mathscr{T}_1\\\\ \Rightarrow& \mathscr{T}_2^{-1}(\lambda \mathscr{E}-\mathscr{T}_1)\mathscr{T}_2=\lambda \mathscr{E}+\mathscr{T}_1 =-(-\lambda\mathscr{E}-\mathscr{T}_1). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle -\lambda$ 于是 $\displaystyle \mathscr{T}_1$ 的特征值. 由
$$\begin{aligned} \alpha\in V_\lambda\Rightarrow&\mathscr{T}_1\mathscr{T}_2\alpha=-\mathscr{T}_2\mathscr{T}_1\alpha=(-\lambda)\mathscr{T}_2\alpha \Rightarrow \mathscr{T}_2\alpha\in V_{-\lambda} \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \mathscr{T}_2|_{V_\lambda}: V_\lambda\to V_{-\lambda}$. 由 $\displaystyle \mathscr{T}_2$ 是单射知 $\displaystyle \mathscr{T}_2|_{V_\lambda}$ 也是单射. 又由
$$\begin{aligned} \alpha\in V_{-\lambda}\Rightarrow&\mathscr{T}_1\alpha=-\lambda \mathscr{T}_1\alpha\\\\ \Rightarrow&\mathscr{T}_1\mathscr{T}_2^{-1}\alpha =-\mathscr{T}_2^{-1}\mathscr{T}_1\alpha=\lambda \mathscr{T}_2^{-1}\alpha \Rightarrow \mathscr{T}_2^{-1}\alpha\in V_\lambda \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \mathscr{T}_2|_{V_\lambda}$ 是满射, $\displaystyle \mathscr{T}_2|_{V_\lambda}: V_\lambda\to V_{-\lambda}$ 是可逆的. 又由
$$\begin{aligned} \alpha\in V_\lambda\cap V_{-\lambda}\Rightarrow -\lambda \alpha=\mathscr{T}_1\alpha=\lambda \alpha\Rightarrow 2\lambda\alpha=0\stackrel{\lambda\neq 0}{\Rightarrow}\alpha=0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle U=V_1\oplus V_2$ 是直和, $\displaystyle \dim U=2\dim V_\lambda$ 是偶数 $\displaystyle < 2n+1$, 而 $\displaystyle U$ 是 $\displaystyle V$ 的非平凡子空间, 且是 $\displaystyle \mathscr{T}_2$ 的不变子空间. 这与已设矛盾. 故 $\displaystyle \mathscr{T}_2$ 确有非平凡的不变子空间.跟锦数学微信公众号. 在线资料/公众号/
1395、 1、 若 $\displaystyle \alpha_1,\cdots,\alpha_m$ 线性无关, 且 $\displaystyle \beta_1,\cdots,\beta_s$ 与 $\displaystyle \alpha_1,\cdots,\alpha_m$ 满足如下线性关系:
$$\begin{aligned} \left\{\begin{array}{rrrrrrrrrrrrrrrr} \beta_1=&a_{11}\alpha_1+a_{21}\alpha_2+\cdots+a_{m1}\alpha_m,\\\\ \beta_2=&a_{12}\alpha_1+a_{22}\alpha_2+\cdots+a_{m2}\alpha_m,\\\\ \cdot=&\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots,\\\\ \beta_s=&a_{1s}\alpha_1+a_{2s}\alpha_2+\cdots+a_{ms}\alpha_m. \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
有 $\displaystyle m\times s$ 矩阵 $\displaystyle A=\left(\begin{array}{cccccccccccccccccccc}a_{11}&a_{12}&\cdots&a_{1s}\\\\ a_{21}&a_{22}&\cdots&a_{2s}\\\\ \vdots&\vdots&&\vdots\\\\ a_{m1}&a_{m2}&\cdots&a_{ms}\end{array}\right)$. 判断以下命题正误, 并证明: (1)、 当 $\displaystyle \beta_1,\cdots,\beta_s$ 线性相关时, 则有 $\displaystyle m > s$; (2)、 $\displaystyle \beta_1,\cdots,\beta_s$ 线性无关的充要条件是 $\displaystyle \mathrm{rank} A=s$. (哈尔滨工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 错误. 比如 $\displaystyle \beta_i=\alpha_1, 1\leq i\leq m+1$, 则 $\displaystyle A=\left(\begin{array}{cccccccccccccccccccc}1&\cdots&1\\\\ 0&\cdots&0\\\\ \vdots&&\vdots\\\\ 0&\cdots&0\end{array}\right)_{m\times (m+1)}$. 此时, $\displaystyle m < s=m+1$. (2)、 正确.
