## 张祖锦2023年数学专业真题分类70天之第68天
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1542、 7、 (1)、 给定 $\displaystyle \alpha\_1,\alpha\_2$, 求 $\displaystyle \alpha\_3$, 使得 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 是线性空间 $\displaystyle V$ 的一组基. (2)、 给定一个矩阵 $\displaystyle A$, 定义 $\displaystyle (\xi,\eta)=\xi^\mathrm\{T\} A\eta$, 求 $\displaystyle V^\perp$ 的一组基. [题目不全, 跟锦数学微信公众号没法做哦.] (东南大学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) / [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](
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---
1543、 10、 设 $\displaystyle \mathscr\{A\}$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 上的对称变换, 即
\begin\{aligned\} \left(\mathscr\{A\}\alpha,\beta\right)=\left(\alpha,\mathscr\{A\}\beta\right),\quad \forall\ \alpha,\beta\in V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \mathscr\{A\}$ 的特征值全为实数, 且
\begin\{aligned\} \min\_\{0\neq\alpha\in V\}\frac\{\left(\alpha,\mathscr\{A\}\alpha\right)\}\{(\alpha,\alpha)\}=\lambda\_1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \lambda\_1$ 为 $\displaystyle \mathscr\{A\}$ 的最小特征值. (东南大学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 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组标准正交基, $\displaystyle \mathscr\{A\}$ 在该基下的矩阵的矩阵为 $\displaystyle A$, 则由
\begin\{aligned\} &(\mathscr\{A\}\varepsilon\_i,\varepsilon\_j)=\left(\sum\_k a\_\{ki\}\varepsilon\_k,\varepsilon\_j\right)=a\_\{ji\},\\\\ &(\varepsilon\_i,\mathscr\{A\}\varepsilon\_j)=\left(\varepsilon\_i,\sum\_k a\_\{kj\}\varepsilon\_k\right)=a\_\{ij\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a\_\{ij\}=a\_\{ji\}$, $\displaystyle A$ 是实对称矩阵. 从而存在正交矩阵 $\displaystyle P=(\eta\_1,\cdots,\eta\_n)$, 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\in\mathbb\{R\}, \lambda\_1\leq \cdots\leq \lambda\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle (\alpha\_1,\cdots,\alpha\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)P$, 则由 $\displaystyle P$ 正交知 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 是 $\displaystyle V$ 的一组标准正交基, 且
\begin\{aligned\} &\mathscr\{A\}(\alpha\_1,\cdots,\alpha\_n) =\mathscr\{A\}(\varepsilon\_1,\cdots,\varepsilon\_n)P\\\\ =&(\varepsilon\_1,\cdots,\varepsilon\_n)AP =(\alpha\_1,\cdots,\alpha\_n)\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\}
故 $\displaystyle \mathscr\{A\}$ 的特征值为 $\displaystyle \lambda\_i\in\mathbb\{R\}$, 对应的单位特征向量为 $\displaystyle \alpha\_i$. 进一步, 对 $\displaystyle \forall\ 0\neq \alpha=\sum\_i x\_i\varepsilon\_i\in V$,
\begin\{aligned\} (\alpha,\mathscr\{A\}\alpha)=&\left(\sum\_i x\_i\varepsilon\_i, \sum\_j x\_j\mathscr\{A\}\varepsilon\_j\right) =\left(\sum\_i x\_i\varepsilon\_i, \sum\_j x\_j\lambda\_j\varepsilon\_j\right)\\\\ =&\sum\_i \lambda\_i x\_i^2 \geq \sum\_i \lambda\_1 x\_i^2 \geq \lambda\_1 \sum\_i x\_i^2=\lambda\_1(\alpha,\alpha). \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\} (\alpha\_1,\mathscr\{A\}\alpha\_1)=(\alpha\_1,\lambda\_1\alpha\_1)=\lambda\_1(\alpha\_1,\alpha\_1) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \min\_\{0\neq\alpha\in V\}\frac\{\left(\alpha,\mathscr\{A\}\alpha\right)\}\{(\alpha,\alpha)\}=\lambda\_1$.跟锦数学微信公众号. [在线资料](
