张祖锦2023年数学专业真题分类70天之第48天

zhangzujin · 发表于 2023-3-5 09:18:49

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## 张祖锦2023年数学专业真题分类70天之第48天 --- 1082、 (4)、设 $\displaystyle \left\\{A\in\mathbb\{R\}^\{4\times 4\}; A^2=2A+3I\right\\}$ 按其实数域上是否相似来分类, 共分为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ 类. (电子科技大学2023年高等代数考研试题) [矩阵 ] [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) / $\displaystyle A$ 的零化多项式为 $\displaystyle x^2-2x-3=(x+1)(x-3)$ 没有重根, 而可对角化. 从而 $\displaystyle A\sim \mathrm\{diag\} (\lambda\_1,\cdots,\lambda\_4)$, $\displaystyle \lambda\_i\in \left\\{-1,3\right\\}$. 注意实矩阵在 $\displaystyle \mathbb\{C\}$ 中相似则定在 $\displaystyle \mathbb\{R\}$ 中相似, 我们知共有 $\displaystyle 5$ 类, 它们的 Jordan 标准形为 \begin\{aligned\} &\mathrm\{diag\}(-1,-1,-1,-1), \mathrm\{diag\}(-1,-1,-1,3), \mathrm\{diag\}(-1,-1,3,3),\\\\ &\mathrm\{diag\}(-1,3,3,3), \mathrm\{diag\}(3,3,3,3). \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)/ --- 1083、 (2)、设 $\displaystyle A,C$ 都是 $\displaystyle n$ 阶实对称矩阵, $\displaystyle B$ 是 $\displaystyle n$ 阶实矩阵, $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B^\mathrm\{T\}&C\end\{array\}\right)$ 是正定阵. (2-1)、证明: $\displaystyle C-B^\mathrm\{T\} A^\{-1\}B$ 正定; (2-2)、证明: $\displaystyle \left|\begin\{array\}\{cccccccccc\}A&B\\\\ B^\mathrm\{T\}&C\end\{array\}\right|\leq |A|\cdot |C|$. (电子科技大学2023年高等代数考研试题) [矩阵 ] [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 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B^\mathrm\{T\}&C\end\{array\}\right)$ 正定知 $\displaystyle A,C$ 正定, 而可逆. 再由 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -B^\mathrm\{T\} A^\{-1\}&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B^\mathrm\{T\}&C\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A^\{-1\}B\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A&0\\\\ 0&C-B^\mathrm\{T\} A^\{-1\}B\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 $\displaystyle C-B^\mathrm\{T\} A^\{-1\}B$ 正定. 由 $\displaystyle C$ 正定知存在可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\mathrm\{T\} CP=E$. 于是 \begin\{aligned\} P^\mathrm\{T\} (C-B^\mathrm\{T\} A^\{-1\}B)P=E-P^\mathrm\{T\} B^\mathrm\{T\} A^\{-1\}BP \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 正定. 又由 $\displaystyle A^\{-1\}$ 正定 $\displaystyle \Rightarrow P^\mathrm\{T\} B^\mathrm\{T\} A^\{-1\}BP$ 半正定 ($B$ 未必是方阵; 即使是方阵, 也未必可逆) $\displaystyle \Rightarrow P^\mathrm\{T\} B^\mathrm\{T\} A^\{-1\}BP$ 的特征值 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 都 $\displaystyle \geq 0$. 进而 $\displaystyle E-P^\mathrm\{T\} B^\mathrm\{T\} A^\{-1\}BP$ 的特征值 $\displaystyle 1-\lambda\_1,\cdots,1-\lambda\_n$ 都 $\displaystyle \leq 1$. 综上, 我们知 $\displaystyle E-P^\mathrm\{T\} B^\mathrm\{T\} A^\{-1\}BP$ 的特征值 $\displaystyle 1-\lambda\_i\in (0,1]$. 于是 \begin\{aligned\} &\det (E-P^\mathrm\{T\} B^\mathrm\{T\} A^\{-1\}BP)=\prod\_\{i=1\}^n (1-\lambda\_i)\leq 1\\\\ \Rightarrow& \det (C-B^\mathrm\{T\} A^\{-1\}B)\leq \frac\{1\}\{\det P^\mathrm\{T\} \det P\}=\det C\\\\ \Rightarrow&\det\left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B^\mathrm\{T\}&C\end\{array\}\right)=\det A\cdot \det (C-B^\mathrm\{T\} A^\{-1\}B) \leq \det A\cdot \det C . \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)/ --- 1084、 (3)、设数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 阶方阵 $\displaystyle A,B$ 使得 $\displaystyle AB=BA=0$, 且秩 $\displaystyle \mathrm\{rank\} A=n-1$. 设非零向量 $\displaystyle \alpha,\beta$ 分别是 $\displaystyle AX=0$ 与 $\displaystyle A^\mathrm\{T\} X=0$ 的非零解. (3-1)、证明存在数 $\displaystyle k\in\mathbb\{F\}$ 使得 $\displaystyle B=k\alpha\beta^\mathrm\{T\}$; (3-2)、证明存在多项式 $\displaystyle g(x)\in\mathbb\{F\}[x]$, 使得 $\displaystyle B=g(A)$. (电子科技大学2023年高等代数考研试题) [矩阵 ] [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) / (3-1)、由 $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}(A^\mathrm\{T\})=n-1$ 知 $\displaystyle \alpha,\beta$ 分别是 $\displaystyle AX=0, A^\mathrm\{T\} X=0$ 的基础解系. 