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张祖锦2023年数学专业真题分类70天之第69天

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发表于 2023-3-5 13:26:06 | 显示全部楼层 |阅读模式
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## 张祖锦2023年数学专业真题分类70天之第69天 --- 1565、 (4)、 设 $\displaystyle \sigma$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 上的对称变换, 则 $\displaystyle \dim(\mathrm\{im\}\sigma\cap \ker\sigma)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (上海大学2023年高等代数考研试题) [欧氏空间 ] [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\in \mathrm\{im\}\sigma\cap \ker \sigma\Rightarrow \exists\ \beta\in V,\mathrm\{ s.t.\} \alpha=\sigma(\beta); \sigma(\alpha)=0\\\\ \Rightarrow&(\alpha,\alpha)=\left(\sigma(\beta),\alpha\right)=\left(\beta,\sigma(\alpha)\right)=\left(\beta,0\right)=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 $\displaystyle \mathrm\{im\}\sigma\cap \ker\sigma=\left\\{0\right\\}\Rightarrow \dim\left(\mathrm\{im\}\sigma\cap \ker\sigma\right)=0$.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)/ --- 1566、 6、 (20 分) 设 $\displaystyle U,W$ 是欧氏空间 $\displaystyle V$ 的两个子空间, 考虑下列两个等式: \begin\{aligned\} \mbox\{甲\}: (U+W)^\perp=U^\perp+W^\perp, \mbox\{乙\}: (U+W)^\perp=U^\perp\cap W^\perp. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 证明其中成立的等式, 并举例证伪其中不成立的等式. (上海交通大学2023年高等代数考研试题) [欧氏空间 ] [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)、 乙对. 设 $\displaystyle \alpha\in (U+W)^\perp$, 则 \begin\{aligned\} &\alpha\perp \beta, \forall\ \beta\in U+W\\\\ \stackrel\{U,W\subset U+W\}\{\Rightarrow\}& \left\\{\begin\{array\}\{llllllllllll\}\alpha\perp \beta, \forall\ \beta\in U\Rightarrow \alpha\in U^\perp\\\\ \alpha\perp \beta, \forall\ \beta\in W\Rightarrow \alpha\in W^\perp\end\{array\}\right.\\\\ \Rightarrow&\alpha\in U^\perp\cap W^\perp. \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 $\displaystyle (U+W)^\perp\subset U^\perp \cap W^\perp$. 再设 $\displaystyle \alpha\in U^\perp \cap W^\perp$, 则 \begin\{aligned\} &\alpha\perp \beta\_1, \forall\ \beta\_1\in U; \alpha\perp \beta\_2, \forall\ \beta\_2\in W\\\\ \Rightarrow& \alpha\perp(\beta\_1+\beta\_2), \forall\ \beta\_1+\beta\_2\in U+W \Rightarrow \alpha\in (U+W)^\perp. \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 $\displaystyle U^\perp \cap W^\perp\subset (U+W)^\perp$. 综上即知结论成立. (2)、 甲错. 比如对 $\displaystyle V=\mathbb\{R\}^3$, $\displaystyle U=L(e\_1), W=L(e\_2,e\_3)$, 则 \begin\{aligned\} (U+W)^\perp=V^\perp=\left\\{0\right\\}, U^\perp+W^\perp=L(e\_2,e\_3)+L(e\_1)=V. \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)/ --- 1567、 5、 令 \begin\{aligned\} \alpha\_1=&(1,0,-1,2,1), \\\\ \alpha\_2=&(-1,1,0,0,1), \\\\ \alpha\_3=&(-1,2,-1,2,3) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 为欧氏空间 $\displaystyle \mathbb\{R\}^5$ 中的三个向量, $\displaystyle V=L(\alpha\_1,\alpha\_2,\alpha\_3)$. 求 $\displaystyle V$ 的正交补 $\displaystyle W$, 并求 $\displaystyle W$ 的一组标准正交基. (首都师范大学2023年高等代数考研试题) [欧氏空间 ] [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\} &\beta\in W^\perp\Leftrightarrow\alpha\_i\beta^\mathrm\{T\}=0, 1\leq i\leq 3\\\\ \Leftrightarrow& A\beta^\mathrm\{T\}=0, A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1\\\\\alpha\_2\\\\\alpha\_3\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1&2&1\\\\ -1&1&0&0&1\\\\ -1&2&-1&2&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\} 由 \begin\{aligned\} A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1&2&1\\\\ 0&1&-1&2&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\} 知 $\displaystyle Ax=0$ 的基础解系为 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1\\\\0\\\\0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\-2\\\\0\\\\1\\\\0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-2\\\\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\} 将它们标准正交化, 并转置即得 $\displaystyle W^\perp$ 的一组标准正交基 \begin\{aligned\} &\left(\frac\{1\}\{\sqrt\{3\}\},\frac\{1\}\{\sqrt\{3\}\},\frac\{1\}\{\sqrt\{3\}\},0,0\right), \left(-\frac\{2\}\{\sqrt\{3\}\},-\frac\{2\}\{\sqrt\{3\}\},\frac\{4\}\{\sqrt\{3\}\},\frac\{3\}\{\sqrt\{3\}\}\right),\\\\ &\left(\frac\{4\}\{\sqrt\{231\}\},-\frac\{7\}\{\sqrt\{231\}\},\frac\{3\}\{\sqrt\{231\}\},-\frac\{6\}\{\sqrt\{231\}\},\frac\{11\}\{\sqrt\{231\}\}\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)/ --- 1568、 6、 设 $\displaystyle \mathbb\{R\}^n$ 是所有 $\displaystyle n$ 维实列向量组成的线性空间, $\displaystyle M$ 是 $\displaystyle n$ 阶实方阵, 定义函数 $\displaystyle (-,-): \mathbb\{R\}^n\times\mathbb\{R\}^n\to \mathbb\{R\}$ 为 \begin\{aligned\} (X,Y)=X^\mathrm\{T\} AY\in\mathbb\{R\}, \forall\ X, Y\in \mathbb\{R\}^n. \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: $\displaystyle (-,-)$ 是 $\displaystyle \mathbb\{R\}^n$ 上的一个内积当且仅当存在可逆实矩阵 $\displaystyle B$, 使得 $\displaystyle A=B^\mathrm\{T\} B$; (2)、 对任意的 $\displaystyle \alpha\in\mathbb\{R\}^n$, 定义函数 $\displaystyle \rho\_\alpha: \mathbb\{R\}^n\to\mathbb\{R\}$ 为 $\displaystyle \rho\_\alpha(X)=(\alpha,X)$, 定义映射 $\displaystyle \tau: \mathbb\{R\}^n\to (\mathbb\{R\}^n)^\star$ 为 $\displaystyle \alpha\mapsto \rho\_\alpha$, 其中 $\displaystyle (\mathbb\{R\}^n)^\star$ 是 $\displaystyle \mathbb\{R\}^n$ 的对偶空间. 