## 张祖锦2023年数学专业真题分类70天之第65天
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1473、 6、 已知 $\displaystyle M\_3(\mathbb\{C\})$ 是三阶复方阵全体构成的线性空间, $\displaystyle A\in M\_3(\mathbb\{C\})$. (1)、 证明所有与 $\displaystyle A$ 可交换的矩阵构成一个线性子空间, 记为 $\displaystyle C(A)$. (2)、 已知 $\displaystyle A^3=I$, 求 $\displaystyle C(A)$ 的维数. (天津大学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 $\displaystyle \forall\ B,C\in C(A), \forall\ k,l\in\mathbb\{C\}$,
\begin\{aligned\} &A(kB+lC)=kAB+lAC=kBA+lCA=(kB+lC)A\\\\ \Rightarrow& kB+lC\in C(A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle C(A)$ 确为 $\displaystyle M\_3(\mathbb\{C\})$ 的子空间. (2)、 由 $\displaystyle A^3=I$ 知 $\displaystyle A$ 的最小多项式 $\displaystyle \lambda^3-1$ 没有重根, 而对角化. 相同的按一个计算, (2-1)、 如果 $\displaystyle A$ 只有一个特征值, 则 $\displaystyle A$ 就是数量矩阵, $\displaystyle C(A)=M\_3(\mathbb\{C\})\Rightarrow \dim C(A)=9$. (2-2)、 如果 $\displaystyle A$ 有两个特征值, 则注意到与准对角矩阵可交换的还是与之同类型的准对角矩阵知 $\displaystyle \dim C(A)=1+2^2=5$. (2-3)、 如果 $\displaystyle A$ 有三个特征值, 则 $\displaystyle C(A)$ 就是对角矩阵, $\displaystyle \dim C(A)=3$.跟锦数学微信公众号. [在线资料](
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1474、 1、 (15 分) 设 $\displaystyle M\_\{2\times 2\}(\mathbb\{C\})$ 是全体 $\displaystyle 2$ 阶复矩阵组成的线性空间, 线性变换
\begin\{aligned\} \mathscr\{A\} X=MXN, M=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ 1&1\end\{array\}\right), N=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1\\\\ -1&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\}
求一组基使得 $\displaystyle \mathscr\{A\}$ 在该基下的矩阵为 Jordan 阵. (武汉大学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) / 算出
\begin\{aligned\} &\mathscr\{A\}(E\_\{11\})=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1\\\\ 1&-1\end\{array\}\right), \mathscr\{A\}(E\_\{12\})=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&1\\\\-1&1\end\{array\}\right),\\\\ &\mathscr\{A\}(E\_\{21\})=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0\\\\ 1&-1\end\{array\}\right), \mathscr\{A\}(E\_\{22\})=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0\\\\ -1&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\}
我们知
\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\}1&-1&0&0\\\\ -1&1&0&0\\\\ 1&-1&1&-1\\\\ -1&1&-1&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\}
易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,0,0$. 由
\begin\{aligned\} 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&0&0\\\\ 0&1&0&0&\\\\ 0&0&1&1\\\\ 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 $\displaystyle 2$ 的特征向量为
\begin\{aligned\} \xi\_1=(0,0,-1,1)^\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 $\displaystyle \mathrm\{rank\}(2E-A)=3$. 由
\begin\{aligned\} 0 E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&-1&0&0\\\\ 0&0&1&-1\\\\ 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 $\displaystyle 0$ 的特征向量为
\begin\{aligned\} \xi\_3=(1,1,0,0)^\mathrm\{T\}, \xi\_4=(0,0,1,1)^\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 $\displaystyle \mathrm\{rank\}(0E-A)=2$. 故 $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&&\\\\ &2&&\\\\ &&0&\\\\ &&&0\end\{array\}\right)$. 设可逆矩阵 $\displaystyle P=(\xi\_1,\xi\_2,\xi\_3,\xi\_4)$ 使得 $\displaystyle P^\{-1\}AP=J$, 则由
\begin\{aligned\} (A-2E)\xi\_2=\xi\_1, (A-2E,\xi\_1)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&-\frac\{1\}\{2\}\\\\ 0&1&0&0&\frac\{1\}\{2\}\\\\ 0&0&1&1&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\}
知可取 $\displaystyle \xi\_2=\left(-\frac\{1\}\{2\},\frac\{1\}\{2\},0,0\right)^\mathrm\{T\}$. 故 $\displaystyle \sigma$ 在 $\displaystyle M\_\{2\times 2\}(\mathbb\{C\})$ 的基
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}0&0\\\\ -1&1\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{2\}&\frac\{1\}\{2\}\\\\ 0&0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ 0&0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}0&0\\\\ 1&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 Jordan 标准形 $\displaystyle J$.跟锦数学微信公众号. [在线资料](
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---
1475、 3、 设 $\displaystyle M\_\{n\times n\}(\mathbb\{C\})$ 是 $\displaystyle n\times n$ 复矩阵全体在通常的运算下所构成的复数域 $\displaystyle \mathbb\{C\}$ 上的线性空间,
\begin\{aligned\} F=\left(\begin\{array\}\{cccccccccccccccccccc\} 0&&&&-a\_n\\\\ 1&0&&&-a\_\{n-1\}\\\\ &\ddots&\ddots&&\vdots\\\\ &&1&0&-a\_2\\\\ &&&1&-a\_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\}
(1)、 设
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{11\}&a\_\{12\}&\cdots&a\_\{1n\}\\\\ a\_\{21\}&a\_\{22\}&\cdots&a\_\{2n\}\\\\ \vdots&\vdots&\ddots&\vdots\\\\ a\_\{n1\}&a\_\{n2\}&\cdots&a\_\{nn\}\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\}
若 $\displaystyle AF=FA$, 证明:
\begin\{aligned\} A=a\_\{n1\}F^\{n-1\}+a\_\{n-1,1\}F^\{n-2\}+\cdots+a\_\{21\}F+a\_\{11\}E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 求 $\displaystyle M\_\{n\times n\}(\mathbb\{C\})$ 的子空间