$$\begin{aligned} &\quad \ \beta_1,\cdots,\beta_s\mbox{线性无关} \Leftrightarrow \sum_{i=1}^s x_i\beta_i=0\mbox{只有零解}\\\\ &\Leftrightarrow\left(\beta_1,\cdots,\beta_s\right)X=0\mbox{只有零解} \Leftrightarrow \left(\begin{array}{cccccccccccccccccccc}\alpha_1,\cdots,\alpha_m\end{array}\right)AX=0\mbox{只有零解}\\\\ &\stackrel{\alpha_1,\cdots,\alpha_m\mbox{线性无关}}{\Leftrightarrow}AX=0\mbox{只有零解} \Leftrightarrow \mathrm{rank} A=s. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1396、 8、 设 $\displaystyle \mathbb{P}^{n\times n}$ 表示数域 $\displaystyle \mathbb{P}$ 上的 $\displaystyle n$ 阶方阵的集合, $\displaystyle A\in \mathbb{P}^{n\times n}$. 记
$$\begin{aligned} C(A)=\left\{B\in \mathbb{P}^{n\times n}; AB=BA\right\}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
证明: (1)、 $\displaystyle C(A)$ 是 $\displaystyle \mathbb{P}^{n\times n}$ 的线性子空间; (2)、 若 $\displaystyle A=\left(\begin{array}{cccccccccccccccccccc}1&0&0\\\\ 1&1&0\\\\ 0&0&2\end{array}\right)$, 求 $\displaystyle C(A)$ 的维数和一组基. (哈尔滨工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 设 $\displaystyle X_1,X_2\in C(A)$,$k\in\mathbb{P}$, 则
$$\begin{aligned} (kX_1+X_2)A=kX_1A+X_2A=kAX_1+AX_2=A(kX_1+X_2). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
而 $\displaystyle kX_1+X_2\in C(A)$. 于是 $\displaystyle C(A)$ 是 $\displaystyle \mathbb{P}^{n\times n}$ 的一个线性子空间. (2)、 由
$$\begin{aligned} &B\in C(A)\Leftrightarrow AB=BA\Leftrightarrow (A-E)B=B(A-E)\\\\ \Leftrightarrow& \left(\begin{array}{cccccccccccccccccccc}-b_{12}&0&-b_{13}\\\\ b_{11}-b_{22}&b_{12}&b_{13}-b_{23}\\\\ b_{31}-b_{32}&b_{32}&0\end{array}\right)=0\\\\ \Leftrightarrow&b_{12}=b_{13}=b_{23}=b_{31}=b_{32}=0, b_{11}=b_{22} \Leftrightarrow B=\left(\begin{array}{cccccccccccccccccccc}a&&\\\\ b&a&\\\\ &&c\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle C(A)$ 有一组基 $\displaystyle E_{11}+E_{12}, E_{21}, E_{33}$. 从而 $\displaystyle \dim C(A)=3$.跟锦数学微信公众号. 在线资料/公众号/
1397、 1、 向量组
$$\begin{aligned} &\alpha_1=(1,-1,2,-1)^\mathrm{T}, \alpha_2=(-3,4,-1,2)^\mathrm{T}, \alpha_3=(4,-5,3,-3)^\mathrm{T},\\\\ &\alpha_4=(-1,\lambda,3,0)^\mathrm{T}, \beta=(0,\kappa,5,-1)^\mathrm{T}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
且 $\displaystyle \lambda,\kappa\in\mathbb{R}, \mathbb{R}$ 表实数域. (1)、 讨论 $\displaystyle \lambda,\kappa$ 取何值时, $\displaystyle \beta$ 不能由 $\displaystyle \alpha_1,\alpha_2,\alpha_3,\alpha_4$ 线性表示; (2)、 讨论 $\displaystyle \lambda,\kappa$ 取何值时, $\displaystyle \beta$ 可由 $\displaystyle \alpha_1,\alpha_2,\alpha_3,\alpha_4$ 线性表示, 并写出表达式. (合肥工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 /