证明: $\displaystyle \tau$ 是同构映射当且仅当 $\displaystyle A$ 可逆. (3)、 设 $\displaystyle A$ 是反对称的, 对 $\displaystyle \mathbb\{R\}^n$ 的任意子空间 $\displaystyle W$, 设 \begin\{aligned\} W^\perp=\left\\{\alpha\in\mathbb\{R\}^n; (\alpha,\beta)=0, \forall\ \beta\in W\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: \begin\{aligned\} \dim W+\dim W^\perp=n+\dim\left\[W\cap (\mathbb\{R\}^n)^\perp\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (四川大学2023年高等代数考研试题) [欧氏空间 ] [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)、 $\displaystyle (\cdot,\cdot)$ 是双线性函数. 从而 $\displaystyle (\cdot,\cdot)$ 是内积当且仅当 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} &(X,Y)=(Y,X), \forall\ X,Y\Leftrightarrow e\_i^\mathrm\{T\} Ae\_j=e\_jAe\_i\forall\ i,j\Leftrightarrow a\_\{ij\}=a\_\{ji\},\\\\ \mbox\{且\}&\boxed\{(X,X)\geq 0, (X,X)=0\Leftrightarrow X=0\}\Leftrightarrow A\mbox\{正定\},\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\} 进而当且仅当 $\displaystyle A$ 合同对于单位矩阵, 即存在可逆实矩阵 $\displaystyle B$, 使得 $\displaystyle A=B^\mathrm\{T\} B$. (2)、 由 \begin\{aligned\} &\tau(k\alpha+l\beta)(X)=\rho\_\{k\alpha+l\beta\}(X) =(k\alpha+l\beta,X)=k(\alpha,X)+l(\beta,X)\\\\ =&k\rho\_\alpha(X)+l\rho\_\beta(X) =(k\tau\_\alpha+l\tau\_\beta)(X) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle \tau$ 是线性映射. 再者, $\displaystyle \tau$ 是同构 $\displaystyle \Leftrightarrow \tau$ 是单射 $\displaystyle \Leftrightarrow \ker\tau=\left\\{0\right\\}$ \begin\{aligned\} \Leftrightarrow&\mbox\{若\}\rho\_\alpha(X)=0, \forall\ X, \mbox\{则\}\alpha=0 \Leftrightarrow \mbox\{若\}\alpha^\mathrm\{T\} AX=0, \forall\ X, \mbox\{则\}\alpha=0\\\\ \Leftrightarrow&\mbox\{若\}\alpha^\mathrm\{T\} A=0, \mbox\{则\}\alpha=0 \Leftrightarrow\mbox\{若\}A^\mathrm\{T\} \alpha=0, \mbox\{则\}\alpha=0 \Leftrightarrow A^\mathrm\{T\} \mbox\{可逆\}\Leftrightarrow A\mbox\{可逆\}. \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\} W^\perp=&\left\\{\alpha\in\mathbb\{R\}^n; \alpha^\mathrm\{T\} A\beta=0 \forall\ \beta\in W\right\\} =\left\[\mathrm\{im\}(A|\_W)\right\]^\{\perp\_1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [这里 $\displaystyle \perp\_1$ 表示在标准内积下的正交补] 知 \begin\{aligned\} &\dim W+\dim W^\perp=\dim W+\dim \left\[\mathrm\{im\}(A|\_W)\right\]^\{\perp\_1\}\\\\ =&\dim W+n-\dim \mathrm\{im\}(A|\_W) =n+\dim \ker(A|\_W)\\\\ =&n+\dim\left\\{\alpha\in W; A\alpha=0\right\\}\\\\ =&n+\dim\left\\{\alpha\in W; (e\_i,\alpha)=e\_i^\mathrm\{T\} A\alpha=0, \forall\ i\right\\}\\\\ =&n+\dim\left\\{\alpha\in W; \alpha\in (\mathbb\{R\}^n)^\perp\right\\} =n+\dim\left\[W\cap (\mathbb\{R\}^n)^\perp\right\]. \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)/ --- 1569、 7、 内积空间, $\displaystyle p\_n(x)$ 为实系数多项式. 已知 \begin\{aligned\} (p\_i,p\_j)=0\left(i\neq j\right), \deg p\_n=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (1)、 计算 $\displaystyle p\_0,p\_1,p\_2$. (2)、 对任意正整数 $\displaystyle m$, 存在 $\displaystyle \alpha\_m,\beta\_m,\gamma\_m$, 使得 \begin\{aligned\} xp\_m=\alpha\_mp\_\{m+1\}+\beta\_mp\_\{m+1\}+\gamma\_mp\_\{m+1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [题目不全, 跟锦数学微信公众号没法做哦.] (天津大学2023年高等代数考研试题) [欧氏空间 ] [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, 跟锦数学微信公众号没法做哦.]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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1570、 8、 设 $\displaystyle V$ 是欧氏空间, $\displaystyle \alpha\in V$ 为非零单位向量, 定义镜面反射变换 \begin\{aligned\} \phi\_\alpha(v)=v-2(\alpha,v)\alpha,\quad \forall\ v\in V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 求证: (1)、 $\displaystyle \varphi\_\alpha(v)$ 为正交变换. (2)、 若 $\displaystyle \beta,\gamma$ 互异且长度相同, 则存在镜面反射 $\displaystyle \phi\_\alpha$ 使得 $\displaystyle \phi\_\alpha(\beta)=\gamma$. (3)、 $\displaystyle V$ 中任一正交变换可以表示为不超过 $\displaystyle n$ 个镜面发射的乘积. (天津大学2023年高等代数考研试题) [欧氏空间 ] [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)、 将 $\displaystyle \alpha$ 扩充为 $\displaystyle V$ 的一组标准正交 $\displaystyle \alpha,\alpha\_2,\cdots,\alpha\_n$, 则 $\displaystyle \varphi\_\alpha$ 在该标准正交基下的矩阵为 \begin\{aligned\} \mathrm\{diag\}(-1,1,\cdots,1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这是一个正交矩阵, 而 $\displaystyle \varphi\_\alpha$ 是正交变换. (2)、 \begin\{aligned\} &\varphi\_\{\frac\{\gamma-\beta\}\{|\gamma-\beta|\}\}(\beta) =\beta-2\left(\frac\{\gamma-\beta\}\{|\gamma-\beta|\},\beta\right)\frac\{\gamma-\beta\}\{|\gamma-\beta|\}\\\\ =&\beta-2\left(\gamma-\beta,\beta\right)\frac\{\gamma-\beta\}\{|\gamma-\beta|^2\} =\beta-2\left\[(\gamma,\beta)-(\beta,\beta)\right\]\frac\{\gamma-\beta\}\{\left(\gamma-\beta,\gamma-\beta\right)\}\\\\ =&\beta-2\left\[(\gamma,\beta)-1\right\]\frac\{\gamma-\beta\}\{2-2(\gamma,\beta)\} =\beta+(\gamma-\beta) =\gamma. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (3)、 设 $\displaystyle \mathscr\{A\}$ 是 $\displaystyle V$ 的一个正交变换. 取定 $\displaystyle V$ 的一组标准正交基 $\displaystyle \xi\_1,\cdots,\xi\_n$, 则 $\displaystyle \eta\_1=\mathscr\{A\}\xi\_1,\cdots,\eta\_n=\mathscr\{A\} \xi\_n$ 也是 $\displaystyle V$ 的一组标准正交基. (3-1)、 若 $\displaystyle \forall\ i, \eta\_i=\xi\_i$, 则 $\displaystyle \mathscr\{A\}$ 是恒等变换. 作镜面反射 $\displaystyle \varphi\_\{\xi\_1\}(\alpha)=\alpha-2(\alpha,\xi\_1)\xi\_1$, 则 $\displaystyle \mathscr\{A\}=\varphi\_\{\xi\_1\}\circ \varphi\_\{\xi\_1\}$. (3-2)、 若 $\displaystyle \exists\ i\_0,\mathrm\{ s.t.\} \eta\_\{i\_0\}\neq \xi\_\{i\_0\}$, 则不妨设 $\displaystyle i\_0=1$. 因为 $\displaystyle \xi\_1\neq \eta\_1$, 我们可由第 2 步知存在镜面反射 $\displaystyle \varphi\_1$, 将 $\displaystyle \xi\_1$ 变为 $\displaystyle \eta\_1$, 而保持其余 $\displaystyle \xi\_i\ (2\leq i\leq n)$ 不动. 