\begin\{aligned\} C(F)=\left\\{X\in M\_\{n\times n\}(\mathbb\{C\}); FX=XF\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的维数. (武汉大学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
\begin\{aligned\} \beta&=(-a\_n,-a\_\{n-1\},\cdots,-a\_1)^\mathrm\{T\},\\\\ e\_i&=(\underbrace\{0,\cdots,0,1\}\_i,0,\cdots,0)^\mathrm\{T\},\\\\ M&=a\_\{n1\}F^\{n-1\}+a\_\{n-1,1\}F^\{n-2\}+\cdots+a\_\{21\}F+a\_\{11\}E, \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\} F=(e\_2,\cdots,e\_n,\beta), MF^i=F^iM\left(\forall\ i\in\mathbb\{N\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为证
\begin\{aligned\} A=M\Leftrightarrow Ae\_i=Me\_i\left(1\leq i\leq n\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们一个一个去验算 (从第一个就可看出 $\displaystyle M$ 的形状):
\begin\{aligned\} Ae\_1&=\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{11\}\\\\a\_\{21\}\\\\\vdots\\\\a\_\{n1\}\end\{array\}\right)=a\_\{11\}e\_1+a\_\{21\}e\_2+\cdots+a\_\{n1\}e\_n\\\\ &=a\_\{11\}Ee\_1+a\_\{21\}Fe\_1+\cdots+a\_\{n1\}F^\{n-1\}e\_1\\\\ &=Me\_1,\\\\ Ae\_2&=AFe\_1=FAe\_1=FMe\_1=MFe\_1=Me\_2,\\\\ \cdot&=\cdots,\\\\ A\_n&=AF^\{n-1\}e\_1=F^\{n-1\}Ae\_1=F^\{n-1\}Me\_1=MF^\{n-1\}e\_1=Me\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由第 1 步即知
\begin\{aligned\} C(F)=L(E,F,F^2,\cdots,F^\{n-1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
往证 $\displaystyle E,F,F^2,\cdots,F^\{n-1\}$ 线性无关, 而 $\displaystyle \dim C(F)=n$. 事实上,
\begin\{aligned\} &k\_0E+k\_1F+k\_2F^2+\cdots+k\_\{n-1\}F^\{n-1\}=0\\\\ \Rightarrow&k\_0Ee\_1+k\_1Fe\_1+k\_2F^2e\_1+\cdots+k\_\{n-1\}F^\{n-1\}e\_1=0\\\\ \Rightarrow&k\_0e\_1+k\_1e\_2+k\_2e\_3+\cdots+k\_\{n-1\}e\_n=0\\\\ \Rightarrow&k\_0=k\_1=\cdots=k\_\{n-1\}=0. \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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---
1476、 6、 (20 分) 设 $\displaystyle \lambda$ 是复数域 $\displaystyle \mathbb\{C\}$ 上 $\displaystyle n$ 阶方阵 $\displaystyle A$ 的 $\displaystyle r$ 重特征值, 证明子空间
\begin\{aligned\} \left\\{X\in\mathbb\{C\}^n; (\lambda E-A)^rX=0\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的维数为 $\displaystyle r$. (武汉大学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 Jordan 标准形理论, 存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\{-1\}AP=\mathrm\{diag\}\left(R\_1,\cdots,R\_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 $\displaystyle \lambda\_1,\cdots,\lambda\_s$ 是 $\displaystyle A$ 的互异特征值, $\displaystyle R\_i$ 是以 $\displaystyle \lambda\_i$ 为对角元的 $\displaystyle n\_i\times n\_i$ 阶上三角矩阵, 且至多只有 $\displaystyle (i,i+1)$ 元为 $\displaystyle 1$, 其余元为 $\displaystyle 0$. 不妨设 $\displaystyle \lambda=\lambda\_1$, 则 $\displaystyle r=n\_1$. 于是
\begin\{aligned\} &P^\{-1\}(\lambda\_1E-A)^rP=\mathrm\{diag\}\left(0,(\lambda E\_\{n\_2\}-R\_2)^r,\cdots,(\lambda E\_\{n\_s\}-R\_s)^r\right)\\\\ \Rightarrow& \mathrm\{rank\}(\lambda E-A)^r=\mathrm\{rank\}(\lambda\_1E-A)^r=n-n\_1=n-r. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle \dim\left\\{X\in\mathbb\{C\}^n; (\lambda E-A)^rX=0\right\\}=n-(n-r)=r$.跟锦数学微信公众号. [在线资料](
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---
1477、 7、 设线性变换 $\displaystyle \mathscr\{T\}$ 在线性空间 $\displaystyle V$ 的基 $\displaystyle \varepsilon\_1,\varepsilon\_2,\varepsilon\_3$ 下的矩阵为 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&2&-1\\\\ -2&-2&2\\\\ 3&6&-1\end\{array\}\right)$, 并且 $\displaystyle \mathscr\{T\}$ 的特征多项式为 $\displaystyle f(\lambda)=(\lambda-\lambda\_1)^\{s\_1\}(\lambda-\lambda\_2)^\{s\_2\}$, 其中 $\displaystyle s\_1 > s\_2$. 再设
\begin\{aligned\} W=\ker(\mathscr\{T\}-\lambda\_1\mathscr\{I\})^\{s\_1\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \mathscr\{I\}$ 为恒等变换. (1)、 求出具体的 $\displaystyle f(\lambda)$. (2)、 求 $\displaystyle W$ 的基和维数. (武汉理工大学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 $\displaystyle f(\lambda)=|\lambda E-A|=(\lambda-2)^2(\lambda+4)$. (2)、 由
\begin\{aligned\} (A-2E)^2=\left(\begin\{array\}\{cccccccccccccccccccc\}-6&-12&6\\\\ 12&24&-12\\\\ -18&-36&18\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-1\\\\ 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 $\displaystyle (A-2E)^2X=0$ 的基础解系为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\1\end\{array\}\right)$. 故 $\displaystyle W$ 的基为
\begin\{aligned\} -2\varepsilon\_1+\varepsilon\_2, \varepsilon\_1+\varepsilon\_3, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \dim W=2$.跟锦数学微信公众号. [在线资料](
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---
1478、 8、 设 $\displaystyle A$ 为复方阵, $\displaystyle A$ 的特征多项式为 $\displaystyle (\lambda-1)^2(\lambda+1)^2$,