$$\begin{aligned} &(A,\beta)\equiv (\alpha_1,\alpha_2,\alpha_3,\alpha_4,\beta) =\left(\begin{array}{cccccccccccccccccccc}1&-3&4&-1&0\\\\ -1&4&-5&\lambda&\kappa\\\\ 2&-1&3&3&5\\\\ -1&2&-3&0&-1\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&-3&4&-1&0\\\\ 0&1&-1&\lambda-1&\kappa\\\\ 0&5&-5&5&5\\\\ 0&-1&1&-1&-1\end{array}\right)\to\left(\begin{array}{cccccccccccccccccccc}1&0&1&2&3\\\\ 0&0&0&\lambda-2&\kappa-1\\\\ 0&0&0&0&0\\\\ 0&1&-1&1&1\end{array}\right)\\\\ \to&\left(\begin{array}{cccccccccccccccccccc}1&0&1&2&3\\\\ 0&1&-1&1&1\\\\ 0&0&0&\lambda-2&\kappa-1\\\\ 0&0&0&0&0\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 (1)、 当 $\displaystyle \lambda=2, \kappa\neq 1$ 时, $\displaystyle \mathrm{rank} A=2 < 3=\mathrm{rank}(A,\beta)$, $\displaystyle Ax=\beta$ 无解, $\displaystyle \beta$ 不能由 $\displaystyle \alpha_1,\alpha_2,\alpha_3,\alpha_4$ 线性表示. (2)、 当 $\displaystyle \lambda\neq 2$ 时,
$$\begin{aligned} (A,\beta)\to \left(\begin{array}{cccccccccccccccccccc}1&0&1&0&-\frac{4+2\kappa-3\lambda}{\lambda-2}\\\\ 0&1&-1&0&\frac{\lambda-\kappa-1}{\lambda-2}\\\\ 0&0&0&1&\frac{\kappa-1}{\lambda-2}\\\\ 0&0&0&0&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
取 $\displaystyle x_3$ 为自由变量知 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} k(-1,1,1,0)^\mathrm{T}+\left(\begin{array}{cccccccccccccccccccc}-\frac{4+2\kappa-3\lambda}{\lambda-2},\frac{\lambda-\kappa-1}{\lambda-2}, 0,\frac{\kappa-1}{\lambda-2}\end{array}\right)^\mathrm{T},\quad \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
此时, $\displaystyle \beta$ 可由 $\displaystyle \alpha_1,\alpha_2,\alpha_3,\alpha_4$ 线性表示, 且
$$\begin{aligned} \beta=&\left(-k-\frac{4+2\kappa-3\lambda}{\lambda-2}\right)\alpha_1+\left(k+\frac{\lambda-\kappa-1}{\lambda-2}\right)\alpha_2\\\\ &+k\alpha_3+\frac{\kappa-1}{\lambda-2}\alpha_4,\quad \forall\ k. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(3)、 当 $\displaystyle \lambda=2, \kappa=1$ 时,
$$\begin{aligned} (A,\beta)\to\left(\begin{array}{cccccccccccccccccccc}1&0&1&2&3\\\\ 0&1&-1&1&1\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
而 $\displaystyle Ax=\beta$ 的通解为
$$\begin{aligned} k(-1,1,1,0)^\mathrm{T}+l(-2,-1,0,0)^\mathrm{T}+(3,1,0,0)^\mathrm{T}, \quad \forall\ k,l. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
此时, $\displaystyle \beta$ 可由 $\displaystyle \alpha_1,\alpha_2,\alpha_3,\alpha_4$ 线性表示, 且
$$\begin{aligned} \beta=(-k-2l+3)\alpha_1+(k-l+1)\alpha_2+k\alpha_3+l\alpha_4,\quad \forall\ k,l. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1398、 6、 设线性空间
$$\begin{aligned} V=\left\{\left(\begin{array}{cccccccccccccccccccc}x_1&x_2\\\\ x_3&x_4\end{array}\right); x_1+x_4=0, x_i\in\mathbb{R}, i=1,2,3,4\right\}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 给定 $\displaystyle V$ 上的变换 $\displaystyle \sigma$:
$$\begin{aligned} \sigma(X)=X+X^\mathrm{T}, \quad \forall\ X\in V. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
验证 $\displaystyle \sigma$ 为 $\displaystyle V$ 上的线性变换. (2)、 求 $\displaystyle V$ 的一组基及 $\displaystyle \sigma$ 在该基下的矩阵. (3)、 求 $\displaystyle \sigma$ 的全体特征值与特征向量. (合肥工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 对 $\displaystyle \forall\ k,l\in\mathbb{R}, X,Y\in V$,