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1544、 (3)、 若 $\displaystyle 3$ 维欧氏空间中, 内积在基 $\displaystyle \varepsilon\_1,\varepsilon\_2,\varepsilon\_3$ 下的度量矩阵为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ -1&2&0\\\\ 0&0&3\end\{array\}\right)$, 则向量 $\displaystyle \alpha=2\varepsilon\_1+\varepsilon\_2-\varepsilon\_3$ 的模 $\displaystyle |\alpha|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (哈尔滨工程大学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 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ -1&2&0\\\\ 0&0&3\end\{array\}\right), x=(2,1,-1)^\mathrm\{T\}$,
\begin\{aligned\} |\alpha|=&\sqrt\{(\alpha,\alpha)\} =\sqrt\{\left(\sum\_i x\_i\varepsilon\_i,\sum\_j x\_j\varepsilon\_j\right)\}\\\\ =&\sqrt\{\sum\_\{ij\}(\varepsilon\_i,\varepsilon\_j)x\_ix\_j\} =\sqrt\{x^\mathrm\{T\} Ax\}=\sqrt\{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](
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---
1545、 (5)、 设 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 为欧氏空间 $\displaystyle V$ 的一组基,
\begin\{aligned\} \beta\_1=\alpha\_1, \beta\_2=\alpha\_1+\alpha\_2, \beta\_3=\alpha\_1+\alpha\_2+\alpha\_3, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若内积在基 $\displaystyle \alpha\_1, \alpha\_2, \alpha\_3$ 下的度量矩阵为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}4&1&2\\\\ 1&5&3\\\\ 2&3&6\end\{array\}\right)$, 则内积在基 $\displaystyle \beta\_1,\beta\_2,\beta\_3$ 下的度量矩阵为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (哈尔滨工程大学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 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}4&1&2\\\\ 1&5&3\\\\ 2&3&6\end\{array\}\right)$, 则
\begin\{aligned\} (\beta\_1,\beta\_2,\beta\_3)=(\alpha\_1,\alpha\_2,\alpha\_3)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&1&1\\\\ 0&0&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} (\beta\_i,\beta\_j)=\left(\sum\_k t\_\{ki\}\varepsilon\_k, \sum\_l t\_\{lj\}\varepsilon\_l\right) =\sum\_\{kl\}t\_\{ki\}(\varepsilon\_k,\varepsilon\_l)t\_\{lj\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明内积在基 $\displaystyle \beta\_1,\beta\_2,\beta\_3$ 下的度量矩阵为
\begin\{aligned\} T^\mathrm\{T\} AT=\left(\begin\{array\}\{cccccccccccccccccccc\}4&5&7\\\\ 5&11&16\\\\ 7&16&27\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](
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1546、 9、 若 $\displaystyle A$ 为实数域上的方阵, $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 为 $\displaystyle A$ 的列向量组. 证明以下命题等价: (1)、 $\displaystyle A$ 为正交矩阵; (2)、 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 为一组规范正交基; (3)、 对 $\displaystyle \forall\ X,Y\in\mathbb\{R\}^n$, 有 $\displaystyle (AX,AY)=(X,Y)$. (哈尔滨工业大学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)、 $\displaystyle (1)\Leftrightarrow (2)$: 由
\begin\{aligned\} A^\mathrm\{T\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\\vdots\\\\\alpha\_n^\mathrm\{T\}\end\{array\}\right)(\alpha\_1,\cdots,\alpha\_n)=(\alpha\_i^\mathrm\{T\} \alpha\_j)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &A\mbox\{正交\}\Leftrightarrow A^\mathrm\{T\} A=E\_n\Leftrightarrow (\alpha\_i,\alpha\_j)=\alpha\_i^\mathrm\{T\} \alpha\_j=\delta\_\{ij\}=\left\\{\begin\{array\}\{llllllllllll\}1,&i=j\\\\ 0,&i\neq j\end\{array\}\right.\\\\ \Leftrightarrow& \mbox\{$\alpha\_1,\cdots,\alpha\_n$ 为一组规范正交基\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle (1)\Rightarrow (3)$:
\begin\{aligned\} (AX,AY)=(AX)^\mathrm\{T\} (AY)=X^\mathrm\{T\} A^\mathrm\{T\} AY=X^\mathrm\{T\} Y=(X,Y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 $\displaystyle (3)\Rightarrow (1)$: 取 $\displaystyle X=e\_i, Y=e\_j$, 则
\begin\{aligned\} \delta\_\{ij\}=(e\_i,e\_j)=(Ae\_i,Ae\_j)=(\alpha\_i,\alpha\_j)=\alpha\_i^\mathrm\{T\} \alpha\_j. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (I)$ 知 $\displaystyle E\_n=A^\mathrm\{T\} A\Rightarrow A$ 正交.跟锦数学微信公众号. [在线资料](
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---
1547、 10、 设 $\displaystyle \sigma$ 为 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 上的一个线性变换, $\displaystyle V$ 上的线性变换 $\displaystyle \sigma^\star$ 称为 $\displaystyle \sigma$ 的伴随变换, 若
\begin\{aligned\} \left(\sigma(\alpha),\beta\right)=\left(\alpha,\sigma^\star(\beta)\right),\quad \forall\ \alpha,\beta\in V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 设 $\displaystyle \sigma$ 在 $\displaystyle V$ 的一组基标准正交基下的矩阵为 $\displaystyle A$. 证明: $\displaystyle \sigma^\star$ 在此基下的矩阵为 $\displaystyle A^\mathrm\{T\}$. (2)、 证明: $\displaystyle \sigma^\star(V)=\left\[\sigma^\{-1\}(0)\right\]^\perp$, 其中 $\displaystyle \sigma^\star(V)$ 为 $\displaystyle \sigma^\star$ 的值域, $\displaystyle \sigma^\{-1\}(0)$ 为 $\displaystyle \sigma$ 的核. (合肥工业大学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 $\displaystyle \sigma,\sigma^\star$ 在该基下的矩阵分别为 $\displaystyle A,B$, 则
\begin\{aligned\} &\sigma(\varepsilon\_i)=\sum\_k a\_\{ki\}\varepsilon\_k, \sigma^\star(\varepsilon\_j)=\sum\_l b\_\{lj\}\varepsilon\_l,\\\\ &\left(\sigma(\varepsilon\_i),\varepsilon\_j\right)=a\_\{ji\}, \left(\varepsilon\_i,\sigma^\star(\varepsilon\_j)\right)=b\_\{ij\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由题设即知
\begin\{aligned\} &a\_\{ji\}=b\_\{ij\}\Rightarrow B=A^\mathrm\{T\}\\\\ \Rightarrow& \dim \mathrm\{im\} \sigma^\star =\mathrm\{rank\} B=\mathrm\{rank\} A^\mathrm\{T\} =n-\left\[n-\mathrm\{rank\} A\right\]=\dim (\ker \sigma)^\perp. \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\} \alpha\in \mathrm\{im\} \sigma^\star&\Rightarrow \exists\ \beta\in V,\mathrm\{ s.t.\} \alpha=\sigma^\star(\beta)\\\\ &\Rightarrow \forall\ \gamma\in \ker \sigma, (\gamma,\alpha)=\left(\gamma,\sigma^\star(\beta)\right) =\left(\sigma(\gamma),\beta\right) =\left(0,\beta\right)=0\\\\ &\Rightarrow \alpha\in (\ker\sigma)^\perp \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \mathrm\{im\} \sigma^\star\subset (\ker \sigma)^\perp$. 联合
\begin\{aligned\} \dim \mathrm\{im\}\sigma^\star=\dim (\ker\sigma)^\perp \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知 $\displaystyle \mathrm\{im\} \sigma^\star=(\ker \sigma)^\perp$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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1548、 9、 设 $\displaystyle A,B$ 为 $\displaystyle n$ 阶实对称矩阵, 且 $\displaystyle AB=BA$. 求证: $\displaystyle A,B$ 有公共特征向量组成欧氏空间 $\displaystyle \mathbb\{R\}^n$ 的标准正交基. (河北工业大学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 $\displaystyle A$ 实对称知存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}\left(\lambda\_1E\_\{n\_1\},\cdots,\lambda\_sE\_\{n\_s\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \lambda\_i$ 互异. 而