\begin\{aligned\} V\_1=N\left((A-E)^2\right), \quad V\_2=N\left((A+E)^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 $\displaystyle N(A)$ 为 $\displaystyle AX=0$ 的解空间. (1)、 证明 $\displaystyle \dim V\_1=\dim V\_2=2$; (2)、 证明: $\displaystyle \mathbb\{C\}^4=V\_1\oplus V\_2$. (武汉理工大学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 $\displaystyle A$ 的特征多项式的形式知存在可逆复方阵 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)$, 使得
\begin\{aligned\} P^\{-1\}AP=J=&J\_1\equiv \left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&&\\\\ &&-1&\\\\ &&&-1\end\{array\}\right)\mbox\{或\} J\_2\equiv \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &1&&\\\\ &&-1&\\\\ &&&-1\end\{array\}\right)\\\\ &\mbox\{或\} J\_3\equiv \left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&&\\\\ &&-1&1\\\\ &&&-1\end\{array\}\right)\mbox\{或\} J\_4\equiv \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &1&&\\\\ &&-1&1\\\\ &&&-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 $\displaystyle (J\_i\pm E)^2$ 即知
\begin\{aligned\} \dim V\_1=4-\mathrm\{rank\}\left\[(A-E)^2\right\] =4-\mathrm\{rank\}\left\[(J-E)^2\right\]=4-2=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle \dim V\_2=2$. 由 $\displaystyle (J\_i\pm E)^2$ 的形式知 $\displaystyle (J\_i-E)^2Y=0$ 的基础解系为 $\displaystyle e\_1,e\_2$, 而
\begin\{aligned\} &X\in V\_1\Leftrightarrow (A-E)^2X=0\stackrel\{X=PY\}\{\Leftrightarrow\}(J\_i-E)^2Y=0\\\\ \Leftrightarrow& Y\in L(e\_1,e\_2) \Leftrightarrow X\in L(Pe\_1,Pe\_2)=L(\alpha\_1,\alpha\_2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle V\_2=L(\alpha\_3,\alpha\_4)$. 故
\begin\{aligned\} \mathbb\{C\}^4=L(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)=L(\alpha\_1,\alpha\_2)\oplus L(\alpha\_3,\alpha\_4)=V\_1\oplus V\_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](
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---
1479、 6、 (20 分) 已知实矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&-2\\\\ 2&5&-4\\\\ -2&-4&5\end\{array\}\right)$. (1)、 求一个正交矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\}AP$ 为对角矩阵; (2)、 令 $\displaystyle V$ 是所有与 $\displaystyle A$ 可交换的实矩阵全体, 证明: $\displaystyle V$ 是实数域上的一个线性空间, 并确定 $\displaystyle V$ 的维数. (西安电子科技大学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)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 10,1,1$. 由
\begin\{aligned\} 10E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}2&0&1\\\\ 0&1&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 $\displaystyle 10$ 的特征向量为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-2\\\\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\}
由
\begin\{aligned\} 1E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-2\\\\ 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 $\displaystyle 1$ 的特征向量为
\begin\{aligned\} \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\0\end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\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\}
将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$, 并设
\begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{3\}&-\frac\{2\}\{\sqrt\{5\}\}&\frac\{2\}\{3\sqrt\{5\}\}\\\\ -\frac\{2\}\{3\}&\frac\{1\}\{\sqrt\{5\}\}&\frac\{4\}\{3\sqrt\{5\}\}\\\\ \frac\{2\}\{3\}&0&\frac\{\sqrt\{5\}\}\{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 $\displaystyle P$ 正交, 且
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(10,1,1). \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\} &B,C\in V\Rightarrow AB=BA, AC=CA\\\\ \Rightarrow& A(kB+lC)=kAB+lAC=kBA+lCA=(kB+lC)A\\\\ \Rightarrow& kB+lC\in V \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V$ 是线性空间 $\displaystyle \mathbb\{R\}^\{n\times n\}$ 的线性子空间, 而是一个线性空间. 进一步, 设 $\displaystyle \varLambda=\mathrm\{diag\}(10,1,1)$, 则
\begin\{aligned\} B\in V&\Leftrightarrow AB=BA\Leftrightarrow P^\mathrm\{T\} AP\cdot P^\mathrm\{T\} BP=P^\mathrm\{T\} BP\cdot P^\mathrm\{T\} AP\\\\ \stackrel\{C=P^\mathrm\{T\} BP\}\{\Leftrightarrow\}&\varLambda C=C\varLambda \Leftrightarrow C=\left(\begin\{array\}\{cccccccccccccccccccc\}c\_1&&\\\\ &c\_2&c\_3&\\\\ &c\_4&c\_5\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 $\displaystyle \dim V=5$, 且 $\displaystyle V$ 有一组基
\begin\{aligned\} PE\_\{11\}P^\mathrm\{T\}, PE\_\{22\}P^\mathrm\{T\}, PE\_\{23\}P^\mathrm\{T\}, PE\_\{32\}P^\mathrm\{T\}, PE\_\{33\}P^\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](
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---
1480、 7、 (15 分) 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{P\}$ 上的有限维线性空间, $\displaystyle \lambda\_1,\cdots,\lambda\_m$ 是 $\displaystyle \sigma$ 的不同特征值, 而 $\displaystyle \alpha\_i$ 是属于 $\displaystyle \lambda\_i\ (i=1,\cdots,m)$ 的特征值, $\displaystyle W$ 是 $\displaystyle \sigma$ 的一个不变子空间. 试证: 如果 $\displaystyle \alpha\_1+\cdots+\alpha\_m=\alpha\in W$, 则有 $\displaystyle \alpha\_i\in W, i=1,\cdots,m$. (西安电子科技大学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 $\displaystyle \alpha\_1+\cdots+\alpha\_m=\alpha\in W$, 则