如果 $\displaystyle \xi\_i=\eta\_i, 2\leq i\leq n$, 则 $\displaystyle \mathscr\{A\}=\varphi\_1$. 否则, 再作镜面反射 $\displaystyle \varphi\_2$, 将 $\displaystyle \xi\_2$ 变为 $\displaystyle \eta\_2$, 保持其余向量不动. 如此继续下去, 即知 $\displaystyle \mathscr\{A\}$ 是至多 $\displaystyle n$ 个镜面反射的乘积.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)/ --- 1571、 5、 (20 分) $\displaystyle V$ 是实 $\displaystyle n$ 阶实对称矩阵构成的线性空间, \begin\{aligned\} (A,B)=\mathrm\{tr\}(AB),\forall\ A,B\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 $\displaystyle \mathrm\{tr\}(AB)$ 为 $\displaystyle AB$ 的迹. (1)、 证明 $\displaystyle V$ 构成一欧氏空间; (2)、 求 $\displaystyle S=\left\\{A\in V; \mathrm\{tr\} A=0\right\\}$ 与 $\displaystyle S^\perp$ 的维数. (武汉大学2023年高等代数考研试题) [欧氏空间 ] [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 (1-1)、 $\displaystyle (A,B)=\mathrm\{tr\}(AB)=\mathrm\{tr\}(BA)=(B,A)$, (1-2)、 $\displaystyle (kA,B)=\mathrm\{tr\}(kAB)=k\mathrm\{tr\}(AB)=k(A,B)$, (1-3)、 $\displaystyle (A+B,C)=\mathrm\{tr\}\left((A+B)C\right) =\mathrm\{tr\}(AC)+\mathrm\{tr\}(BC)=(A,C)+(B,C)$, (1-4)、 $\displaystyle (A,A)=\mathrm\{tr\}(AA)=\mathrm\{tr\}(A^\mathrm\{T\} A)=\sum\_\{i,j=1\}^n a\_\{ij\}^2\geq 0$, 且 \begin\{aligned\} (A,A)=0\Leftrightarrow a\_\{ij\}=0, \forall\ i,j\Leftrightarrow A=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle V$ 关于 $\displaystyle (A,B)$ 构成欧氏空间. (2)、 解关于 $\displaystyle a\_\{ij\}, 1\leq i\leq j\leq n$ 的方程 $\displaystyle a\_\{11\}+\cdots+a\_\{nn\}=0$, 取 $\displaystyle a\_\{ij\}, i < j; a\_\{ii\}, 2\leq i\leq n$ 为自由变量知 \begin\{aligned\} S=L(E\_\{ij\}+E\_\{ji\}, 1\leq i < j\leq n; -E\_\{11\}+E\_\{ii\}, 2\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\} 从而 \begin\{aligned\} \dim S=&\sum\_\{i=1\}^n \sum\_\{j=i+1\}^n 1+(n-1)=\frac\{(n+2)(n-1)\}\{2\},\\\\ \dim S^\perp=&\frac\{n(n+1)\}\{2\}-\dim S=1. \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)/ --- 1572、 6、 在实值函数线性空间中, 定义内积 \begin\{aligned\} \left(f(x),g(x)\right)=\int\_\{-1\}^1 f(x)g(x)\mathrm\{ d\} x,\quad \forall\ f(x),g(x)\in C[-1,1]. \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\} V\_1=\mathrm\{span\}\left\\{2,x+1,\sin^2x\right\\}, V\_2=\mathrm\{span\}\left\\{1,x\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 $\displaystyle V\_1$ 和 $\displaystyle V\_2$ 的基和维数. (2)、 设 $\displaystyle W=V\_1\cap V\_2$, 求 $\displaystyle W$ 的一组标准正交基和维数. (3)、 设 $\displaystyle f(x)=\sin x$, 并且 $\displaystyle f(x)=\alpha\_1+\alpha\_2$, 其中 $\displaystyle \alpha\_1\in W, \alpha\_2\perp W$. 求 $\displaystyle \alpha\_1,\alpha\_2$. (武汉理工大学2023年高等代数考研试题) [欧氏空间 ] [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)、 设 $\displaystyle k\_1\cdot 2+k\_2\cdot (x+1)+k\_3\cdot \sin^2x=0$, 则取 $\displaystyle x=0,-1,1$ 后知 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}2&1&0\\\\ 2&0&\sin^21\\\\ 2&2&\sin^21\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}k\_1\\\\k\_2\\\\k\_3\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}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\} 由系数矩阵行列式 $\displaystyle =-4\sin^21\neq 0$ 知 $\displaystyle k\_i=0$, 而 $\displaystyle 2,x+1,\sin^2x$ 线性无关, 是 $\displaystyle V\_1$ 的一组基, $\displaystyle \dim V\_1=3$. (1-2)、 设 $\displaystyle k\_1\cdot 1+k\_2\cdot x=0$, 则令 $\displaystyle x=0$ 知 $\displaystyle k\_1=0\Rightarrow k\_2x=0\Rightarrow k\_2=0$, 而 $\displaystyle 1,x$ 线性无关, 是 $\displaystyle V\_2$ 的一组基, $\displaystyle \dim V\_2=2$. (2)、 由 $\displaystyle 1=\frac\{1\}\{2\}\cdot 2, x=(x+1)-\frac\{1\}\{2\}\cdot 2$ 知 $\displaystyle V\_2\subset V\_1$, $\displaystyle W=V\_2=L(1,x)$. 注意到 $\displaystyle \int\_\{-1\}^1 1\cdot x\mathrm\{ d\} x=0$, 我们知 \begin\{aligned\} \eta\_1=\frac\{1\}\{|1|\}=\frac\{1\}\{\sqrt\{\displaystyle\int\_\{-1\}^1 1^2\mathrm\{ d\} x\}\}=\frac\{1\}\{\sqrt\{2\}\}, \eta\_2=\frac\{x\}\{|x|\}=\frac\{x\}\{\sqrt\{\displaystyle\int\_\{-1\}^1 x^2\mathrm\{ d\} x\}\}=\sqrt\{\frac\{3\}\{2\}\}x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 是 $\displaystyle W$ 的一组标准正交基, 且 $\displaystyle \dim W=2$. (3)、 设 $\displaystyle \alpha\_1=k\cdot 1+l\cdot x$, 则 \begin\{aligned\} 0=&(\sin x-\alpha\_1,1)=\int\_\{-1\}^1 (\sin x-k-lx)\mathrm\{ d\} x=-2k,\\\\ 0=&(\sin x-\alpha\_1,x)=\int\_\{-1\}^1 (\sin x-k-lx)x\mathrm\{ d\} x=-\frac\{2\}\{3\}(l+3\cos 1-3\sin 1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle k=0, l=3(\sin 1-\cos 1)$, \begin\{aligned\} \alpha\_1=3(\sin 1-\cos 1)x, \alpha\_2=\sin x-\alpha\_1=\sin x-3(\sin 1-\cos 1)x. \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)/ --- 1573、 10、 (15 分) 设 $\displaystyle V$ 为 $\displaystyle n$ 维欧氏空间, $\displaystyle \sigma,\tau$ 为 $\displaystyle V$ 上的两个变换, 若 $\displaystyle \sigma$ 为正交变换, 且对任意的 $\displaystyle \alpha,\beta\in V$, 有 \begin\{aligned\} \left(\sigma(\alpha),\beta\right)=\left(\alpha,\tau(\beta)\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: $\displaystyle \tau$ 也是 $\displaystyle V$ 上的正交变换. (西安电子科技大学2023年高等代数考研试题) [欧氏空间 ] [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) / 取定 $\displaystyle V$ 的一组标准正交基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 并设 $\displaystyle \sigma$ 在该基下的矩阵为 $\displaystyle A$, $\displaystyle \tau$ 在该基下的矩阵为 $\displaystyle B$, 则 \begin\{aligned\} \sigma(\varepsilon\_1,\cdots,\varepsilon\_n)&=(\varepsilon\_1,\cdots,\varepsilon\_n)A,\\\\ \tau(\varepsilon\_1,\cdots,\varepsilon\_n)&=(\varepsilon\_1,\cdots,\varepsilon\_n)B. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle \sigma$ 是正交变换知 $\displaystyle A$ 是正交矩阵. 又由 \begin\{aligned\} \left(\sigma(\varepsilon\_i),\varepsilon\_j\right)&=(a\_\{1i\}\varepsilon\_1+\cdots+a\_\{ni\}\varepsilon\_n,\varepsilon\_j)=a\_\{ji\},\\\\ \left(\varepsilon\_i,\tau(\varepsilon\_j)\right)&=(\varepsilon\_i,b\_\{1j\}\varepsilon\_1+\cdots+b\_\{nj\}\varepsilon\_n)=b\_\{ij\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 及 $\displaystyle \left(\sigma(\alpha),\beta\right)=\left(\alpha,\tau(\beta)\right), \forall\ \alpha,\beta\in V$ 知 \begin\{aligned\} a\_\{ji\}=b\_\{ij\}\left(\forall\ 1\leq i,j\leq n\right) \Rightarrow B=A^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle A$ 是正交矩阵知 $\displaystyle B$ 也是正交矩阵. 这就证明了 $\displaystyle \tau$ 是 $\displaystyle V$ 上的正交变换.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)/ --- 1574、 11、 (10 分) 设 $\displaystyle \sigma,\tau$ 为欧氏空间 $\displaystyle V$ 上的两个线性变换, 对任意的 $\displaystyle \alpha,\beta\in V$, 有 \begin\{aligned\} \left(\sigma(\alpha),\sigma(\beta)\right)=\left(\tau(\alpha),\tau(\beta)\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: $\displaystyle V\_1=\sigma V$ 与 $\displaystyle V\_2=\tau V$ 同构. (西安电子科技大学2023年高等代数考研试题) [欧氏空间 ] [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\in\ker \sigma\Leftrightarrow& \sigma(\alpha)=0\Leftrightarrow 0=\left(\sigma(\alpha),\sigma(\alpha)\right)=\left(\tau(\alpha),\tau(\alpha)\right)\\\\ \Leftrightarrow& \tau(\alpha)=0\Leftrightarrow \ker \tau \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=\ker\tau\Rightarrow \dim \ker \sigma=\dim \ker \tau \Rightarrow \dim \mathrm\{im\}\sigma=\dim\mathrm\{im\} \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 从而 $\displaystyle V\_1=\sigma V$ 与 $\displaystyle V\_2=\tau V$ 同构.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)/ --- 1575、 6、 (20 分) 已知 $\displaystyle \sigma$ 为欧氏空间 $\displaystyle V$ 上的对称变换, 若对任意的非零向量 $\displaystyle \alpha\in V$, 有 $\displaystyle \left(\sigma(\alpha),\alpha\right) > 0$, 则称 $\displaystyle \sigma$ 为正定的对称变换. 证明: 对称变换 $\displaystyle \sigma$ 是正定的当且仅当 $\displaystyle \sigma$ 的特征值全大于零. (西南财经大学2023年高等代数考研试题) [欧氏空间 ] [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) / 取定 $\displaystyle V$ 的一组标准正交基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 设 $\displaystyle \sigma$ 在该基下的矩阵为 $\displaystyle A$, 则由 $\displaystyle \sigma$ 是对称变换知 $\displaystyle A$ 实对称. 再者, $\displaystyle \forall\ \alpha=\sum\_i x\_i\varepsilon\_i\in V,$ \begin\{aligned\} &\left(\sigma(\alpha),\alpha\right)=\left(\sum\_i x\_i\sigma(\varepsilon\_i),\sum\_j x\_j\varepsilon\_j\right) =\sum\_\{i,j\}x\_ix\_j \left(\sigma(\varepsilon\_i),\varepsilon\_j\right)\\\\ =&\sum\_\{i,j\}x\_ix\_j\left(\sum\_k a\_\{ki\}\varepsilon\_k,\varepsilon\_j\right) =\sum\_\{i,j\}x\_ix\_ja\_\{ji\}=x^\mathrm\{T\} Ax. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \sigma$ 正定 $\displaystyle \Leftrightarrow A$ 正定 $\displaystyle \Leftrightarrow A$ 的特征值全大于 $\displaystyle 0\Leftrightarrow \sigma$ 的特征值全大于 $\displaystyle 0$.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)/ --- 1576、 3、 (10 分) 设 $\displaystyle V$ 为有限维欧氏空间, 对任意 $\displaystyle \alpha,\beta\in V$, 内积记为 $\displaystyle (\alpha,\beta)$, $\displaystyle \mathscr\{A\}$ 是 $\displaystyle V$ 上的正交变换. 令 \begin\{aligned\} V\_1=\left\\{\alpha\in V; \mathscr\{A\}\alpha=\alpha\right\\}, V\_2=\left\\{\beta-\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: $\displaystyle V\_1,V\_2$ 为 $\displaystyle V$ 的子空间, 且 $\displaystyle V=V\_1\oplus V\_2$. (西南交通大学2023年高等代数考研试题) [欧氏空间 ] [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 $\displaystyle \forall\ k,l\in\mathbb\{R\}, \alpha,\beta\in V\_1$, \begin\{aligned\} \mathscr\{A\}(k\alpha+l\beta)=k\mathscr\{A\}\alpha+l\mathscr\{A\}\beta=k\alpha+l\beta\Rightarrow k\alpha+l\beta\in V\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle V\_1$ 是 $\displaystyle V$ 的子空间. 再者, 对 $\displaystyle \forall\ \alpha,\beta\in V$, \begin\{aligned\} &(k\alpha+l\beta)-\mathscr\{A\}(k\alpha+l\beta) =k(\alpha-\mathscr\{A\}\alpha)+l(\beta-\mathscr\{A\}\beta)\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\} 故 $\displaystyle V\_2$ 也是 $\displaystyle V$ 的子空间. 由 \begin\{aligned\} &\alpha\in V\_1\cap V\_2\Rightarrow \mathscr\{A\}\alpha=\alpha; \exists\ \beta\in V,\mathrm\{ s.t.\} \alpha=\beta-\mathscr\{A\}\beta\\\\ \Rightarrow&\mathscr\{A\}\alpha=\alpha, \mathscr\{A\}\beta=\beta-\alpha \Rightarrow (\alpha,\beta)=\left(\mathscr\{A\}\alpha,\mathscr\{A\}\beta\right)=(\alpha,\beta-\alpha)\\\\ \Rightarrow& (\alpha,\alpha)=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 $\displaystyle V\_1\cap V\_2=\left\\{0\right\\}$. 从而 [$\mathscr\{E\}$ 是恒等变换] \begin\{aligned\} \dim(V\_1+V\_2)=&\dim V\_1+\dim V\_2-\dim(V\_1\cap V\_2)\\\\ =&\dim \ker(\mathscr\{E\}-\mathscr\{A\})+\dim \mathrm\{im\} (\mathscr\{E\}-\mathscr\{A\})-0=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 $\displaystyle V\_1+V\_2=V$. 联合 $\displaystyle V\_1\cap V\_2=\left\\{0\right\\}$ 即知 $\displaystyle V=V\_1\oplus V\_2$.