## 张祖锦2023年数学专业真题分类70天之第61天
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \dim V\_2-\dim(V\_1\cap V\_2)=\dim(V\_1+V\_2)-\dim V\_1=1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \varphi$ 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](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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\}
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
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](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 13:20:51