\begin\{aligned\} &\lambda\_1\alpha\_1+\cdots+\lambda\_m\alpha\_m=\mathscr\{A\}\alpha\in W, \cdots,\\\\ &\lambda\_1^\{m-1\}\alpha\_1+\cdots+\lambda\_m^\{m-1\}\alpha\_m=\mathscr\{A\}^\{m-1\}\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\}
于是
\begin\{aligned\} &(\alpha\_1,\cdots,\alpha\_m)\left(\begin\{array\}\{cccccccccccccccccccc\}1&\lambda\_1&\cdots&\lambda\_1^\{m-1\}\\\\ 1&\lambda\_2&\cdots&\lambda\_2^\{m-1\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&\lambda\_m&\cdots&\lambda\_m^\{m-1\}\end\{array\}\right)=(\alpha,\mathscr\{A\}\alpha,\cdots,\mathscr\{A\}^\{m-1\}\alpha)\\\\ \Rightarrow&(\alpha\_1,\cdots,\alpha\_m)=(\alpha,\mathscr\{A\}\alpha,\cdots,\mathscr\{A\}^\{m-1\}\alpha) \left(\begin\{array\}\{cccccccccccccccccccc\}1&\lambda\_1&\cdots&\lambda\_1^\{m-1\}\\\\ 1&\lambda\_2&\cdots&\lambda\_2^\{m-1\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&\lambda\_m&\cdots&\lambda\_m^\{m-1\}\end\{array\}\right)^\{-1\}. \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 $\displaystyle \alpha\_1,\cdots,\alpha\_m$ 是 $\displaystyle \alpha,\mathscr\{A\}\alpha,\cdots,\mathscr\{A\}^\{m-1\}\alpha\in W$ 的线性组合, 而 $\displaystyle \alpha\_1,\cdots,\alpha\_m\in W$.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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---
1481、 3、 设 $\displaystyle V$ 是一个 $\displaystyle n$ 维线性空间, $\displaystyle L(V)$ 是 $\displaystyle V$ 上所有线性变换构成的线性空间, $\displaystyle v\in V$,
\begin\{aligned\} V=\left\\{\mathscr\{T\}\in L(V); \mathscr\{T\} v=0\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 证明: $\displaystyle E$ 是 $\displaystyle L(V)$ 的子空间; (2)、 若 $\displaystyle v\neq 0$, 求 $\displaystyle \dim E$. (西安交通大学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 $\displaystyle \forall\ k,l\in\mathbb\{F\}, \mathscr\{S\},\mathscr\{T\}\in E$,
\begin\{aligned\} (k\mathscr\{S\}+l\mathscr\{T\})(v)=k\mathscr\{S\}(v)+l\mathscr\{T\}(v)=k0+l0=0\Rightarrow k\mathscr\{S\}+l\mathscr\{T\}\in E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle E$ 是 $\displaystyle L(V)$ 的子空间. (2)、 设 $\displaystyle \varepsilon\_1=v$, 将其扩充为 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 则对 $\displaystyle \forall\ \mathscr\{T\}\in E$,
\begin\{aligned\} \mathscr\{T\}(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}0&a\_\{12\}&\cdots&a\_\{1n\}\\\\ \vdots&\vdots&&\vdots\\\\ 0&a\_\{n2\}&\cdots&a\_\{nn\}\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\}
设
\begin\{aligned\} \mathscr\{E\}\_\{ij\}(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)E\_\{ij\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle E\_\{ij\}$ 表示第 $\displaystyle i$ 行第 $\displaystyle j$ 列元素为 $\displaystyle 1$, 其余元素均为零的 $\displaystyle n$ 阶方阵. 则由 $\displaystyle E\_\{ij\}, 1\leq i,j\leq n$ 线性无关知 $\displaystyle \mathscr\{E\}\_\{ij\}, 1\leq i,j\leq n$ 线性无关, 且
\begin\{aligned\} \mathscr\{T\}=\sum\_\{i=1\}^n \sum\_\{j=2\}^n a\_\{ij\}\mathscr\{E\}\_\{ij\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathscr\{E\}\_\{ij\}, 1\leq i\leq n, 2\leq j\leq n$ 是 $\displaystyle E$ 的一组基, $\displaystyle \dim E=n(n-1)$.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1482、 9、 设 $\displaystyle V$ 是全体 $\displaystyle n$ 阶矩阵按照矩阵的加法和数量乘法构成的线性空间, $\displaystyle A\in V$, 且 $\displaystyle A$ 有 $\displaystyle n$ 个互异的特征值, 定义 $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{T\} X=AX-XA$, 证明: 存在 $\displaystyle V$ 的一组基, 使得 $\displaystyle \mathscr\{T\}$ 在此基下的矩阵为对角矩阵. (西安交通大学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, $\displaystyle A$ 可对角化, 即存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\{-1\}AP=\varLambda=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle B\_\{ij\}=PE\_\{ij\}P^\{-1\}, 1\leq i, \leq n$, 则它是 $\displaystyle M\_n(\mathbb\{C\})$ 的一组基, 且
\begin\{aligned\} P^\{-1\}\mathscr\{T\}(B)P&=P^\{-1\}(AB\_\{ij\}-B\_\{ij\}A)P =P^\{-1\}AP E\_\{ij\}-E\_\{ij\}P^\{-1\}AP\\\\ &=\varLambda E\_\{ij\}-E\_\{ij\}\varLambda =(\lambda\_i-\lambda\_j)E\_\{ij\}. \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\} \mathscr\{T\}(B\_\{ij\})=(\lambda\_i-\lambda\_j)PE\_\{ij\}P^\{-1\}=(\lambda\_i-\lambda\_j)B\_\{ij\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \mathscr\{T\}$ 在 $\displaystyle M\_n(\mathbb\{C\})$ 的基 $\displaystyle B\_\{ij\}=PE\_\{ij\}P^\{-1\}, 1\leq i, \leq n$ 下的矩阵为对角矩阵.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)/
---
1483、 4、 设向量组 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 线性无关, $\displaystyle k$ 为常数. 试问: $\displaystyle k$ 取何值时, 向量组
\begin\{aligned\} &k\alpha\_1-\alpha\_2-\alpha\_3-\cdots-\alpha\_n, -\alpha\_1+k\alpha\_2-\alpha\_3-\cdots-\alpha\_n,\\\\ &\cdots, -\alpha\_1-\alpha\_2-\cdots-\alpha\_\{n-1\}+k\alpha\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是线性无关的? $\displaystyle k$ 取何值时, 该向量组线性相关? (西北大学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) / 设题中向量组为 $\displaystyle \beta\_1,\cdots,\beta\_n$, 则