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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1577、 11、 (15 分) 设 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_5$ 是 $\displaystyle 5$ 维欧氏空间 $\displaystyle V$ 的一个标准正交基, $\displaystyle V\_1=L(\alpha\_1,\alpha\_2,\alpha\_3)$, 其中 \begin\{aligned\} \alpha\_1=\varepsilon\_1+\varepsilon\_5, \alpha\_2=\varepsilon\_1-\varepsilon\_2+\varepsilon\_4, \alpha\_3=2\varepsilon\_1+\varepsilon\_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\} 求 $\displaystyle V\_1$ 的一组标准正交基. (新疆大学2023年高等代数考研试题) [欧氏空间 ] [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\} (\alpha\_1,\alpha\_2,\alpha\_3)=(\varepsilon\_1,\cdots,\varepsilon\_5)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&2\\\\ 0&-1&1\\\\ 0&0&1\\\\ 0&1&0\\\\ 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\} 对 $\displaystyle A$ 的列向量施行 Gram-Schmidt 标准正交化过程得 \begin\{aligned\} \beta\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{2\}\}\\\\0\\\\0\\\\0\\\\\frac\{1\}\{\sqrt\{2\}\}\end\{array\}\right), \beta\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{10\}\}\\\\ -\frac\{2\}\{\sqrt\{10\}\}\\\\0\\\\\frac\{2\}\{\sqrt\{10\}\}\\\\-\frac\{1\}\{\sqrt\{10\}\}\end\{array\}\right), \beta\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{2\}\\\\\frac\{1\}\{2\}\\\\\frac\{1\}\{2\}\\\\0\\\\-\frac\{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\} 故 $\displaystyle V\_1$ 有一组标准正交基 \begin\{aligned\} \frac\{1\}\{\sqrt\{2\}\}\varepsilon\_1+\frac\{1\}\{\sqrt\{2\}\}\varepsilon\_5, \frac\{1\}\{\sqrt\{10\}\}\varepsilon\_1 -\frac\{2\}\{\sqrt\{10\}\}\varepsilon\_2+\frac\{2\}\{\sqrt\{10\}\}\varepsilon\_4-\frac\{1\}\{\sqrt\{10\}\}\varepsilon\_5, \frac\{1\}\{2\}\varepsilon\_1+\frac\{1\}\{2\}\varepsilon\_2+\frac\{1\}\{2\}\varepsilon\_3-\frac\{1\}\{2\}\varepsilon\_5. \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)/ --- 1578、 1、 填空题 (每题 6 分, 共 30 分). (1)、 在标准欧氏空间 $\displaystyle \mathbb\{R\}^3$ 中, $\displaystyle (1,0,-1)$ 与 $\displaystyle (1,1,0)$ 的夹角为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年高等代数考研试题) [欧氏空间 ] [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 $\displaystyle \alpha=(1,0,-1), \beta=(1,1,0)$, 则 $\displaystyle \alpha,\beta$ 的夹角 $\displaystyle \theta$ 满足 \begin\{aligned\} \cos\theta=\frac\{\left(\alpha,\beta\right)\}\{\sqrt\{(\alpha,\alpha)\}\sqrt\{(\beta,\beta)\}\}=\frac\{1\}\{\sqrt\{2\}\cdot \sqrt\{2\}\} =\frac\{1\}\{2\}\Rightarrow \theta=\frac\{\pi\}\{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)/ --- 1579、 5、 (15 分) 若欧氏空间 $\displaystyle \mathbb\{R\}[x]\_3=\left\\{\sum\_\{i=0\}^2 a\_ix^i; a\_i\in\mathbb\{R\}\right\\}$ 上的内积 $\displaystyle (\cdot,\cdot)$ 为 \begin\{aligned\} \left(f(x),g(x)\right)=\int\_\{-1\}^1 \frac\{f(x)g(x)\}\{\sqrt\{1-x^2\}\}\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 求 $\displaystyle \mathbb\{R\}[x]\_3$ 的一组标准正交基. (云南大学2023年高等代数考研试题) [欧氏空间 ] [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) / 先算出 $\displaystyle \mathbb\{R\}[x]\_3$ 的基 $\displaystyle 1,x,x^2$ 的度量矩阵 \begin\{aligned\} G=\left((x^\{i-1\},x^\{j-1\})\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\pi&0&\frac\{\pi\}\{2\}\\\\ 0&\frac\{\pi\}\{2\}&0\\\\ \frac\{\pi\}\{2\}&0&\frac\{3\pi\}\{8\}\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 $\displaystyle f(x)=a\_0+a\_1+a\_2x^2$, $\displaystyle g(x)=b\_0+b\_1x+b\_2x^2$ 的内积为 \begin\{aligned\} \left(f,g\right)=&(a\_0,a\_1,a\_2)G\left(\begin\{array\}\{cccccccccccccccccccc\}b\_0\\\\b\_1\\\\b\_2\end\{array\}\right)\\\\ =&\frac\{(8a\_0b\_0+4a\_2b\_0+4a\_1b\_1+4a\_0b\_2+3a\_2b\_2)\pi\}\{8\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 如此就会稍微更简单的对 $\displaystyle 1,x,x^2$ 施行 Gram-Schmidt 标准正交化过程得到 $\displaystyle \mathbb\{R\}[x]\_3$ 的一组标准正交基 [当然也要利用对称法求定积分] \begin\{aligned\} \frac\{1\}\{\sqrt\{\pi\}\},\quad \sqrt\{\frac\{2\}\{\pi\}\}x,\quad 2\sqrt\{\frac\{2\}\{\pi\}\} \left(x^2-\frac\{1\}\{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](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1580、 9、 (15 分) 设 $\displaystyle \sigma$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 上的线性变换. 证明: 存在 $\displaystyle V$ 上的正交变换 $\displaystyle \tau\_1,\tau\_2$ 使得线性变换 $\displaystyle \phi=\tau\_1\sigma\tau\_2$ 满足 $\displaystyle \phi(V)^\perp=\phi^\{-1\}(0)$. (云南大学2023年高等代数考研试题) [欧氏空间 ] [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)、 我们先给出实矩阵的奇异值分解. 设 $\displaystyle A$ 是 $\displaystyle m\times n$ 阶实矩阵. 则存在 $\displaystyle m\times n$ 阶非负实对角矩阵 (即所有行标和列标相同的元素均非负, 其余元素均为 $\displaystyle 0$) 与两个正交矩阵 $\displaystyle U,V$, 使得 $\displaystyle A=UDV$ (即矩阵 $\displaystyle A$ 的奇异值分解). 事实上, 对 $\displaystyle m+n$ 作数学归纳法. 当 $\displaystyle m+n=2$ 时, \begin\{aligned\} \exists\ a\in\mathbb\{R\},\mathrm\{ s.t.\} A=a \Rightarrow \left\\{\begin\{array\}\{llllllllllll\} a\geq 0\Rightarrow U=V=1, D=a,\\\\ \mbox\{或\}a < 0\Rightarrow U=-1, V=1, D=-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\} 设结论对阶数之和 $\displaystyle < m+n$ 的矩阵都成立. 设 $\displaystyle A$ 是 $\displaystyle m+n$ 阶矩阵. 当 $\displaystyle A=0$ 时, 结论自明. 当 $\displaystyle A\neq 0$ 时, 在有界闭集 $\displaystyle \left\\{x\in\mathbb\{R\}^n; x^\mathrm\{T\} x=\left|x\right|^2=1\right\\}$ 上的连续函数 $\displaystyle \sqrt\{x^\mathrm\{T\} A^\mathrm\{T\} Ax\}=\left|Ax\right|$ 有最大值 $\displaystyle M > 0$, 设在 $\displaystyle x\_1$ 处取得. 