\begin\{aligned\} AB=BA\Rightarrow& P^\mathrm\{T\} AP\cdot P^\mathrm\{T\} BP=P^\mathrm\{T\} BP\cdot P^\mathrm\{T\} AP\\\\ \Rightarrow& \mathrm\{diag\}\left(\lambda\_1E\_\{n\_1\},\cdots,\lambda\_sE\_\{n\_s\}\right) \tilde\{B\}=\tilde\{B\}\mathrm\{diag\}\left(\lambda\_1E\_\{n\_1\},\cdots,\lambda\_sE\_\{n\_s\}\right)\\\\ &\left(\tilde\{B\}=P^\mathrm\{T\} BP=(\tilde\{B\}\_\{ij\}),\mbox\{分块与 $\displaystyle \mathrm\{diag\}\left(\lambda\_1E\_\{n\_1\},\cdots,\lambda\_sE\_\{n\_s\}\right)$ 相同\}\right)\\\\ \Rightarrow&\lambda\_i\tilde\{B\}\_\{ij\}=\tilde\{B\}\_\{ij\}\lambda\_j \Rightarrow\tilde\{B\}\_\{ij\}=0\left(\forall\ i\neq j\right)\\\\ \Rightarrow&\tilde\{B\}=\mathrm\{diag\}(\tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进一步,
\begin\{aligned\} B\mbox\{实对称\}&\Rightarrow \tilde\{B\}=P^\mathrm\{T\} BP\mbox\{实对称\} \Rightarrow \tilde\{B\}\_\{ii\}\mbox\{实对称\}\\\\ &\Rightarrow \mbox\{存在正交阵 $\displaystyle Q\_i$, 使得 $\displaystyle Q\_i^\mathrm\{T\} \tilde\{B\}\_\{ii\}Q\_i=D\_i,D\_i$ 为对角阵\}. \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\} U=P\mathrm\{diag\}(Q\_1,\cdots,Q\_s)=(\eta\_1,\cdots,\eta\_n), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle U$ 正交, 且
\begin\{aligned\} U^\{-1\}AU&=\mathrm\{diag\}\left(\lambda\_1E\_\{n\_1\},\cdots,\lambda\_sE\_\{n\_s\}\right),\\\\ U^\{-1\}BU&=\mathrm\{diag\}(D\_1,\cdots,D\_s). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle \eta\_1,\cdots,\eta\_n$ 是 $\displaystyle A,B$ 的公共特征向量, 且是 $\displaystyle \mathbb\{R\}^n$ 的标准正交基.跟锦数学微信公众号. [在线资料](
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---
1549、 4、 设 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 是 $\displaystyle n$ 维欧氏空间的向量组, 向量 $\displaystyle \alpha\_i$ 与 $\displaystyle \alpha\_j$ 的内积记为 $\displaystyle (\alpha\_i,\alpha\_j)$. 证明: $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 线性无关的充要条件是矩阵 $\displaystyle A=(a\_\{ij\})\_\{s\times s\}$ 可逆. (湖南大学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)、 $\displaystyle \Leftarrow$: 设
\begin\{aligned\} k\_1\alpha\_1+\cdots+k\_s\alpha\_s=0, \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\} 0&=\alpha\_1^\mathrm\{T\} (k\_1\alpha\_1+\cdots+k\_s\alpha\_s),\\\\ 0&=\cdots,\\\\ 0&=\alpha\_s^\mathrm\{T\} (k\_1\alpha\_1+\cdots+k\_s\alpha\_s). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
也即
\begin\{aligned\} A\left(\begin\{array\}\{cccccccccccccccccccc\}k\_1\\\\\vdots\\\\k\_s\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\\vdots\\\\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 $\displaystyle A$ 可逆知 $\displaystyle k\_1=\cdots=k\_s=0$. (2)、 $\displaystyle \Rightarrow$: 设 $\displaystyle A x=0$, 则
\begin\{aligned\} 0&=x^\mathrm\{T\} A x=x^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\ \vdots \\\\ \alpha\_s^\mathrm\{T\}\end\{array\}\right)\left(\alpha\_1,\cdots, \alpha\_s\right)x\\\\ &=y^\mathrm\{T\} y\left(y=(\alpha\_1,\cdots,\alpha\_s)x\right)\\\\ &=|y|^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} &0=y=(\alpha\_1,\cdots,\alpha\_s)\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\\vdots\\\\x\_s\end\{array\}\right)=x\_1\alpha\_1+\cdots+x\_s\alpha\_s\\\\ \Rightarrow&x\_1=\cdots=x\_s=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle A x=0$ 只有零解, 而 $\displaystyle \mathrm\{rank\} A=s$, $\displaystyle A$ 可逆.跟锦数学微信公众号. [在线资料](