\begin\{aligned\} (\beta\_1,\cdots,\beta\_n)=(\alpha\_1,\cdots,\alpha\_n)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}k&-1&\cdots&-1\\\\ -1&k&\cdots&-1\\\\ \vdots&\vdots&&\vdots\\\\ -1&-1&\cdots&k\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\}
由
\begin\{aligned\} |T|=&\left|\begin\{array\}\{cccccccccc\}k-(n-1)&-1&\cdots&-1\\\\ k-(n-1)&k&\cdots&-1\\\\ \vdots&\vdots&&\vdots\\\\ k-(n-1)&-1&\cdots&k\end\{array\}\right|=[k-(n-1)]\left|\begin\{array\}\{cccccccccc\}1&-1&\cdots&-1\\\\ 1&k&\cdots&-1\\\\ \vdots&\vdots&&\vdots\\\\ 1&-1&\cdots&k\end\{array\}\right|\\\\ =&[k-(n-1)]\left|\begin\{array\}\{cccccccccc\}1&-1&\cdots&-1\\\\ &k+1&&\\\\ &&\ddots&\\\\ &&&k+1\end\{array\}\right|=[k-(n-1)] (k+1)^\{n-1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当 $\displaystyle k\neq n-1\mbox\{且\} k\neq -1$ 时, $\displaystyle \beta\_1,\cdots,\beta\_n$ 线性无关.跟锦数学微信公众号. [在线资料](
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---
1484、 8、 设 $\displaystyle A$ 是元素全为 $\displaystyle 1$ 的 $\displaystyle n$ 阶方阵, $\displaystyle E$ 是 $\displaystyle n$ 阶单位阵. (1)、 求 $\displaystyle |aE+bA|$, 其中 $\displaystyle a,b$ 是实常数; (2)、 已知 $\displaystyle 1 < \mathrm\{rank\}(aE+bA) < n$, 试确定 $\displaystyle a,b$ 满足的条件, 并求子空间
\begin\{aligned\} W=\left\\{X\in\mathbb\{R\}^n; (aE+bA)X=0\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的维数. (西北大学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 $\displaystyle \mathrm\{rank\} A=1$ 知 $\displaystyle Ax=0$ 的基础解系有 $\displaystyle n-1$ 个向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-1\}$. 设 $\displaystyle \eta\_n=(1,\cdots,1)^\mathrm\{T\}$, 则 $\displaystyle A\eta\_n=n\eta\_n$. 于是
\begin\{aligned\} &P\equiv(\eta\_1,\cdots,\eta\_n)\Rightarrow AP=P\left(\begin\{array\}\{cccccccccccccccccccc\}0\_\{n-1\}&\\\\ &n\end\{array\}\right)\\\\ \Rightarrow& P^\{-1\}(aE+bA)P=\left(\begin\{array\}\{cccccccccccccccccccc\}aE\_\{n-1\}&\\\\ a+nb\end\{array\}\right)\Rightarrow |aE+bA|=a^\{n-1\}(a+nb). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由第 1 步知 $\displaystyle a=0\mbox\{或\} a=-nb$. 但 $\displaystyle a=0\Rightarrow \mathrm\{rank\} A\leq 1$, 与题设矛盾. 故$a=-nb\neq 0$. 此时, 第 $\displaystyle i$ 行加到第 $\displaystyle n$ 行, $\displaystyle 1\leq i\leq n-1$; 第 $\displaystyle 1$ 行 $\displaystyle \cdot (-1)$ 加到第 $\displaystyle i$ 行, $\displaystyle 2\leq i\leq n-1$, 得
\begin\{aligned\} &aE+bA=\left(\begin\{array\}\{cccccccccccccccccccc\}(1-n)b&b&\cdots&b\\\\ b&(1-n)b&\cdots&b\\\\ \vdots&\vdots&&\vdots\\\\ b&b&\cdots&(1-n)b\end\{array\}\right)\\\\ \to& \left(\begin\{array\}\{cccccccccccccccccccc\}1-n&1&\cdots&1&1\\\\ 1&1-n&\cdots&1&1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ 1&1&\cdots&1-n&1\\\\ 0&0&\cdots&0&0\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}1-n&1&\cdots&1&1\\\\ n&-n&&&\\\\ \vdots&&\ddots&&\\\\ n&&&-n&\\\\ 0&0&\cdots&0&0\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}-1&&&&1\\\\ 1&-1&&&\\\\ \vdots&&\ddots&&\\\\ 1&&&-1&\\\\ 0&0&\cdots&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 $\displaystyle \mathrm\{rank\}(aE+bA)=n-1\Rightarrow \dim W=n-(n-1)=1$.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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---
1485、 9、 设 $\displaystyle f(x)$ 和 $\displaystyle g(x)$ 是数域 $\displaystyle \mathbb\{P\}$ 上互素的一元多项式, $\displaystyle A$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n$ 阶方阵. 证明: $\displaystyle n$ 元齐次线性方程组 $\displaystyle f(A)g(A)X=0$ 的解空间 $\displaystyle V$ 是 $\displaystyle f(A)X=0$ 的解空间 $\displaystyle V\_1$ 与 $\displaystyle g(A)X=0$ 的解空间 $\displaystyle V\_2$ 的直和, 其中 $\displaystyle X=(x\_1,\cdots,x\_n)^\mathrm\{T\}$. (西北大学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) / 显然 $\displaystyle V\_1,V\_2\subset V$. 由 $\displaystyle (f(x),g(x))=1$ 知存在多项式 $\displaystyle u(x),v(x)$ 使得
\begin\{aligned\} u(x)f(x)+v(x)g(x)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是对 $\displaystyle \forall\ X\in \mathbb\{F\}^n$,
\begin\{aligned\} X=v(A)g(A)X+u(A)f(A)X. \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\} f(A)[v(A)g(A)X]=0,\quad g(A)[u(A)f(A)X]=0 \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\} v(A)g(A)X\in V\_1,\quad u(A)f(A)X\in V\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle V=V\_1+V\_2$. 又由
\begin\{aligned\} X\in V\_1\cap V\_2\Rightarrow& f(A)X=g(A)X=0\\\\ \Rightarrow& X=u(A)f(A)X+v(A)g(A)X=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V=V\_1\oplus V\_2$.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](
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---
1486、 10、 设 $\displaystyle V$ 是 $\displaystyle n$ 维线性空间, $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组基, $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{T\}$ 定义为