令 $\displaystyle y\_1=\frac\{Ax\_1\}\{M\}$, 则 \begin\{aligned\} \left|x\_1\right|=\left|y\_1\right|=1, Ax\_1=My\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 将 $\displaystyle x\_1$ 扩充为 $\displaystyle \mathbb\{R\}^n$ 的一组标准正交基 $\displaystyle x\_1, x\_2,\cdots, x\_n$, 将 $\displaystyle y\_1$ 扩充为 $\displaystyle y\_1,\cdots,y\_n$, 并令 \begin\{aligned\} U\_1=(x\_1,\cdots, x\_n), \quad V\_1=(y\_1,\cdots,y\_n), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle U\_1,V\_1$ 都是正交矩阵, 且 \begin\{aligned\} \exists\ \alpha\in \mathbb\{R\}^\{(n-1)\times 1\}, B\in \mathbb\{R\}^\{(n-1)\times (n-1)\},\mathrm\{ s.t.\} AU\_1=V\_1\left(\begin\{array\}\{cccccccccccccccccccc\}M&\alpha^\mathrm\{T\}\\\\ 0&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\} 注意到正交矩阵保持向量长度不变, 对 $\displaystyle \beta=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\ \alpha\end\{array\}\right)$, 我们有 $\displaystyle \left|U\_1 \frac\{\beta\}\{\left|\beta\right|\}\right|=1$, 而 \begin\{aligned\} M&\geq \left|AU\_1 \frac\{\beta\}\{\left|\beta\right|\}\right| =\left|V\_1\left(\begin\{array\}\{cccccccccccccccccccc\}M&\alpha^\mathrm\{T\}\\\\ 0&B\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\ \alpha\end\{array\}\right)\right| =\left|\left(\begin\{array\}\{cccccccccccccccccccc\}M&\alpha^\mathrm\{T\}\\\\ 0&B\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\ \alpha\end\{array\}\right)\right|\\\\ &=\left|\left(\begin\{array\}\{cccccccccccccccccccc\}M+\alpha^\mathrm\{T\} \alpha\\\\ B\alpha\end\{array\}\right)\right| \geq M+\alpha^\mathrm\{T\}\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \alpha^\mathrm\{T\}\alpha=0\Rightarrow \alpha=0\Rightarrow U\_1AV\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}M&0\\\\ 0&B\end\{array\}\right)$. 据归纳假设, 存在正交阵 $\displaystyle U\_2,V\_2$, 使得 $\displaystyle B=U\_2DV\_2$. 令 $\displaystyle U=U\_1^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&\\\\ &U\_2\end\{array\}\right), V=\left(\begin\{array\}\{cccccccccccccccccccc\}1&\\\\ 0&V\_2\end\{array\}\right)V\_1^\mathrm\{T\},$ 则 $\displaystyle U,V$ 都是正交阵, 且 \begin\{aligned\} A=U\_1^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}M&\\\\ &B\end\{array\}\right)V\_1^\mathrm\{T\}=U\left(\begin\{array\}\{cccccccccccccccccccc\}M&\\\\ &D\end\{array\}\right)V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 回到题目. 设 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组标准正交基. 设 \begin\{aligned\} \sigma(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)A . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由第 1 步知存在正交阵 $\displaystyle Q\_1,Q\_2$ 使得 \begin\{aligned\} Q\_1AQ\_2=\varLambda=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_r,0,\cdots,0), r=\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\} 取 \begin\{aligned\} \tau\_1(\varepsilon\_1,\cdots,\varepsilon\_n)=&(\varepsilon\_1,\cdots,\varepsilon\_n)Q\_1,\\\\ \tau\_2(\varepsilon\_1,\cdots,\varepsilon\_n)=&(\varepsilon\_1,\cdots,\varepsilon\_n)Q\_2,\\\\ \phi(\varepsilon\_1,\cdots,\varepsilon\_n)=&(\varepsilon\_1,\cdots,\varepsilon\_n)\varLambda, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle \phi=\tau\_1\sigma\tau\_2$, 且 \begin\{aligned\} &\phi(\varepsilon\_i)=\lambda\_i\varepsilon\_i, 1\leq i\leq r; \phi(\varepsilon\_i)=0, r+1\leq i\leq n\\\\ \Rightarrow&\mathrm\{im\} \phi=L(\varepsilon\_1,\cdots,\varepsilon\_r), \ker \phi=L(\varepsilon\_\{r+1\},\cdots,\varepsilon\_n)\\\\ \Rightarrow&\left(\mathrm\{im\}\phi\right)^\perp=\left\[L(\varepsilon\_1,\cdots,\varepsilon\_r)\right\]^\perp=L(\varepsilon\_\{r+1\},\cdots,\varepsilon\_n)=\ker\phi. \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)/ --- 1581、 6、 (10 分) 设 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 为欧氏空间 $\displaystyle V$ 的正交向量组, 证明: (0-15)、 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 线性无关; (0-16)、 若 $\displaystyle \alpha\_1,\cdots,\alpha\_s,\beta$ 线性相关, 且 $\displaystyle (\beta,\alpha\_i)=0, i=1,\cdots,s$, 则 $\displaystyle \beta=0$. (长安大学2023年高等代数考研试题) [欧氏空间 ] [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 $\displaystyle \beta=\sum\_\{i=1\}^s x\_i\alpha\_i$, 则 \begin\{aligned\} (\beta,\beta)=\left(\beta,\sum\_\{i=1\}^s x\_i\alpha\_i\right) =\sum\_\{i=1\}^s x\_i(\beta,\alpha\_i)=\sum\_\{i=1\}^s x\_i0=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\} 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)/ --- 1582、 10、 (20 分) 设 $\displaystyle V$ 是实数域上所有 $\displaystyle n$ 阶对称矩阵按矩阵加法与数乘运算构成的线性空间, 对任意的 $\displaystyle A,B\in V$, 定义 $\displaystyle (A,B)=\mathrm\{tr\}(AB)$, 其中 $\displaystyle (A,B)$ 表示 $\displaystyle A$ 与 $\displaystyle B$ 的内积, $\displaystyle \mathrm\{tr\}(AB)$ 表示 $\displaystyle AB$ 的迹. 