由 $\displaystyle AB=0$ 知 $\displaystyle B$ 的列向量都是 $\displaystyle \alpha$ 的线性组合: \begin\{aligned\} B=(k\_1\alpha,\cdots, k\_n\alpha)=\alpha(k\_1,\cdots,k\_n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle \alpha\neq 0$ 知 $\displaystyle B$ 的行向量组都是 $\displaystyle (k\_1,\cdots,k\_n)$ 的倍数. 又由 $\displaystyle BA=0\Rightarrow A^\mathrm\{T\} B^\mathrm\{T\}=0$ 知 $\displaystyle B^\mathrm\{T\}$ 的列向量组 (就是 $\displaystyle B$ 的行向量组的转置) 都是 $\displaystyle A^\mathrm\{T\} X=0$ 的解. 从而 \begin\{aligned\} \exists\ k,\mathrm\{ s.t.\} \left(\begin\{array\}\{cccccccccccccccccccc\}k\_1\\\\\vdots\\\\k\_n\end\{array\}\right)=k\beta\Rightarrow B=\alpha(k\_1,\cdots,k\_n)=\alpha \cdot k\beta^\mathrm\{T\} =k\alpha\beta^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (3-2)、由 $\displaystyle |A|=0$ 知 $\displaystyle A$ 以 $\displaystyle 0$ 为特征值, 可设 $\displaystyle A$ 的最小多项式为 $\displaystyle m(\lambda)=\lambda f(\lambda)$, 其中 $\displaystyle \partial(f)\leq n-1$. 由最小多项式的定义知 \begin\{aligned\} &f(A)\neq 0\Rightarrow \mathrm\{rank\} f(A)\geq 1,\\\\ &0=m(A)=Af(A)=f(A)A\Rightarrow \mathrm\{rank\} A+\mathrm\{rank\} f(A)\leq n\\\\ &\Rightarrow \mathrm\{rank\} f(A)\leq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \mathrm\{rank\} f(A)=1$. 再由 $\displaystyle Af(A)=f(A)A=0$ 及第 i 步知 $\displaystyle f(A)=l\alpha\beta^\mathrm\{T\}, l\neq 0$. 于是 \begin\{aligned\} B=k\alpha\beta^\mathrm\{T\}=\frac\{k\}\{l\}f(A), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 确为 $\displaystyle A$ 的多项式.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)/ --- 1085、 3、设矩阵 \begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&3&-1\\\\ 0&1&-1&1\\\\ 1&2&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\} (1)、求 $\displaystyle A^\mathrm\{T\} A$ 的秩; (2)、求矩阵 $\displaystyle B$, 使得 $\displaystyle AB=E$. (东北大学2023年高等代数考研试题) [矩阵 ] [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\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&6\\\\ 0&1&0&-4\\\\ 0&0&1&-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\} 知 $\displaystyle \mathrm\{rank\}(A^\mathrm\{T\} A)=\mathrm\{rank\} A=3$. 又由 \begin\{aligned\} (A,E)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&6&2&6&-1\\\\ 0&1&0&-4&-1&-3&1\\\\ 0&0&1&-5&-1&-4&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 $\displaystyle Ax=e\_i, i=1,2,3$ 有特解 \begin\{aligned\} \xi\_1=(2,-1,-1,0)^\mathrm\{T\}, \xi\_2=(6,-3,-4,0)^\mathrm\{T\}, \xi\_3=(-1,1,1,0)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle B=(\xi\_1,\xi\_2,\xi\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}2&6&-1\\\\ -1&-3&1\\\\ -1&-4&1\\\\ 0&0&0\end\{array\}\right)$, 则 $\displaystyle AB=E\_3$.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)/ --- 1086、 8、设 $\displaystyle A$ 为 $\displaystyle m\times n$ 矩阵, $\displaystyle B$ 为 $\displaystyle n\times s$ 矩阵, 证明: 若 $\displaystyle AB=0$, 则 \begin\{aligned\} \mathrm\{rank\} A+\mathrm\{rank\} B\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (用三种方法.) (东北大学2023年高等代数考研试题) [矩阵 ] [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 $\displaystyle B=(\beta\_1,\cdots,\beta\_l)$, 它的极大无关组为 $\displaystyle \beta\_\{i\_1\},\cdots,\beta\_\{i\_r\}$, 其中 $\displaystyle r=\mathrm\{rank\} B$. 由 $\displaystyle AB=0$ 知 $\displaystyle \beta\_\{i\_j\}\left(1\leq j\leq r\right)$ 是 $\displaystyle Ax=0$ 的解. 于是 \begin\{aligned\} \left\\{\beta\_\{i\_1\},\cdots,\beta\_\{i\_r\}\right\\}\subset \left\\{x; Ax=0\right\\}\equiv V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 $\displaystyle \dim V=n-\mathrm\{rank\} A$, 我们知 $\displaystyle r\leq n-\mathrm\{rank\} A\Rightarrow \mathrm\{rank\} A+\mathrm\{rank\} B\leq n$. (2)、设 \begin\{aligned\} \mathrm\{im\} B=\left\\{Bx; x\in\mathbb\{F\}^s\right\\}, \ker A=\left\\{x\in \mathbb\{F\}^n; Ax=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 $\displaystyle AB=0\Rightarrow \mathrm\{im\} B\subset \ker A$, 而 \begin\{aligned\} n=\dim \ker A+\dim \mathrm\{im\} A\geq \dim \mathrm\{im\} B+\dim \mathrm\{im\} A=\mathrm\{rank\} B+\mathrm\{rank\} A . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (3)、由 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}A&0\\\\ E\_n&B\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&-AB\\\\ E\_n&B\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0\\\\ E\_n&B\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&0\\\\ E\_n&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 \begin\{aligned\} n=\mathrm\{rank\} E\_n=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&0\\\\ E\_n&B\end\{array\}\right)\geq \mathrm\{rank\} A+\mathrm\{rank\} B . \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)/ --- 1087、 10、设 $\displaystyle A$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n\times n$ 矩阵, 满足 $\displaystyle A^2=A$. (1)、证明: $\displaystyle A$ 相似于一个对角矩阵 (用两种方法). (2)、证明: $\displaystyle \mathrm\{tr\} A=\mathrm\{rank\} A$. (东北大学2023年高等代数考研试题) [矩阵 ] [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)、 (1-1)、由 $\displaystyle A$ 的零化多项式 $\displaystyle \lambda^2-\lambda$ 没有重根知 $\displaystyle A$ 的极小多项式也没有重根, 而在 $\displaystyle \mathbb\{C\}$ 中可对角化. 由 $\displaystyle A$ 的特征值为 $\displaystyle 0$ 或 $\displaystyle 1$ 知 $\displaystyle A$ 在 $\displaystyle \mathbb\{C\}$ 中相似于 $\displaystyle \mathrm\{diag\}(E\_r,0)$. 从而在 $\displaystyle \mathbb\{P\}$ 中相似与 $\displaystyle \mathrm\{diag\}(E\_r,0)$. (1-2)、设 \begin\{aligned\} V\_1=\left\\{\alpha\in\mathbb\{P\}^n; A\alpha=\alpha\right\\}, V\_0=\left\\{\alpha\in\mathbb\{P\}^n; 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\} 则 $\displaystyle \dim V\_0=n-\mathrm\{rank\} A\equiv r$. 由 \begin\{aligned\} \mathbb\{P\}^n\ni \alpha=&A\alpha+(\alpha-A\alpha)\in V\_1+V\_0,\\\\ \alpha\in V\_1\cap V\_0\Rightarrow&\alpha=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\} 知 $\displaystyle \mathbb\{P\}^n=V\_1\oplus V\_0$. 由 \begin\{aligned\} \dim V\_0=n-\mathrm\{rank\} A=n-r \Rightarrow \dim V\_1=n-\dim V\_0=r \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 $\displaystyle V\_1$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$, $\displaystyle V\_0$ 的一组基 $\displaystyle \varepsilon\_\{r+1\},\cdots,\varepsilon\_n$, 则 $\displaystyle T=(\varepsilon\_1,\cdots,\varepsilon\_n)$ 可逆, 且 \begin\{aligned\} &T^\{-1\}AT=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)\Rightarrow A\sim \mathrm\{diag\}(E\_r,0). \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 步的第 2 个证明即知 $\displaystyle \mathrm\{tr\} A=r=\mathrm\{rank\} A$.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)/ --- 1088、 4、 (15 分) 证明: 秩等于 $\displaystyle r$ 的矩阵可以表示为 $\displaystyle r$ 个秩等于 $\displaystyle 1$ 的矩阵之和, 但不能表示为少于 $\displaystyle r$ 个秩等于 $\displaystyle 1$ 的矩阵之和. (东北师范大学2023年高等代数与解析几何考研试题) [矩阵 ] [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 $\displaystyle \mathrm\{rank\} A=r$, 则存在可逆矩阵 $\displaystyle P,Q$ 使得 \begin\{aligned\} A=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)Q=\sum\_\{i=1\}^r PE\_\{ii\}Q, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 是 $\displaystyle r$ 个秩等于 $\displaystyle 1$ 的矩阵之和. 往用反证法证明 $\displaystyle A$ 不能表示为少于 $\displaystyle r$ 个秩等于 $\displaystyle 1$ 的矩阵之和. 若不然, \begin\{aligned\} A=\sum\_\{i=1\}^s A\_i, s < r, \mathrm\{rank\} A\_i=1\Rightarrow \mathrm\{rank\} A\leq \sum\_\{i=1\}^s \mathrm\{rank\} A\_i=s < r. \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.