\begin\{aligned\} \mathscr\{T\}\varepsilon\_i=\left\\{\begin\{array\}\{llllllllllll\}\varepsilon\_\{i+1\},&1\leq i\leq n-1,\\\\ 0,&i=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\}
(1)、 求 $\displaystyle \mathscr\{T\}$ 在基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 下的矩阵 $\displaystyle A$; (2)、 求 $\displaystyle \mathscr\{T\}$ 的值域与核的维数; (3)、 证明: $\displaystyle \mathscr\{T\}^n=\mathscr\{O\}, \mathscr\{T\}^\{n-1\}\neq\mathscr\{O\}$, 这里 $\displaystyle \mathscr\{O\}$ 表示零变换. (西北大学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\} \mathscr\{T\}(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&&&\\\\ 1&\ddots&&\\\\ &\ddots&0&\\\\ &&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\}
这表明 $\displaystyle \mathrm\{im\} \mathscr\{T\}=L(\varepsilon\_2,\cdots,\varepsilon\_n), \ker \mathscr\{T\}=L(\varepsilon\_n)$. 再者,
\begin\{aligned\} A^\{n-1\}=E\_\{n1\}, A^n=0\Rightarrow \mathscr\{T\}^n=\mathscr\{O\}, \mathscr\{T\}^\{n-1\}\neq\mathscr\{O\}. \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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---
1487、 3、 (20 分) 设向量组
\begin\{aligned\} \alpha\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}\star\\\\ \star\\\\ \star\end\{array\}\right), \alpha\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}\star\\\\ \star\\\\ \star\end\{array\}\right), \alpha\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}\star\\\\ \star\\\\ \star\end\{array\}\right)\left(\mbox\{具体数据未知\}\right), \beta=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\2k\\\\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\}
讨论 $\displaystyle k$ 的取值, 使得 $\displaystyle \beta$ 不可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出; $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 唯一线性表出; $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出, 且表示方法不唯一, 并求出一般表达式. [题目不全, 张祖锦没法做哦.] (西南财经大学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) / [题目不全, 张祖锦没法做哦.]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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---
1488、 7、 (20 分) 设 $\displaystyle \sigma,\tau$ 是线性空间 $\displaystyle V$ 上的线性变换, 且 $\displaystyle \sigma\tau=\tau\sigma$, 记
\begin\{aligned\} V^\{\lambda\_0\}=\left\\{\alpha\in V; \exists\ m\in \mathbb\{Z\}\_+,\mathrm\{ s.t.\} (\sigma-\lambda\_0\mathscr\{E\})^m\alpha=0\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 $\displaystyle \mathscr\{E\}$ 为恒等变换. 证明: $\displaystyle V^\{\lambda\_0\}$ 为 $\displaystyle V$ 的子空间, 且 $\displaystyle V^\{\lambda\_0\}$ 是 $\displaystyle \sigma$ 与 $\displaystyle \tau$ 的不变子空间. (西南财经大学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) / 对 $\displaystyle \forall\ \alpha,\beta\in V^\{\lambda\_0\}, k,l\in \mathbb\{F\}$,
\begin\{aligned\} &(\sigma-\lambda\_0\mathscr\{E\})^m(k\alpha+l\beta)=k(\sigma-\lambda\_0\mathscr\{E\})^m\alpha+l(\sigma-\lambda\_0\mathscr\{E\})^m\beta=k0+l0=0\\\\ \Rightarrow&k\alpha+l\beta\in V^\{\lambda\_0\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle V^\{\lambda\_0\}$ 是 $\displaystyle V$ 的子空间. 再者,
\begin\{aligned\} \alpha\in V^\{\lambda\_0\}\Rightarrow (\sigma-\lambda\_0\mathscr\{E\})^m \left\[\sigma(\alpha)\right\] =\sigma(\sigma-\lambda\_0\mathscr\{E\})^m\alpha=\sigma 0=0\Rightarrow \sigma(\alpha)\in V^\{\lambda\_0\}, \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\in V^\{\lambda\_0\}\Rightarrow (\sigma-\lambda\_0\mathscr\{E\})^m\left\[\tau(\alpha)\right\] =\tau (\sigma-\lambda\_0\mathscr\{E\})^m\alpha=\tau(0)=0\Rightarrow \tau(\alpha)\in V^\{\lambda\_0\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle V^\{\lambda\_0\}$ 是 $\displaystyle \sigma$ 与 $\displaystyle \tau$ 的不变子空间.跟锦数学微信公众号. [在线资料](
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---
1489、 8、 (15 分) 设 $\displaystyle A$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n$ 阶方阵, $\displaystyle f\_1(x), f\_2(x)\in\mathbb\{P\}[x]$, $\displaystyle f\_1(x), f\_2(x)$ 互素, $\displaystyle V$ 为齐次线性方程组 $\displaystyle f\_1(A)f\_2(A)X=0$ 的解空间, $\displaystyle V\_1$ 为 $\displaystyle f\_1(A)X=0$ 的解空间, $\displaystyle V\_2$ 为 $\displaystyle f\_2(A)X=0$ 的解空间. 求证: $\displaystyle V=V\_1\oplus V\_2$. (西南财经大学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,
\begin\{aligned\} V=&\left\\{X;f\_1(A)f\_2(A)X=0\right\\},\\\\ V\_1=&\left\\{X;f\_1(A)X=0\right\\},\quad V\_2=\left\\{X;f\_2(A)X=0\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle V\_1,V\_2$ 是 $\displaystyle V$ 的子空间. 又由 $\displaystyle f\_1,f\_2$ 互素知存在多项式 $\displaystyle u,v$ 使得
\begin\{aligned\} u(x)f\_1(x)+v(x)f\_2(x)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 (1)、 $\displaystyle X\in V\_1\cap V\_2\Rightarrow X=u(A)f\_1(A)X+v(A)f\_2(A)X=0$. (2)、 由 $\displaystyle f\_2(A)u(A)f\_1(A)X=u(A)f\_1(A)f\_2(A)X=0\Rightarrow u(A)f\_1(A)X\in V\_2$ 知
\begin\{aligned\} X\in V\Rightarrow X=u(A)f\_1(A)X+v(A)f\_2(A)X\in V\_2+ V\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle V=V\_1\oplus V\_2$.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](