证明: (0-17)、 $\displaystyle V$ 按上述定义的内积构成欧氏空间; (0-18)、 对于任意 $\displaystyle A,B\in V$, 有 $\displaystyle \mathrm\{tr\}\left\[(AB)^2\right\]\leq \mathrm\{tr\}(A^2)\mathrm\{tr\}(B^2)$; (0-19)、 若 $\displaystyle W=\left\\{A\in V; \mathrm\{tr\} A=0\right\\}$, 则 $\displaystyle \dim(W^\perp)=1$. (长安大学2023年高等代数考研试题) [欧氏空间 ] [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) / (0-20)、 由 (0-1)、 $\displaystyle (A,B)=\mathrm\{tr\}(AB)=\mathrm\{tr\}(BA)=(B,A)$, (0-2)、 $\displaystyle (kA,B)=\mathrm\{tr\}(kAB)=k\mathrm\{tr\}(AB)=k(A,B)$, (0-3)、 $\displaystyle (A+B,C)=\mathrm\{tr\}\left((A+B)C\right) =\mathrm\{tr\}(AC)+\mathrm\{tr\}(BC)=(A,C)+(B,C)$, (0-4)、 $\displaystyle (A,A)=\mathrm\{tr\}(AA)=\mathrm\{tr\}(A^\mathrm\{T\} A)=\sum\_\{i,j=1\}^n a\_\{ij\}^2\geq 0$, 且 \begin\{aligned\} (A,A)=0\Leftrightarrow a\_\{ij\}=0, \forall\ i,j\Leftrightarrow A=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle V$ 关于 $\displaystyle (A,B)$ 构成欧氏空间. (1)、 令 $\displaystyle C=AB-BA$, 则由 $\displaystyle A,B$ 实对称知 \begin\{aligned\} 0\leq&\mathrm\{tr\}(C^\mathrm\{T\} C) =\mathrm\{tr\}\left\[(AB-BA)^\mathrm\{T\} (AB-BA)\right\]\\\\ =&\mathrm\{tr\}\left\[(BA-AB)(AB-BA)\right\]\\\\ =&\mathrm\{tr\}\left(B\cdot A^2B-B\cdot ABA -ABAB+AB^2\cdot A\right)\\\\ =&\mathrm\{tr\}(A^2B^2)-\mathrm\{tr\}(ABAB)-\mathrm\{tr\}(ABAB)+\mathrm\{tr\}(A^2B^2)\\\\ =&2\left\[\mathrm\{tr\}(A^2B^2)-\mathrm\{tr\}(ABAB)\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle \dim V=n+(n-1)+\cdots+1=\frac\{n(n+1)\}\{2\}$, 而通过解 $\displaystyle a\_\{11\}+\cdots+a\_\{nn\}=0$ 知 $\displaystyle W$ 有一组基 \begin\{aligned\} -E\_\{11\}+E\_\{ii\}\left(2\leq i\leq n\right), E\_\{ij\}+E\_\{ji\}\left(1\leq i < j\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\} 而 $\displaystyle \dim W=(n-1)+\sum\_\{i=1\}^\{n-1\}(n-i)=\frac\{(n+2)(n-1)\}\{2\}$. 故 \begin\{aligned\} \dim W^\perp=\dim V-\dim W=\frac\{n(n+1)\}\{2\}-\frac\{(n+2)(n-1)\}\{2\}=1. \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)/ --- 1583、 (5)、 设 \begin\{aligned\} \alpha\_1=(1,2,1), \alpha\_2=(2,5,3), \alpha\_3=(1,4,3) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 为欧氏空间 $\displaystyle \mathbb\{R\}^3$ 中的向量, $\displaystyle W=\left < \alpha\_1,\alpha\_2,\alpha\_3\right > $ 为向量组 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 生成的线性子空间, 则 $\displaystyle W$ 的维数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$, $\displaystyle W$ 有一组标准正交基为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国科学技术大学2023年高等代数考研试题) [欧氏空间 ] [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^\mathrm\{T\}, \alpha\_2^\mathrm\{T\}, \alpha\_3^\mathrm\{T\})=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&1\\\\ 2&5&4\\\\ 1&3&3\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-3\\\\ 0&1&2\\\\ 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 $\displaystyle \dim W=2$, 且 $\displaystyle \alpha\_1,\alpha\_2$ 为 $\displaystyle W$ 的一组基, 将其标准正交化为 \begin\{aligned\} \left(\frac\{1\}\{\sqrt\{6\}\},\frac\{2\}\{\sqrt\{6\}\},\frac\{1\}\{\sqrt\{6\}\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{\sqrt\{2\}\},0,\frac\{1\}\{\sqrt\{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\} 就是 $\displaystyle W$ 的一组标准正交基.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)/ --- 1584、 (4)、 (40 分) 考虑标准欧氏空间 $\displaystyle \mathbb\{R\}^4$ 中的向量 $\displaystyle \beta=(4,-3,4,-1)$ 以及向量组 \begin\{aligned\} T=\left\\{ \begin\{array\}\{cc\}\alpha\_1=(0,1,2,1), &\alpha\_2=(1,-1,0,2), \\\\ \alpha\_3=(2,-1,-1,2), &\alpha\_4=(1,1,1,1), \alpha\_5=(2,0,1,3)\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\} 记向量组 $\displaystyle T$ 生成的线性子空间为 $\displaystyle V$. (4-1)、 试求 $\displaystyle T$ 的所有极大线性无关组. (4-2)、 试求 $\displaystyle V$ 中向量 $\displaystyle \alpha$ 使得向量 $\displaystyle \alpha-\beta$ 最短. (4-3)、 试证明: 存在 $\displaystyle \mathbb\{R\}^4$ 上非恒等的正交变换 $\displaystyle \mathscr\{A\}$, 使得对于 $\displaystyle V$ 中任一向量 $\displaystyle \gamma$, $\displaystyle \mathscr\{A\}(\gamma)=\gamma$ 总成立. (4-4)、 判断并讨论: 满足上述条件的 $\displaystyle \mathscr\{A\}$ 是否唯一? (中国科学技术大学2023年高等代数考研试题) [欧氏空间 ] [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) / (4-1)、 \begin\{aligned\} (\alpha\_1^\mathrm\{T\},\cdots,\alpha\_5^\mathrm\{T\})=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&2&1&2\\\\ 1&-1&-1&1&0\\\\ 2&0&-1&1&1\\\\ 1&2&2&1&3\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1&1\\\\ 0&1&0&-1&0\\\\ 0&0&1&1&1\\\\ 0&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\} $\displaystyle (I)$ 的前三行任取三列 ($C\_5^3=10$ 钟), 只要它的行列式 $\displaystyle \neq 0$, 则它对应的列就是一个极大无关组. 于是 $\displaystyle T$ 的所有极大线性无关组有 $\displaystyle 8$ 组: \begin\{aligned\} &\alpha\_1,\alpha\_2,\alpha\_3; \alpha\_1,\alpha\_2,\alpha\_4; \alpha\_1,\alpha\_2,\alpha\_5;\\\\ &\alpha\_1,\alpha\_3,\alpha\_4; \alpha\_1,\alpha\_4,\alpha\_5; \alpha\_2,\alpha\_3,\alpha\_4;\\\\ &\alpha\_2,\alpha\_3,\alpha\_5; \alpha\_3,\alpha\_4,\alpha\_5. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (4-2)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1\\\\\alpha\_2\\\\\alpha\_3\end\{array\}\right)$, $\displaystyle \alpha=\sum\_\{i=1\}^3 x\_i\alpha\_i=xA$ 为所求, 其中 $\displaystyle x=(x\_1,x\_2,x\_3)$, 则 \begin\{aligned\} &(\beta-\alpha,\alpha\_i)=0, 1\leq i\leq 3 \Leftrightarrow (\beta-\alpha)\alpha\_i^\mathrm\{T\}=0, 1\leq i\leq 3\\\\ \Leftrightarrow&(\beta-xA)A^\mathrm\{T\}=0\Leftrightarrow xAA^\mathrm\{T\}=\beta A^\mathrm\{T\}\\\\ \Leftrightarrow&AA^\mathrm\{T\} x^\mathrm\{T\} =A\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\} 由 \begin\{aligned\} (AA^\mathrm\{T\}, A\beta^\mathrm\{T\})\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&\frac\{2\}\{3\}\\\\ 0&1&0&\frac\{1\}\{3\}\\\\ 0&0&1&\frac\{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\} 知 \begin\{aligned\} x=\left(\frac\{2\}\{3\},\frac\{1\}\{3\},\frac\{1\}\{3\}\right), \alpha=xA=(1,0,1,2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (4-3)、 取一 $\displaystyle \beta-\alpha=(3,-3,3,-3)$ 为法向量的镜面反射 \begin\{aligned\} \mathscr\{A\}(\gamma)=\gamma-2(\gamma,\eta)\eta,\quad \forall\ \gamma\in\mathbb\{R\}^4, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle \eta=\frac\{\beta-\alpha\}\{|\beta-\alpha|\}$, 则 $\displaystyle \mathscr\{A\}$ 保持 $\displaystyle V$ 中元素不动. (4-4)、 满足上述条件的 $\displaystyle \mathscr\{A\}$ 唯一. 