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)/ --- 1089、 5、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实满秩矩阵, 证明: 存在正交矩阵 $\displaystyle P\_1,P\_2$, 使得 \begin\{aligned\} P\_1^\{-1\}AP\_2=\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\} 其中 $\displaystyle \lambda\_i > 0, i=1,\cdots,n$. (东北师范大学2023年高等代数与解析几何考研试题) [矩阵 ] [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 $\displaystyle \mathrm\{rank\} A=n$ 知 $\displaystyle Ax=0$ 只有零解, 而 \begin\{aligned\} 0\neq x\in\mathbb\{R\}^n\Rightarrow y=Ax\neq 0\Rightarrow x^\mathrm\{T\} (A^\mathrm\{T\} A)x=y^\mathrm\{T\} y > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A^\mathrm\{T\} A$ 正定, 存在正交阵 $\displaystyle P\_2$ 使得 \begin\{aligned\} P\_2^\mathrm\{T\} (A^\mathrm\{T\} A)P\_2=\mathrm\{diag\}(\mu\_1,\cdots,\lambda\_n), \mu\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 记 \begin\{aligned\} \lambda\_i=\sqrt\{\mu\_i\}, P\_1=AP\_2\mathrm\{diag\}(\lambda\_1^\{-1\},\cdots,\lambda\_n^\{-1\}), \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\} P\_1^\{-1\}AP\_2=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)P\_2^\{-1\}A^\{-1\}\cdot AP\_2=\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\} 且 \begin\{aligned\} P\_1^\mathrm\{T\} P\_1=\mathrm\{diag\}(\lambda\_1^\{-1\},\cdots,\lambda\_n^\{-1\})P\_2^\mathrm\{T\} A^\mathrm\{T\} \cdot AP\_2\mathrm\{diag\}(\lambda\_1^\{-1\},\cdots,\lambda\_n^\{-1\})=E\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 蕴含 $\displaystyle P\_1$ 是正交矩阵.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)/ --- 1090、 2、设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 1&-2&0\\\\ 1&2&-1\end\{array\}\right)$. 证明: 存在正交阵 $\displaystyle Q$, 使得 $\displaystyle QA$ 为上三角阵且对角元全为正数. (数字是什么不重要, 只要 $\displaystyle A$ 可逆就行) (东南大学2023年高等代数考研试题) [矩阵 ] [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) / 这就是可逆矩阵的 $\displaystyle QR$ 分解. 设 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_n)$, 则对 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 施行 Gram-Schmidt 标准正交化过程得 \begin\{aligned\} &\beta\_1=\frac\{\alpha\_1\}\{k\_1\}, \beta\_2=\frac\{\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1\}\{k\_2\},\cdots,\\\\ &\beta\_n=\frac\{\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}\}\{k\_n\},\\\\ &k\_1=|\alpha\_1| > 0, k\_2=|\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1| > 0, \cdots,\\\\ &k\_n=\left|\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}\right| > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 注意: $\displaystyle A$ 可逆蕴含 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 线性无关, 而各 $\displaystyle k\_i > 0$. 于是 \begin\{aligned\} &\alpha\_1=k\_1\beta\_1, \alpha\_2=(\alpha\_2,\beta\_1)\beta\_1+k\_2\beta\_2,\cdots,\\\\ &\alpha\_n=(\alpha\_n,\beta\_1)\beta\_1+\cdots+(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}+k\_n\beta\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 写成矩阵形式就是 \begin\{aligned\} A=(\alpha\_1,\cdots,\alpha\_n)=(\beta\_1,\cdots,\beta\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}k\_1&(\alpha\_2,\beta\_1)&\cdots&(\alpha\_n,\beta\_1)\\\\ &k\_2&\cdots&(\alpha\_n,\beta\_2)\\\\ &&\ddots&\vdots\\\\ &&&k\_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\} 恰为一个正交矩阵与上三角矩阵的乘积. 令 $\displaystyle Q=(\beta\_1,\cdots,\beta\_n)^\mathrm\{T\}$, 则 $\displaystyle Q$ 正交, 且 $\displaystyle QA$ 为上三角阵且对角元全为正数.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)/ --- 1091、 3、设 $\displaystyle A,B$ 都是二阶实对称矩阵, 行列式都为负数. 证明: 存在可逆矩阵 $\displaystyle C$, 使得 $\displaystyle C^\mathrm\{T\} AC=B$. (东南大学2023年高等代数考研试题) [矩阵 ] [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 $\displaystyle A,B$ 都是二阶实对称矩阵知它们的特征值都是实数. 又由它们的行列式都为负数知它们的特征值一正一负. 从而它们的正负惯性指数都为 $\displaystyle 1$, 而 $\displaystyle A,B$ 都合同于 $\displaystyle \mathrm\{diag\}(1,-1)$. 进而 $\displaystyle A,B$ 合同, 即存在可逆矩阵 $\displaystyle C$, 使得 $\displaystyle C^\mathrm\{T\} AC=B$.