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---
1490、 (3)、 (15 分) 设 $\displaystyle V$ 为数域 $\displaystyle \mathbb\{K\}$ 上四维线性空爱你, $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ 为 $\displaystyle V$ 的一组基, $\displaystyle \mathscr\{A\}$ 为 $\displaystyle V$ 上的线性变换, 且
\begin\{aligned\} \mathscr\{A\}\alpha\_1=\alpha\_2, \mathscr\{A\}\alpha\_2=\alpha\_2, \mathscr\{A\}\alpha\_3=\alpha\_2, \mathscr\{A\}\alpha\_4=\alpha\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令
\begin\{aligned\} \mathrm\{im\} \mathscr\{A\}=&\left\\{\beta\in V; \beta=\mathscr\{A\} \alpha, \forall\ \alpha\in V\right\\},\\\\ \ker \mathscr\{A\}=&\left\\{\alpha\in V; \mathscr\{A\}\alpha=0\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
分别为 $\displaystyle \mathscr\{A\}$ 的像空间与核. 求 $\displaystyle \mathrm\{im\}\mathscr\{A\}, \ker\mathscr\{A\}, \ker \mathscr\{A\}+\mathrm\{im\} \mathscr\{A\}, \mathrm\{im\}\mathscr\{A\}\cap \ker \mathscr\{A\}$ 的一组基和维数. (西南交通大学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\} \mathscr\{A\}(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)=(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&0&1\\\\ 1&1&1&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\}
(3-1)、 取 $\displaystyle x\_2,x\_3$ 为自由变量知 $\displaystyle Ax=0$ 的基础解系为 $\displaystyle (-1,1,0,0)^\mathrm\{T\}, (-1,0,1,0)^\mathrm\{T\}$. 故
\begin\{aligned\} \ker\mathscr\{A\}=L(-\alpha\_1+\alpha\_2,-\alpha\_1+\alpha\_3)\Rightarrow \dim \ker \mathscr\{A\}=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3-2)、
\begin\{aligned\} \mathrm\{im\}\mathscr\{A\}=&L(\mathscr\{A\}\alpha\_1,\mathscr\{A\}\alpha\_2,\mathscr\{A\}\alpha\_3,\mathscr\{A\}\alpha\_4) =L(\alpha\_2,\alpha\_2,\alpha\_2,\alpha\_1)=L(\alpha\_1,\alpha\_2), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle \dim \mathrm\{im\}\mathscr\{A\}=2$. (3-3)、 由
\begin\{aligned\} &(-\alpha\_1+\alpha\_2,-\alpha\_1+\alpha\_3,\alpha\_1,\alpha\_2)=(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4)B,\\\\ &B=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&-1&1&0\\\\ 1&0&0&1\\\\ 0&1&0&0\\\\ 0&0&0&0\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 0&1&0&0\\\\ 0&0&1&1\\\\ 0&0&0&0\end\{array\}\right)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle -\alpha\_1+\alpha\_2,-\alpha\_1+\alpha\_3,\alpha\_1$ 线性无关, 是 $\displaystyle \ker\mathscr\{A\}+\mathrm\{im\}\mathscr\{A\}$ 的一组基. 当然也有更为简单的一组基 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$. 从而 $\displaystyle \dim(\ker\mathscr\{A\}+\mathrm\{im\}\mathscr\{A\})=3$. (3-4)、 由
\begin\{aligned\} &\alpha\in \ker \mathscr\{A\}\cap \mathrm\{im\}\mathscr\{A\}\\\\ \Leftrightarrow& \alpha=x\_1(-\alpha\_1+\alpha\_2)+x\_2(-\alpha\_1+\alpha\_3)=-x\_3\alpha\_1-x\_4\alpha\_2\\\\ \Leftrightarrow&\alpha=-x\_3\alpha\_1-x\_4\alpha\_2, Bx=0\\\\ \stackrel\{(I)\}\{\Leftrightarrow\}&\alpha=-x\_3\alpha\_1-x\_4\alpha\_2, x=k\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\-1\\\\1\end\{array\}\right) \Leftrightarrow\alpha=k(\alpha\_1-\alpha\_2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \ker \mathscr\{A\}\cap \mathrm\{im\}\mathscr\{A\}=L(\alpha\_1-\alpha\_2), \dim (\ker \mathscr\{A\}\cap \mathrm\{im\}\mathscr\{A\})=1$.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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---
1491、 4、 (10 分) 设 $\displaystyle V$ 为实数域上的全体 $\displaystyle n$ 元二次型组成的集合. 证明: $\displaystyle V$ 对于多项式的加法和数域与多项式的乘法构成线性空间, 并给出 $\displaystyle V$ 的维数和一组基. (西南交通大学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 $\displaystyle f(x)=x^\mathrm\{T\} Ax, g(x)=x^\mathrm\{T\} Bx\in V$, 其中 $\displaystyle A,B\in\mathbb\{R\}^\{n\times n\}$ 是对称矩阵, 则
\begin\{aligned\} kf(x)+lg(x)=x^\mathrm\{T\} (kA+lB)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\}
故 $\displaystyle V$ 是 $\displaystyle \mathbb\{P\}[x\_1,\cdots,x\_n]$ 的子空间, 而是线性空间. 再者, 由 $\displaystyle f\mapsto A$ 是线性同构知
\begin\{aligned\} x^\mathrm\{T\} E\_\{ii\}x=x\_i^2, x^\mathrm\{T\}\frac\{E\_\{ij\}+E\_\{ji\}\}\{2\}x=x\_ix\_j, 1\leq i\leq n, 1\leq j\neq 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\}
是 $\displaystyle V$ 的一组基, 而 $\displaystyle \dim V=\frac\{n(n+1)\}\{2\}$.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](
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---
1492、 6、 设 $\displaystyle A=(a\_\{ij\}))\{m\times n\}$ 是实矩阵, $\displaystyle V\_1$ 为 $\displaystyle Ax=0$ 的解空间. 令
\begin\{aligned\} \alpha\_i=(a\_\{i1\},a\_\{i2\},\cdots,a\_\{in\})^\mathrm\{T\}, i=1,2,\cdots,m, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle V\_2=L(\alpha\_1,\cdots,\alpha\_m)$. 证明: $\displaystyle \mathbb\{R\}^n=V\_1\oplus V\_2$. (湘潭大学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 $\displaystyle \mathrm\{rank\} A=r$, 则 $\displaystyle \dim V\_1=n-r, \dim V\_2=r$. 由题设, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\\vdots\\\\\alpha\_m^\mathrm\{T\}\end\{array\}\right)$. 故