$$\begin{aligned} &\sigma(kX+lY)=(kX+lY)+(kX+lY)^\mathrm{T}\\\\ =&k(X+X^\mathrm{T})+l(Y+Y^\mathrm{T})=k\sigma(X)+l\sigma(Y). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \sigma$ 是 $\displaystyle V$ 上的线性变换. (2)、 求解关于 $\displaystyle x_1,x_2,x_3,x_4$ 的线性方程组 $\displaystyle x_1+x_4=0$. 取 $\displaystyle x_2,x_3,x_4$ 为自由变量即知 $\displaystyle V$ 有一组基
$$\begin{aligned} \varepsilon_1=\left(\begin{array}{cccccccccccccccccccc}0&1\\\\ 0&0\end{array}\right), \varepsilon_2=\left(\begin{array}{cccccccccccccccccccc}0&0\\\\ 1&0\end{array}\right), \varepsilon_3=\left(\begin{array}{cccccccccccccccccccc}-1&0\\\\ 0&1\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
算出
$$\begin{aligned} &\sigma(\varepsilon_1)=\left(\begin{array}{cccccccccccccccccccc}0&1\\\\ 1&0\end{array}\right)=\varepsilon_1+\varepsilon_2, \sigma(\varepsilon_2)=\left(\begin{array}{cccccccccccccccccccc}0&1\\\\ 1&0\end{array}\right)=\varepsilon_2+\varepsilon_2,\\\\ &\sigma(\varepsilon_3)=\left(\begin{array}{cccccccccccccccccccc}-2&0\\\\ 0&2\end{array}\right)=2\varepsilon_3 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
即知
$$\begin{aligned} \sigma(\varepsilon_1,\varepsilon_2,\varepsilon_3)=(\varepsilon_1,\varepsilon_2,\varepsilon_3)A, A=\left(\begin{array}{cccccccccccccccccccc}1&1&0\\\\ 1&1&0\\\\ 0&0&2\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(3)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,0$. 由
$$\begin{aligned} 2E-A\to\left(\begin{array}{cccccccccccccccccccc}1&-1&0\\\\ 0&0&0\\\\ 0&0&0\end{array}\right), 0E-A\to \left(\begin{array}{cccccccccccccccccccc}1&1&0\\\\ 0&0&1\\\\ 0&0&0\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle A$ 的属于特征值 $\displaystyle 2,0$ 的特征向量分别为
$$\begin{aligned} \xi_1=\left(\begin{array}{cccccccccccccccccccc}1\\\\1\\\\0\end{array}\right), \xi_2=\left(\begin{array}{cccccccccccccccccccc}0\\\\0\\\\1\end{array}\right);\quad \xi_3=\left(\begin{array}{cccccccccccccccccccc}-1\\\\1\\\\0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \sigma$ 的属于特征值 $\displaystyle 2,0$ 的特征向量分别为
$$\begin{aligned} &(\varepsilon_1,\varepsilon_2,\varepsilon_3)\xi_1=\left(\begin{array}{cccccccccccccccccccc}0&1\\\\ 1&0\end{array}\right), (\varepsilon_1,\varepsilon_2,\varepsilon_3)\xi_2=\left(\begin{array}{cccccccccccccccccccc}-1&0\\\\ 0&1\end{array}\right);\\\\ &(\varepsilon_1,\varepsilon_2,\varepsilon_3)\xi_3=\left(\begin{array}{cccccccccccccccccccc}0&-1\\\\ 1&0\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1399、 7、 在 $\displaystyle V=\mathbb{P}[x]_{n+1}$ 上定义线性变换 $\displaystyle \sigma$:
$$\begin{aligned} \sigma\left(f(x)\right)=xf'(x)-f(x),\quad \forall\ f(x)\in V. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 求 $\displaystyle \sigma$ 的核 $\displaystyle \sigma^{-1}(0)$ 与值域 $\displaystyle \sigma(V)$. (2)、 问 $\displaystyle V=\sigma^{-1}(0)\oplus \sigma(V)$ 是否成立并说明理由. (合肥工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 由