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)/ --- 1092、 4、 $\displaystyle A$ 为元素全为有理数的 $\displaystyle n$ 阶方阵, 且 $\displaystyle \sqrt\{3\}$ 为 $\displaystyle A$ 的一个特征值. (1)、证明: $\displaystyle -\sqrt\{3\}$ 也是 $\displaystyle A$ 的一个特征值. (2)、证明: $\displaystyle n=3$ 时, 存在实可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\} AP$ 为对角阵. (东南大学2023年高等代数考研试题) [矩阵 ] [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, $\displaystyle f(\lambda)=|\lambda E\_n-A|$ 是 $\displaystyle n$ 次有理系数多项式. 易知 $\displaystyle g(\lambda)=\lambda^2-3$ 在 $\displaystyle \mathbb\{Q\}$ 中不可约, 而 \begin\{aligned\} \left(f(\lambda),g(\lambda)\right)=1\mbox\{或\} g(\lambda)\mid f(\lambda). \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,g)=1\Rightarrow&\exists\ u,v\in\mathbb\{Q\}[\lambda],\mathrm\{ s.t.\} u(\lambda)f(\lambda)+v(\lambda)g(\lambda)=1\\\\ \stackrel\{\lambda=\sqrt\{3\}\}\{\Rightarrow\}&0=1,\mbox\{矛盾\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle g(\lambda)\mid f(\lambda)$, $\displaystyle f(-\sqrt\{3\})=0$, $\displaystyle -\sqrt\{3\}$ 也是 $\displaystyle A$ 的一个特征值. (2)、由第 1 步知 $\displaystyle (\lambda^2-3)\mid f(\lambda)$, 而 \begin\{aligned\} \exists\ m\in\mathbb\{Z\},\mathrm\{ s.t.\} f(\lambda)=(\lambda^2-3)(\lambda-m). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 $\displaystyle A$ 的特征值为 $\displaystyle \sqrt\{3\},-\sqrt\{3\}, m$. 它们是互异的. 设 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 为对应的特征向量, 则它们线性无关. 设 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3)$, 则 $\displaystyle P$ 可逆, 且 \begin\{aligned\} AP=P\mathrm\{diag\}(\sqrt\{3\},-\sqrt\{3\}, m)\Rightarrow P^\{-1\}AP=\mathrm\{diag\}(\sqrt\{3\},-\sqrt\{3\}, m). \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)/ --- 1093、 6、设 $\displaystyle A$ 是一个 $\displaystyle m\times n$ 实矩阵. 证明: (1)、 $\displaystyle \mathrm\{rank\}(A^\mathrm\{T\} A)=\mathrm\{rank\} A$; (2)、 $\displaystyle A^\mathrm\{T\} A$ 是半正定矩阵, 并且 $\displaystyle A^\mathrm\{T\} A$ 是正定矩阵当且仅当 $\displaystyle \mathrm\{rank\} A=n$. (东南大学2023年高等代数考研试题) [矩阵 ] [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\} V=\left\\{\alpha\in\mathbb\{R\}^n; A\alpha=0\right\\}, W=\left\\{\alpha\in\mathbb\{R\}^n; A^\mathrm\{T\} 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\} 则显然有 $\displaystyle V\subset W$. 又由 \begin\{aligned\} \alpha\in W&\Rightarrow A^\mathrm\{T\} A\alpha=0\Rightarrow \alpha^\mathrm\{T\} A^\mathrm\{T\} A\alpha=0 \stackrel\{\beta=A\alpha\}\{\Rightarrow\}\beta^\mathrm\{T\} \beta=0\\\\ &\Rightarrow \beta=0 \Rightarrow A\alpha=0\Rightarrow \alpha\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 $\displaystyle W\subset V$. 于是 \begin\{aligned\} V=W&\Rightarrow n-\mathrm\{rank\} A=\dim V=\dim W=n-\mathrm\{rank\}(A^\mathrm\{T\} A)\\\\ &\Rightarrow \mathrm\{rank\} A=\mathrm\{rank\}(A^\mathrm\{T\} A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、对 $\displaystyle \forall\ \alpha\in\mathbb\{R\}^n$, 设 $\displaystyle \beta=A\alpha\in\mathbb\{R\}^n$, 则 \begin\{aligned\} \alpha^\mathrm\{T\} (A^\mathrm\{T\} A)\alpha=\beta^\mathrm\{T\} \beta\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A^\mathrm\{T\} A$ 半正定. 往证 $\displaystyle n$ 阶矩阵 $\displaystyle A^\mathrm\{T\} A$ 是正定矩阵当且仅当 $\displaystyle \mathrm\{rank\} A=n$. (2-1)、 $\displaystyle \Rightarrow$: 由 $\displaystyle A^\mathrm\{T\} A$ 正定知 \begin\{aligned\} n=\mathrm\{rank\}(A^\mathrm\{T\} A)\overset\{\tiny\mbox\{第1步\}\}\{=\} \mathrm\{rank\} A . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-2)、 $\displaystyle \Leftarrow$: 由 $\displaystyle A^\mathrm\{T\} A$ 半正定知存在正交阵 $\displaystyle P$ 使得 \begin\{aligned\} P^\mathrm\{T\} (A^\mathrm\{T\} A) P=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_r,0,\cdots,0), \lambda\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 $\displaystyle \mathrm\{rank\} A=n$ 及第 1 步知 $\displaystyle r=\mathrm\{rank\}(A^\mathrm\{T\} A)=n$. 