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---
1550、 8、 设 $\displaystyle V=C[0,1]$ 表示闭区间 $\displaystyle [0,1]$ 上的连续实值函数全体构成的线性空间. 定义 $\displaystyle V$ 上二元函数如下:
\begin\{aligned\} \left(f(x),g(x)\right)=\int\_0^1 f(x)g(x)x^2\mathrm\{ d\} x, \forall\ f(x),g(x)\in V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 证明: 上述二元函数是线性空间 $\displaystyle V$ 的一个内积; (2)、 证明: $\displaystyle V$ 是无限维线性空间; (3)、 对任意 $\displaystyle 1\leq i\leq j\leq n$, 令 $\displaystyle a\_\{ij\}=\frac\{1\}\{i+j+1\}$, 证明: $\displaystyle A=(a\_\{ij\})\_\{n\times n\}$ 为正定矩阵. (湖南大学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)、 由
\begin\{aligned\} &\left(f,f\right)=\int\_0^1 f^2(x)x^2\mathrm\{ d\} x\geq 0,\\\\ &\left(f,f\right)=0\Leftrightarrow f^2(x)x^2\equiv 0\Leftrightarrow f^2(x)\equiv 0\Leftrightarrow f(x)\equiv 0; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \left(f,g\right)=\left(g,f\right)$;
\begin\{aligned\} &\left(kf+k\_1f\_1,g\right)=\int\_0^1 [kf(x)+k\_1f\_1(x)]g(x)x^2\mathrm\{ d\} x\\\\ =&k\int\_0^1 f(x)g(x)x^2\mathrm\{ d\} x+k\_1 \int\_0^1 f\_1(x)g(x)x^2\mathrm\{ d\} x\\\\ =&k\left(f,g\right)+k\_1\left(f\_1,g\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知题中二元函数是线性空间 $\displaystyle V$ 的一个内积. (2)、 先证第 3 问. 对 $\displaystyle \forall\ 0\neq x=(x\_1,\cdots,x\_n)^\mathrm\{T\}\in\mathbb\{R\}^n$, 多项式 $\displaystyle \sum\_i x\_i t^i\neq 0$, 而
\begin\{aligned\} x^\mathrm\{T\} Ax=&\sum\_\{i,j\}\frac\{1\}\{i+j+1\}x\_ix\_j =\sum\_\{i,j\}x\_ix\_j\int\_0^1 t^\{i+j\}\mathrm\{ d\} t\\\\ &=\int\_0^1 \sum\_i x\_it^i\cdot \sum\_j x\_jt^j\mathrm\{ d\} t =\int\_0^1 \left|\sum\_i x\_it^i\right|^2\mathrm\{ d\} t > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 正定. (3)、 为证第 2 问. 我们先给出一个结论. 设 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 是 $\displaystyle V$ 中向量, 且矩阵 $\displaystyle A=\left((\alpha\_i,\alpha\_j)\right)\_\{s\times s\}$ 可逆, 则 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 线性无关. 事实上, 设
\begin\{aligned\} k\_1\alpha\_1+\cdots+k\_s\alpha\_s=0, \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\} 0&=\alpha\_1^\mathrm\{T\} (k\_1\alpha\_1+\cdots+k\_s\alpha\_s),\\\\ 0&=\cdots,\\\\ 0&=\alpha\_s^\mathrm\{T\} (k\_1\alpha\_1+\cdots+k\_s\alpha\_s). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
也即
\begin\{aligned\} A\left(\begin\{array\}\{cccccccccccccccccccc\}k\_1\\\\\vdots\\\\k\_s\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\\vdots\\\\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 $\displaystyle A$ 可逆知 $\displaystyle k\_1=\cdots=k\_s=0$. (4)、 往证第 2 问. 对任意的 $\displaystyle n$, 考虑 $\displaystyle V$ 中的向量组 $\displaystyle 1,x,\cdots,x^\{n-1\}$, 由
\begin\{aligned\} (x^\{i-1\},x^\{j-1\})=\int\_0^1 x^\{i-1\}x^\{j-1\}x^2\mathrm\{ d\} x=\frac\{1\}\{i+j+1\}=a\_\{ij\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及第 2 步知 $\displaystyle \left((x^\{i-1\},x^\{j-1\})\right)\_\{1\leq i,j\leq n\}$ 正定, 而可逆. 据第 3 步知 $\displaystyle 1,x,\cdots,x^\{n-1\}$ 在 $\displaystyle V$ 中线性无关, $\displaystyle \dim V\geq n$. 由 $\displaystyle n$ 的任意性知 $\displaystyle \dim V=+\infty$.跟锦数学微信公众号. [在线资料](