$$\begin{aligned} f\in \sigma^{-1}(0)&\Leftrightarrow xf'(x)=f(x)\\\\ &\Leftrightarrow f\equiv 0\mbox{或}\frac{\mathrm{ d} f}{f}=\frac{1}{x}\\\\ &\Leftrightarrow f(x)\equiv cx \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} \ker\sigma=L(x). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
又由
$$\begin{aligned} \sigma x^k&=x\cdot kx^{k-1}-x^k\\\\ &=(k-1)x^k\left(0\leq k\leq n-1\right)\\\\ &=\left\{\begin{array}{llllllllllll} (k-1)x^k,&k\neq 1\\\\ 0,&k=1 \end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} \mathrm{im}\sigma&=L(\sigma (1),\sigma (x),\cdots,\sigma (x^{n-1}))\\\\ &=L(-1,0,x^2,\cdots,(n-1)x^{n-1})\\\\ &=L(1,x^2,\cdots,x^{n-1}). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、
$$\begin{aligned} V&=\mathbb{P}[x]_n=L(1,x,x^2,\cdots,x^{n-1})\\\\ &=L(x)\oplus L(1,x^2,\cdots,x^{n-1})=\ker\sigma\oplus \mathrm{im}\sigma. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1400、 9、 设 $\displaystyle \mathbb{R}^3$ 上的线性变换 $\displaystyle \sigma$ 定义为:
$$\begin{aligned} \sigma(x,y,z)^\mathrm{T}=(2x+y,y-z,2y+4z)^\mathrm{T}, \forall\ (x,y,z)^\mathrm{T}\in\mathbb{R}^3. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(1)、 求 $\displaystyle \sigma$ 所有的特征子空间及它们的和 $\displaystyle U$. (2)、 问 $\displaystyle \sigma$ 能否对角化 (即能否找到 $\displaystyle \mathbb{R}^3$ 中的一组基, 使 $\displaystyle \sigma$ 在该基下矩阵为对角阵), 并说明理由. (3)、 证明: $\displaystyle \sigma|_U$ 可对角化, 并求出 $\displaystyle U$ 的一组基, 使 $\displaystyle \sigma|_U$ 在该基下的矩阵为对角阵并写出此对角阵. (合肥工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 由
$$\begin{aligned} \sigma\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0\end{array}\right)=\left(\begin{array}{cccccccccccccccccccc}2\\\\0\\\\0\end{array}\right), \sigma\left(\begin{array}{cccccccccccccccccccc}0\\\\1\\\\0\end{array}\right)=\left(\begin{array}{cccccccccccccccccccc}1\\\\1\\\\2\end{array}\right), \sigma\left(\begin{array}{cccccccccccccccccccc}0\\\\0\\\\1\end{array}\right)=\left(\begin{array}{cccccccccccccccccccc}0\\\\-1\\\\4\end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知
$$\begin{aligned} \sigma(e_1,e_2,e_3)=(e_1,e_2,e_3)A, A=\left(\begin{array}{cccccccccccccccccccc}2&1&0\\\\ 0&1&-1\\\\ 0&2&4\end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
易知 $\displaystyle A$ 的特征值为 $\displaystyle 3,2,2$. 由
$$\begin{aligned} 3E-A\to\left(\begin{array}{cccccccccccccccccccc} 1&0&\frac{1}{2}\\\\ 0&1&\frac{1}{2}\\\\ 0&0&0 \end{array}\right), 2E-A\to\left(\begin{array}{cccccccccccccccccccc} 0&1&0\\\\ 0&0&1\\\\ 0&0&0 \end{array}\right) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle A$ 的属于特征值 $\displaystyle 3,2$ 的特征向量分别为
$$\begin{aligned} \xi_1=\left(\begin{array}{cccccccccccccccccccc}-1\\\\-1\\\\2 \end{array}\right);\quad \xi_2=\left(\begin{array}{cccccccccccccccccccc}1\\\\0\\\\0 \end{array}\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \sigma$ 的特征子空间为