因为 $\displaystyle \dim V=3$, $\displaystyle \dim V^\perp=1$, $\displaystyle \mathscr\{A\}$ 保持 $\displaystyle V$ 中元素不动, 不是恒同映射, 而将 $\displaystyle V^\perp$ 中的向量反向, 只能是第 iii 问中的那个镜面反射.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)/ --- 1585、 10、 设 $\displaystyle V$ 是内积空间, $\displaystyle f$ 是 $\displaystyle V$ 上一线性变换且保持向量的夹角不变. 证明: 存在 $\displaystyle \lambda > 0$, 使得 $\displaystyle \lambda f$ 是正交变换. (中国科学院大学2023年高等代数考研试题) [欧氏空间 ] [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 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组标准正交基, 则当 $\displaystyle i\neq j$ 时, \begin\{aligned\} \angle \left(f \varepsilon\_i,f \varepsilon\_j\right)=\angle \left(\varepsilon\_i,\varepsilon\_j\right)=0\Rightarrow \left < f \varepsilon\_i,f \varepsilon\_j\right > =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\} \left\Vert f \varepsilon\_i\right\Vert ^2 =&\left < f \varepsilon\_i,f \varepsilon\_i\right > \\\\ =&\left|f \varepsilon\_i,f (\varepsilon\_i+\varepsilon\_j)\right|\\\\ =&\left\Vert f \varepsilon\_i\right\Vert \cdot \left\Vert f (\varepsilon\_i+\varepsilon\_j)\right\Vert \cdot \cos\angle \left(f \varepsilon\_i,f (\varepsilon\_i+\varepsilon\_j)\right)\\\\ =&\left\Vert f \varepsilon\_i\right\Vert \cdot \left\Vert f \varepsilon\_i+f \varepsilon\_j\right\Vert \cdot \cos\angle \left(\varepsilon\_i,\varepsilon\_i+\varepsilon\_j\right)\\\\ =&\left\Vert f \varepsilon\_i\right\Vert \cdot \sqrt\{\left\Vert f \varepsilon\_i\right\Vert ^2+\left\Vert f \varepsilon\_j\right\Vert ^2\} \cdot\frac\{\left < \varepsilon\_i,\varepsilon\_i+\varepsilon\_j\right > \}\{\left\Vert \varepsilon\_i\right\Vert \cdot \left\Vert \varepsilon\_i+\varepsilon\_j\right\Vert \}\\\\ =&\left\Vert f \varepsilon\_i\right\Vert \cdot \sqrt\{\left\Vert f \varepsilon\_i\right\Vert ^2+\left\Vert f \varepsilon\_j\right\Vert ^2\}\cdot\frac\{1\}\{\sqrt\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 整理后发现 \begin\{aligned\} \left\Vert f \varepsilon\_i\right\Vert =\left\Vert f \varepsilon\_j\right\Vert . \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\} \left(\begin\{array\}\{cccccccccccccccccccc\}(f\varepsilon\_i,f\varepsilon\_j)\end\{array\}\right)=\frac\{1\}\{\lambda^2\}E,\quad \lambda=\frac\{1\}\{\left\Vert f \varepsilon\_i\right\Vert \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这说明 $\displaystyle (\lambda f \varepsilon\_1,\cdots,\lambda f \varepsilon\_n)$ 是正交阵, 而 $\displaystyle \lambda f $ 为正交变换.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)/ --- 1586、 10、 设 $\displaystyle C[-\pi,\pi]$ 表示闭区间 $\displaystyle [-\pi,\pi]$ 上的所有连续函数组成的实线性空间, 其上定义内积如下: \begin\{aligned\} \left(f(x),g(x)\right)=\int\_\{-\pi\}^\pi f(x)g(x)\mathrm\{ d\} x, \forall\ f(x), g(x)\in C[-\pi,\pi]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 如此, $\displaystyle C[-\pi,\pi]$ 是一个实内积空间. 给定向量组 \begin\{aligned\} 1,\cos x, \sin x, \cos 2x, \sin 2x, \cdots,\cos nx, \sin nx. \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)、 上述向量组是正交向量组; (2)、 求由上述向量组生成的子空间的一组标准正交基. (中国矿业大学(北京)2023年高等代数考研试题) [欧氏空间 ] [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 $\displaystyle \forall\ k,l\in\mathbb\{N\}, k\neq l$, 有 $\displaystyle k+l > 0$, 而 \begin\{aligned\} \int\_\{-\pi\}^\pi \cos kx\cos lx\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_\{-\pi\}^\pi [\cos(k+l)x+\cos(k-l)x]\mathrm\{ d\} x=0,\\\\ \int\_\{-\pi\}^\pi \cos kx\sin lx\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_\{-\pi\}^\pi [\sin(k+l)x-\sin(k-l)x]\mathrm\{ d\} x=0,\\\\ \int\_\{-\pi\}^\pi \sin kx\sin lx\mathrm\{ d\} x =-\frac\{1\}\{2\}\int\_\{-\pi\}^\pi [\cos(k+l)x-\cos(k-l)x]\mathrm\{ d\} x=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故题中向量组是正交向量组. (2)、 由 $\displaystyle \int\_\{-\pi\}^\pi 1^2\mathrm\{ d\} x=2\pi$, \begin\{aligned\} \int\_\{-\pi\}^\pi \cos^2kx\mathrm\{ d\} x=&\int\_\{-\pi\}^\pi \frac\{1+\cos 2kx\}\{2\}\mathrm\{ d\} x=\pi,\\\\ \int\_\{-\pi\}^\pi \sin^2kx\mathrm\{ d\} x=&\int\_\{-\pi\}^\pi \frac\{1-\cos2kx\}\{2\}\mathrm\{ d\} x=\pi \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\} \frac\{1\}\{\sqrt\{2\pi\}\}, \frac\{\cos x\}\{\sqrt\{\pi\}\}, \frac\{\sin x\}\{\sqrt\{\pi\}\}, \frac\{\cos 2x\}\{\sqrt\{\pi\}\}, \frac\{\sin 2x\}\{\sqrt\{\pi\}\}, \cdots, \frac\{\cos nx\}\{\sqrt\{\pi\}\}, \frac\{\sin nx\}\{\sqrt\{\pi\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 是题中向量组生成的子空间的一组标准正交基.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)/ --- 1587、 (9)、 设 $\displaystyle V$ 是一个 $\displaystyle 8$ 维欧氏空间, $\displaystyle \alpha$ 是 $\displaystyle V$ 中的一个固定非零向量, 则 $\displaystyle V$ 中与 $\displaystyle \alpha$ 正交的所有向量组成的子空间的维数是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国矿业大学(徐州)2023年高等代数考研试题) [欧氏空间 ] [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) / 将 $\displaystyle \eta\_1=\frac\{\alpha\}\{|\alpha|\}$ 扩充为 $\displaystyle V$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_8$, 则 \begin\{aligned\} \left\\{\alpha\right\\}^\perp=L(\eta\_2,\cdots,\eta\_8)\Rightarrow \dim \left\\{\alpha\right\\}^\perp=7. \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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