这就证明了 $\displaystyle A^\mathrm\{T\} A$ 的特征值都是正数, $\displaystyle A^\mathrm\{T\} A$ 正定.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)/ --- 1094、 8、设实矩阵 $\displaystyle A$ 的特征多项式和最小多项式都是 $\displaystyle (\lambda^2+p\lambda+q)^2$, $\displaystyle a+b\mathrm\{ i\}$ ($b\neq 0$) 为 $\displaystyle A$ 的一个特征值. 证明: 存在实可逆矩阵 $\displaystyle C$, 使得 \begin\{aligned\} C^\{-1\}AC=\left(\begin\{array\}\{cccccccccccccccccccc\}a&-b&0&0\\\\ b&a&0&0\\\\ 1&0&a&-b\\\\ 0&1&b&a\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\} (东南大学2023年高等代数考研试题) [矩阵 ] [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, $\displaystyle f(\lambda)=|\lambda E-A|=(\lambda^2+p\lambda+q)^2$, 而 $\displaystyle A$ 是 $\displaystyle 4$ 阶实矩阵. 又由 $\displaystyle a+b\mathrm\{ i\}$ 是 $\displaystyle f(\lambda)=0$ 的根, 实多项式 $\displaystyle f(\lambda)$ 的虚根成对出现知 $\displaystyle a-b\mathrm\{ i\}$ 也是 $\displaystyle f(\lambda)=0$ 的根. 故 \begin\{aligned\} f(\lambda)=\left\[\lambda-(a+b\mathrm\{ i\})\right\]^2\left\[\lambda-(a-b\mathrm\{ i\})\right\]^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 $\displaystyle A$ 的极小多项式也是 $\displaystyle f(\lambda)$ 知 $\displaystyle A$ 的 Jordan 标准形为 \begin\{aligned\} J=\left(\begin\{array\}\{cccccccccccccccccccc\}a+b\mathrm\{ i\}&1&&\\\\ &a+b\mathrm\{ i\}&&\\\\ &&a-b\mathrm\{ i\}&1\\\\ &&&a-b\mathrm\{ i\}\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\} 于是存在复可逆矩阵 $\displaystyle P=(\xi\_1,\xi\_2,\bar\{\xi\}\_1,\bar\{\xi\_2\})$, 使得 \begin\{aligned\} &AP=PJ=P\left(\begin\{array\}\{cccccccccccccccccccc\}a+b\mathrm\{ i\}&1&&\\\\ &a+b\mathrm\{ i\}&&\\\\ &&a-b\mathrm\{ i\}&1\\\\ &&&a-b\mathrm\{ i\}\end\{array\}\right)\\\\ \Rightarrow& A\xi\_1=(a+b\mathrm\{ i\})\xi\_1, A\xi\_2=\xi\_1+(a+b\mathrm\{ i\})\xi\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \xi\_1=\alpha+\beta\mathrm\{ i\}, \xi\_2=\gamma+\delta\mathrm\{ i\}, \alpha,\beta,\gamma,\delta\in\mathbb\{R\}^n$, 则 \begin\{aligned\} &A(\alpha+\beta\mathrm\{ i\})=(a+b\mathrm\{ i\})(\alpha+\beta\mathrm\{ i\})\\\\ \Rightarrow&A\alpha=a\alpha-b\beta, A\beta=b\alpha+a\beta,\\\\ &A(\gamma+\delta\mathrm\{ i\})=\alpha+\beta \mathrm\{ i\}+(a+b\mathrm\{ i\})(\gamma+\delta\mathrm\{ i\})\\\\ \Rightarrow&A\gamma=\alpha+a\gamma-b\delta, A\delta=\beta+b\gamma+a\delta.\qquad(I) \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\} 0\neq&|P|=|\xi\_1,\xi\_2,\bar\{\xi\}\_1,\bar\{\xi\}\_2|\\\\ =&|\alpha+\beta\mathrm\{ i\}, \gamma+\delta\mathrm\{ i\}, \alpha-\beta\mathrm\{ i\}, \gamma-\delta\mathrm\{ i\}| =|2\alpha,2\gamma,\alpha-\beta\mathrm\{ i\}, \gamma-\delta\mathrm\{ i\}|\\\\ =&2^2|\alpha,\gamma, -\beta\mathrm\{ i\},-\delta\mathrm\{ i\}| =2^2(-\mathrm\{ i\})^2|\alpha,\gamma,\beta,\delta| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle C=(\delta,\gamma,\beta,\alpha)$ 是实可逆矩阵, 且 \begin\{aligned\} (I)\Rightarrow AC=C\left(\begin\{array\}\{cccccccccccccccccccc\}a&-b&0&0\\\\ b&a&0&0\\\\ 1&0&a&-b\\\\ 0&1&b&a\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\} 结论得证.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)/ --- 1095、 1、填空题 (每小题 5 分, 共 25 分). (1)、设 $\displaystyle n$ 阶方阵 $\displaystyle A$ 满足 $\displaystyle A^2+2A-3E=0$, $\displaystyle E$ 为单位阵, 则 $\displaystyle (A+4E)^\{-1\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (福州大学2023年高等代数考研试题) [矩阵 ] [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) / 由带余除法知 $\displaystyle x^2+2x-3=(x-2)(x+4)+5$, 而 \begin\{aligned\} 0=(A-2E)(A+4E)+5E\Rightarrow (A+4E)^\{-1\}=-\frac\{1\}\{5\}(A-2E). \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)/ --- 1096、 (5)、设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}6&-10\\\\ 2&-3\end\{array\}\right)$, 则 $\displaystyle \mathrm\{tr\}(A^\{2023\})=\underline\{\ \ \ \ \ \ \ \ \ \ \}$, 其中 $\displaystyle \mathrm\{tr\} A$ 表示矩阵 $\displaystyle A$ 的迹. (福州大学2023年高等代数考研试题) [矩阵 ] [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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,2$, 而 \begin\{aligned\} A\sim\mathrm\{diag\}(1,2)\Rightarrow A^\{2023\}\sim\mathrm\{diag\}(1,2^\{2023\})\Rightarrow \mathrm\{tr\}(A^\{2023\})=1+2^\{2023\}. \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)/ --- 1097、 2、简答题 (每题 5 分, 共 25 分). (1)、设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1\\\\ 1&4\end\{array\}\right)$, 矩阵 $\displaystyle B$ 满足 $\displaystyle BA=B+2E$, 求 $\displaystyle \det B$. (福州大学2023年高等代数考研试题) [矩阵 ] [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\} B(A-E)=2E\Rightarrow \det B\cdot \det (A-2E)=2^2\Rightarrow \det B=\frac\{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\} 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)/ --- 1098、 (5)、求矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&2&-5\\\\ 2&6&-10\\\\ 1&2&-3\end\{array\}\right)$ 的不变因子组, 初等因子组和 Jordan 标准形. (福州大学2023年高等代数考研试题) [矩阵 ] [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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,2$. 由 $\displaystyle A-2E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-5\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)$ 知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 [严格上三角部分若有两个 $\displaystyle 1$, 则 $\displaystyle \mathrm\{rank\}(J-2E)=2$; 若没有 $\displaystyle 1$, 则 $\displaystyle \mathrm\{rank\}(J-2E)=0$. 都得到矛盾] \begin\{aligned\} \mathrm\{rank\}(J-2E)=\mathrm\{rank\}(A-2E)=1\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}2&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\} $\displaystyle A$ 的初等因子为 $\displaystyle \lambda-2,(\lambda-2)^2$, 不变因子为 $\displaystyle 1,\lambda-2,(\lambda-2)^2$.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)/ --- 1099、 (2)、 (12 分) 设 $\displaystyle A$ 为实矩阵, 证明: $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}(A^\mathrm\{T\} A)$. (福州大学2023年高等代数考研试题) [矩阵 ] [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) / 这只要证明 \begin\{aligned\} &\left\\{\alpha\in\mathbb\{R\}^n; A\alpha=0\right\\}=\left\\{\alpha\in\mathbb\{R\}^n; A^\mathrm\{T\} A\alpha=0\right\\}\\\\ \Rightarrow&n-\mathrm\{rank\} A=\dim \mbox\{左端\}=\dim \mbox\{右端\}=n-\mathrm\{rank\}(A^\mathrm\{T\} A). \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, $\displaystyle A\alpha=0\Rightarrow A^\mathrm\{T\} A\alpha=0$, \begin\{aligned\} A^\mathrm\{T\} A\alpha=0\Rightarrow 0=\alpha^\mathrm\{T\} A^\mathrm\{T\} A\alpha\stackrel\{\beta=A\alpha\}\{=\}\beta^\mathrm\{T\} \beta \Rightarrow \beta=0\Rightarrow 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\} 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)/ --- 1100、 (5)、 (12 分) 设 $\displaystyle A$ 为 $\displaystyle 3$ 阶实对称矩阵, 且 $\displaystyle A$ 的秩为 $\displaystyle 2$, $\displaystyle \lambda\_1=\lambda\_2=2$ 是 $\displaystyle A$ 的特征值, 且对应的特征向量为 \begin\{aligned\} \alpha\_1=(1,1,0)^\mathrm\{T\}, \alpha\_2=(2,1,1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 求另一个特征值及对应的特征向量, 并求出矩阵 $\displaystyle A$. (福州大学2023年高等代数考研试题) [矩阵 ] [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 $\displaystyle \mathrm\{rank\} A=2\Rightarrow 0=|A|=|A-0E|$ 知 $\displaystyle 0$ 是 $\displaystyle A$ 的另一个特征值. 又由实对称矩阵属于不同特征值的特征向量正交知 $\displaystyle A$ 的属于特征值 $\displaystyle 0$ 的特征向量 $\displaystyle \alpha\_3$ 满足 \begin\{aligned\} \alpha\_i^\mathrm\{T\} \alpha\_3=0, i=1,2\Leftrightarrow B\alpha\_3=0, B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 2&1&1\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 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\} 故可取 $\displaystyle \alpha\_3=(-1,1,1)^\mathrm\{T\}$. 