$$\begin{aligned} V_3=L(\xi_1), V_2=L(\xi_2)\Rightarrow U=V_3+V_2=L(\xi_1,\xi_2). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 由 $\displaystyle \mathrm{rank}(2E-A)=2$ 知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 的严格上三角部分有一个 $\displaystyle 1$ (若不然, $\displaystyle \mathrm{rank}(2E-J)=1$! 矛盾). 故在 $\displaystyle A$ 在 $\displaystyle \mathbb{C}$ 中不可对角化, 即不相似于 $\displaystyle \varLambda=\mathrm{diag}(3,2,2)$. 由 $\displaystyle A\in\mathbb{R}^{3\times 3}$ 知 $\displaystyle A$ 在实数域上中不相似于 $\displaystyle \varLambda$, 不可对角化, 从而 $\displaystyle \sigma$ 不可对角化. 另证如下. 用反证法. 若 $\displaystyle \sigma$ 可对角化, 则 $\displaystyle \sigma$ 有 $\displaystyle 3$ 个线性无关的特征向量. 这与第 1 步的结果’$\sigma$ 只有两个线性无关的特征向量'矛盾. (3)、 由 $\displaystyle \sigma(\xi_1)=3\xi_1, \sigma(\xi_2)=2\xi_2$ 知在 $\displaystyle U$ 的基 $\displaystyle \xi_1,\xi_2$ 下, $\displaystyle \sigma$ 的矩阵为对角矩阵, 即
$$\begin{aligned} \sigma(\xi_1,\xi_2)=(\xi_1,\xi_2)\mathrm{diag}(3,2). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
1401、 11、 $\displaystyle W$ 为 $\displaystyle n$ 维线性空间 $\displaystyle V$ 的子空间. 证明: 存在线性变换 $\displaystyle \sigma_1, \sigma_2\in L(V)$, 使得
$$\begin{aligned} \sigma_1(V)=W=\sigma_2^{-1}(0). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(合肥工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / 设 $\displaystyle \dim W=r$, 取定 $\displaystyle W$ 的一组基 $\displaystyle \varepsilon_1,\cdots,\varepsilon_r$, 将其扩充为 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon_1,\cdots,\varepsilon_n$. 再取 $\displaystyle \sigma_1,\sigma_2\in L(V)$ 使得
$$\begin{aligned} \sigma_1(\varepsilon_i)=\left\{\begin{array}{llllllllllll}\varepsilon_i,&1\leq i\leq r,\\\\ 0,&r+1\leq i\leq n;\end{array}\right.\quad \sigma_2(\varepsilon_i)=\left\{\begin{array}{llllllllllll}0,&1\leq i\leq r,\\\\ \varepsilon_i,&r+1\leq i\leq n.\end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则由
$$\begin{aligned} &\alpha\in \mathrm{im}\sigma_1\Leftrightarrow\exists\ \beta=\sum_{i=1}^n x_i\varepsilon_i,\mathrm{ s.t.} \alpha=\sigma(\beta)\\\\ \Leftrightarrow&\alpha=\sigma\left(\sum_{i=1}^n x_i\varepsilon_i\right) =\sum_{i=1}^n x_i\sigma(\varepsilon_i) \sum_{i=1}^r x_i\varepsilon_i\Leftrightarrow \alpha\in W, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
$$\begin{aligned} &\alpha=\sum_{i=1}^n x_i\varepsilon_i\in \ker\sigma_2\Leftrightarrow 0=\sigma_2(\alpha)=\sum_{i=1}^n x_i\sigma(\varepsilon_i)=\sum_{i=r+1}^n x_i\varepsilon_i\\\\ \Leftrightarrow&x_{r+1}=\cdots=x_n=0\Leftrightarrow \alpha=\sum_{i=1}^r x_i\varepsilon_i\in W \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \mathrm{im}\sigma_1=W=\ker \sigma_2$.跟锦数学微信公众号. 在线资料/公众号/
1402、 3、 向量组 $\displaystyle \alpha_1,\cdots,\alpha_r$ 中每个向量都可由 $\displaystyle \beta_1,\cdots,\beta_s$ 线性表示, 且 $\displaystyle r > s$. 证明: $\displaystyle \alpha_1,\cdots,\alpha_r$ 线性相关. (河北工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / 由题设, 存在矩阵 $\displaystyle A$, 使得