\begin\{aligned\} &\beta\in V\_1\cap V\_2\Rightarrow A\beta=0, \beta=\sum\_i x\_i\alpha\_i\\\\ \Rightarrow&\alpha\_i^\mathrm\{T\} \beta=0, \forall\ 1\leq i\leq m, \beta=\sum\_i x\_i\alpha\_i\\\\ \Rightarrow&\beta^\mathrm\{T\} \beta=\sum\_ix\_i\alpha\_i^\mathrm\{T\} \beta=0\Rightarrow \beta=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} \dim(V\_1+V\_2)=&\dim V\_1+\dim V\_2-\dim(V\_1\cap V\_2)=(n-r)+r-0=n. \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 $\displaystyle V\_1+V\_2=\mathbb\{R\}^n$. 联合 $\displaystyle V\_1\cap V\_2=\left\\{0\right\\}$ 知 $\displaystyle \mathbb\{R\}^n=V\_1\oplus V\_2$.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](
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---
1493、 9、 设 $\displaystyle \mathscr\{T\}$ 是线性空间 $\displaystyle V$ 上的线性变换, $\displaystyle x\_1,\cdots,x\_k$ 是 $\displaystyle \mathscr\{T\}$ 的 $\displaystyle k$ 个不用的特征值, 且 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 分别为其对应的特征向量. 证明: 若 $\displaystyle \alpha\_1+\cdots+\alpha\_k\in W$, 且 $\displaystyle W$ 是 $\displaystyle \mathscr\{T\}$ 的不变子空间, 则 $\displaystyle \dim W\geq k$. (湘潭大学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 $\displaystyle \alpha\_1+\cdots+\alpha\_k=\alpha\in W$, 则
\begin\{aligned\} x\_1\alpha\_1+\cdots+x\_k\alpha\_k=\mathscr\{A\}\alpha\in W, \cdots, x\_1^\{k-1\}\alpha\_1+\cdots+x\_k^\{k-1\}\alpha\_k=\mathscr\{A\}^\{k-1\}\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\}
于是
\begin\{aligned\} &(\alpha\_1,\cdots,\alpha\_k)\left(\begin\{array\}\{cccccccccccccccccccc\}1&x\_1&\cdots&x\_1^\{k-1\}\\\\ 1&x\_2&\cdots&x\_2^\{k-1\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&x\_k&\cdots&x\_k^\{k-1\}\end\{array\}\right)=(\alpha,\mathscr\{A\}\alpha,\cdots,\mathscr\{A\}^\{k-1\}\alpha)\\\\ \Rightarrow&(\alpha\_1,\cdots,\alpha\_k)=(\alpha,\mathscr\{A\}\alpha,\cdots,\mathscr\{A\}^\{k-1\}\alpha) \left(\begin\{array\}\{cccccccccccccccccccc\}1&x\_1&\cdots&x\_1^\{k-1\}\\\\ 1&x\_2&\cdots&x\_2^\{k-1\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&x\_k&\cdots&x\_k^\{k-1\}\end\{array\}\right)^\{-1\}. \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 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 是 $\displaystyle \alpha,\mathscr\{A\}\alpha,\cdots,\mathscr\{A\}^\{k-1\}\alpha\in W$ 的线性组合, 而 $\displaystyle \alpha\_1,\cdots,\alpha\_k\in W$. 又由 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 是 $\displaystyle \mathscr\{A\}$ 属于不同特征值的特征向量知它们线性无关. 于是
\begin\{aligned\} \dim W\geq \dim L\left(\alpha\_1,\cdots,\alpha\_k\right)=k. \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)/
---
1494、 4、 (10 分) 设 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 的秩为 $\displaystyle r$, $\displaystyle \alpha\_\{i\_1\},\cdots,\alpha\_\{i\_r\}$ 是 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 的 $\displaystyle r$ 个向量, 使得 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 中的每个向量都可以经它们线性表示. 证明: $\displaystyle \alpha\_\{i\_1\},\cdots,\alpha\_\{i\_r\}$ 是 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 的一个极大线性无关组. (新疆大学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 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_s), B=(\alpha\_\{i\_1\},\cdots,\alpha\_\{i\_r\})$, 则由题设,
\begin\{aligned\} \mathrm\{rank\} A=r; \exists\ C,\mathrm\{ s.t.\} A=BC\Rightarrow \mathrm\{rank\} B\geq \mathrm\{rank\} A=r. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\} B=r$, 而 $\displaystyle \alpha\_\{i\_1\},\cdots,\alpha\_\{i\_r\}$ 线性无关, 是 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 中 $\displaystyle r$ 个线性无关的向量组, 而是 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 的一个极大无关组.
\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)/
---
1495、 8、 (15 分) 证明: 在 $\displaystyle \mathbb\{P\}[x]\_n$ 中, 多项式
\begin\{aligned\} f\_1(x)=&(x-a\_2)(x-a\_3)\cdots(x-a\_n),\\\\ f\_i(x)=&(x-a\_1)\cdots(x-a\_\{i-1\})(x-a\_\{i+1\})\cdots (x-a\_n)\left(i=2,\cdots,n-1\right),\\\\ f\_n(x)=&(x-a\_1)(x-a\_2)\cdots(x-a\_\{n-1\}) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是一组基, 其中 $\displaystyle a\_1,\cdots,a\_n$ 是互不相同的数. (新疆大学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, $\displaystyle f\_i(a\_j)=0, j\neq i$;
\begin\{aligned\} f\_i(a\_i)=(a\_i-a\_1)\cdots(a\_i-a\_\{i-1\})(a\_i-a\_\{i+1\})\cdots (a\_i-a\_n)\neq 0. \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\} \sum\_\{i=1\}^n k\_if\_i(x)=0\Rightarrow 0=\sum\_\{i=1\}^n k\_if\_i(a\_j)=k\_jf\_j(a\_j)\Rightarrow k\_j=0, \forall\ j. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle f\_1,\cdots,f\_n$ 线性无关, 而是 $\displaystyle n$ 维线性空间 $\displaystyle \mathbb\{P\}[x]\_n$ 的一组基.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 13:24:19