令 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-1\\\\ 1&1&1\\\\ 0&1&1\end\{array\}\right)$, 则 \begin\{aligned\} AP=P\mathrm\{diag\}(2,2,0)\Rightarrow A=\frac\{1\}\{3\}\left(\begin\{array\}\{cccccccccccccccccccc\}4&2&2\\\\ 2&4&-2\\\\ 2&-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\} 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)/ --- 1101、 (8)、 (14 分) 已知 $\displaystyle A$ 为实对称矩阵, $\displaystyle B$ 为正定矩阵, $\displaystyle A-B$ 为半正定矩阵. 证明: (8-1)、若 $\displaystyle |A-\lambda B|=0$, 则 $\displaystyle \lambda\geq 1$; (8-2)、 $\displaystyle |A|\geq |B|$. (福州大学2023年高等代数考研试题) [矩阵 ] [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) / (8-1)、由 $\displaystyle B$ 正定知存在可逆矩阵 $\displaystyle P$ 使得 $\displaystyle P^\mathrm\{T\} BP=E$. 于是 \begin\{aligned\} A-B\mbox\{半正定\}\Rightarrow P^\mathrm\{T\} AP-E\mbox\{半正定\} \Leftrightarrow P^\mathrm\{T\} AP\mbox\{的特征值 \}\geq 1.\qquad(\star) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \lambda$ 是 $\displaystyle |A-\lambda B|=0$ 的根, 则 \begin\{aligned\} 0=|P^\mathrm\{T\}|\cdot|A-\lambda B|\cdot |P| =|P^\mathrm\{T\} AP-\lambda E| =(-1)^n |\lambda E-P^\mathrm\{T\} AP|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle (\star)$ 即知 $\displaystyle \lambda\geq 1$. (8-2)、由 $\displaystyle (\star)$ 知 $\displaystyle P^\mathrm\{T\} AP$ 的所有特征值 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 都 $\displaystyle \geq 1$, 而 \begin\{aligned\} |P^\mathrm\{T\} AP|=\lambda\_1\cdots \lambda\_n\geq 1 \Rightarrow |A|\geq \frac\{1\}\{|P|^2\}=|B|. \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)/ --- 1102、注: 内容包括高等代数, 近世代数, 不含微分几何 (第一大题填空题每题 5 分, 高等代数部分试题每题 20 分, 近世代数部分试题每题 15 分). 1、填空题. (1)、给定实对称矩阵 \begin\{aligned\} H=\left(\begin\{array\}\{cccccccccccccccccccc\}2-t&1-t&1-t\\\\ 1-t&2-t&1-t\\\\ 1-t&1-t&2-t\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\} 其中 $\displaystyle t$ 是参数, 则令 $\displaystyle H$ 是正定矩阵的 $\displaystyle t$ 的范围是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (复旦大学2023年代数(第4,6,7,8题没做)考研试题) [矩阵 ] [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) / $\displaystyle H$ 正定等价于 $\displaystyle H$ 的顺序主子式 \begin\{aligned\} 2-t > 0, (2-t)^2-(1-t)^2 > 0, \det H=4-3t > 0 \Leftrightarrow t < \frac\{4\}\{3\}. \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)/ --- 1103、 (2)、已知 $\displaystyle T$ 是一个 $\displaystyle 2022$ 阶复方阵组成的集合, 满足对任意矩阵 $\displaystyle C\in T$, 有 \begin\{aligned\} C^2=2022C, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 且 $\displaystyle T$ 中的矩阵互不相似, 则 $\displaystyle T$ 中至多有 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ 个元素. (复旦大学2023年代数(第4,6,7,8题没做)考研试题) [矩阵 ] [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 $\displaystyle C$ 的零化多项式 $\displaystyle \lambda^2-2022\lambda=\lambda(\lambda-2022)$ 没有重根知 $\displaystyle C$ 可对角化, 且特征值为 $\displaystyle 0$ 或 $\displaystyle 2022$. 故 $\displaystyle C$ 的 Jordan 标准形为 \begin\{aligned\} J=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_\{2022\}), \lambda\_i\in\left\\{0,2022\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 共有 $\displaystyle 2^\{2022\}$ 个.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)/ --- 1104、 (3)、已知 $\displaystyle 10$ 阶实方阵 $\displaystyle B$ 满足 \begin\{aligned\} B^5=0, \quad B^4\neq 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle B$ 可能的秩是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (复旦大学2023年代数(第4,6,7,8题没做)考研试题) [矩阵 ] [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, $\displaystyle B$ 的最小多项式为 $\displaystyle \lambda^5$, 而 $\displaystyle B$ 的 Jordan 标准形 $\displaystyle J$ 有一个 $\displaystyle 5$ 阶 Jordan 块 $\displaystyle J\_5(0)$, 其上有 $\displaystyle 4$ 个 $\displaystyle 1$. 剩下的那些 Jordan 块可能含有 $\displaystyle 1$ 的个数可能为 $\displaystyle 0,1,2,3,4$, 而 \begin\{aligned\} \mathrm\{rank\} B\in\left\\{4,5,6,7,8\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](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/

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