$$\begin{aligned} &(\alpha_1,\cdots,\alpha_r)=(\beta_1,\cdots,\beta_s)A\\\\ \Rightarrow&\mathrm{rank}(\alpha_1,\cdots,\alpha_r)\leq \mathrm{rank}(\beta_1,\cdots,\beta_s)\leq s < r. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \alpha_1,\cdots,\alpha_r$ 线性相关.跟锦数学微信公众号. 在线资料/公众号/
1403、 6、 设 $\displaystyle \mathscr{T}_1,\mathscr{T}_2$ 是 $\displaystyle n$ 维线性空间 $\displaystyle V$ 中的两个线性变换. 证明: $\displaystyle \mathrm{im} \mathscr{T}_2\subset \mathrm{im}\mathscr{T}_1$ 的充要条件是存在线性变换 $\displaystyle \mathscr{T}$, 使得 $\displaystyle \mathscr{T}_2=\mathscr{T}_1\mathscr{T}$. (河北工业大学2023年高等代数考研试题) [线性空间与线性变换 ]
纸质资料/答疑/pdf1/pdf2 / (1)、 $\displaystyle \Leftarrow$: 由 $\displaystyle \mathscr{T}_2=\mathscr{T}_1\mathscr{T}$ 知 $\displaystyle \forall\ \alpha\in V$,
$$\begin{aligned} \mathscr{T}_2\alpha=\mathscr{T}_1\mathscr{T}\alpha\in \mathrm{im}\mathscr{T}_1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \mathrm{im} \mathscr{T}_2\subset \mathrm{im}\mathscr{T}_1$. (2)、 $\displaystyle \Rightarrow$: 设 $\displaystyle \alpha_1,\cdots,\alpha_r$ 是 $\displaystyle \mathrm{im}\mathscr{T}_2$ 的一组基, 则
$$\begin{aligned} \exists\ \gamma_i,\mathrm{ s.t.} \alpha_i=\mathscr{T}_2\gamma_i. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
由
$$\begin{aligned} \sum_{i=1}^r x_i\gamma_i=0\Rightarrow&0=\mathscr{T}_2\left(\sum_{i=1}^r x_i\gamma_i\right) =\sum_{i=1}^r x_i\alpha_i=0\Rightarrow x_i=0, \forall\ 1\leq i\leq r \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \gamma_1,\cdots,\gamma_r$ 线性无关. 又由 $\displaystyle \dim \ker \mathscr{T}_2+\dim\mathrm{im}\mathscr{T}_2=n$ 知可取 $\displaystyle \ker\mathscr{T}_2$ 的一组基 $\displaystyle \gamma_{r+1},\cdots,\gamma_n$, 则由
$$\begin{aligned} &\sum_{i=1}^n x_i\gamma_i=0\Rightarrow 0=\sum_{i=1}^r x_i\mathscr{T}_2\gamma_i =\sum_{i=1}^r x_i\alpha_i=0\Rightarrow x_i=0, \forall\ 1\leq i\leq r\\\\ \Rightarrow& \sum_{i=r+1}^n x_i\gamma_i=0\Rightarrow x_i=0, \forall\ r+1\leq i\leq n \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \gamma_1,\cdots,\gamma_n$ 线性无关, 是 $\displaystyle V$ 的一组基. 此外, 由
$$\begin{aligned} 1\leq i\leq r\Rightarrow \alpha_i\in\mathrm{im}\mathscr{T}_2\subset \mathrm{im}\mathscr{T}_1\Rightarrow \exists\ \beta_i,\mathrm{ s.t.} \alpha_i=\mathscr{T}_1\beta_i \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知取 $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr{T}$ 使得
$$\begin{aligned} \mathscr{T}\gamma_i=\left\{\begin{array}{llllllllllll}\beta_i, &1\leq i\leq r,\\\\ 0,&r+1\leq i\leq n\end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
后,
$$\begin{aligned} \mathscr{T}_1\mathscr{T} \gamma_i=\left\{\begin{array}{llllllllllll}\mathscr{T}_1\beta_i=\alpha_i=\mathscr{T}_2\gamma_i, &1\leq i\leq \gamma,\\\\ \mathscr{T}_10=0=\mathscr{T}_2\gamma_i, &r+1\leq i\leq n.\end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\displaystyle \mathscr{T}_1\mathscr{T}=\mathscr{T}_2$.
$$\begin{aligned} \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. 在线资料/公众号/
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发表于 2023-3-5 13:20:51