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---
1551、 1、 填空题. (1)、 设 $\displaystyle \alpha=(-1,2,0,2), \beta=(1,1,1,1)$ 是欧氏空间 $\displaystyle \mathbb\{R\}^4$ 中的向量, 则 $\displaystyle \alpha,\beta$ 的夹角为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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) / 所求夹角 $\displaystyle \theta$ 满足
\begin\{aligned\} \cos\theta=\frac\{(\alpha,\beta)\}\{|\alpha|\cdot |\beta|\} =\frac\{3\}\{3\cdot 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\}
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---
1552、 (2)、 设 $\displaystyle \alpha=(-1,2,0,2), \beta=(1,1,1,1)$, $\displaystyle W$ 是由它们生成的子空间, 则向量 $\displaystyle \gamma=(0,1,0,1)$ 在 $\displaystyle W$ 中的正交投影为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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) / 设所求为 $\displaystyle \delta$, 则
\begin\{aligned\} &\exists\ k,l,\mathrm\{ s.t.\} \delta=k\alpha+l\beta; (\gamma-\delta,\alpha)=0, (\gamma-\delta,\beta)=0\\\\ \Rightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}(\alpha,\alpha)&(\alpha,\beta)\\\\ (\alpha,\beta)&(\beta,\beta)\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}k\\\\l\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}(\alpha,\gamma)\\\\ (\beta,\gamma)\end\{array\}\right)\\\\ \Leftrightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}9&3\\\\ 3&4\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}k\\\\l\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}4\\\\2\end\{array\}\right)\\\\ \Rightarrow&k=\frac\{10\}\{27\}, l=\frac\{2\}\{9\} \Rightarrow \delta=k\alpha+l\beta=\frac\{1\}\{27\}(-4,26,6,26). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1553、 (2)、 已知 $\displaystyle \mathscr\{A\}: U\to U$ 是酉空间 $\displaystyle U$ 上的线性变换且满足
\begin\{aligned\} \left(v,\mathscr\{A\} v\right)\in \mathbb\{R\}, \quad \forall\ v\in U. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \mathscr\{A\}$ 是 Hermite 变换, 即
\begin\{aligned\} \left(\mathscr\{A\} v, w\right)=\left(v,\mathscr\{A\} w\right),\quad \forall\ v,w\in U. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华东师范大学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) / 对 $\displaystyle \forall\ v,w\in U$,
\begin\{aligned\} &\left(v+w,\mathscr\{A\} (v+w)\right) =(v,\mathscr\{A\} v)+(v,\mathscr\{A\} w)+(w,\mathscr\{A\} v)+(w,\mathscr\{A\} w)\\\\ =&(v,\mathscr\{A\} v)+(v,\mathscr\{A\} w)+\overline\{(\mathscr\{A\} v,w)\}+(w,\mathscr\{A\} w), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} &\left(v+w,\mathscr\{A\} (v+w)\right) =\overline\{(v+w,\mathscr\{A\} (v+w)\} =\left(\mathscr\{A\} (v+w),v+w\right)\\\\ =&(\mathscr\{A\} v,v)+(\mathscr\{A\} v,w)+(\mathscr\{A\} w,v)+(\mathscr\{A\} w,w)\\\\ =&\overline\{\left(\mathscr\{A\} v,v\right)\} +(\mathscr\{A\} v,w) +\overline\{(v,\mathscr\{A\} w)\}+\overline\{(\mathscr\{A\} w,w)\}\\\\ =&(v,\mathscr\{A\} v)+(\mathscr\{A\} v,w)+\overline\{(v,\mathscr\{A\} w)\}+(w,\mathscr\{A\} w). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} &(v,\mathscr\{A\} w)+\overline\{(\mathscr\{A\} v,w)\}=(\mathscr\{A\} v,w)+\overline\{(v,\mathscr\{A\} w)\}\\\\ \Rightarrow& \mathrm\{ Im\} (v,\mathscr\{A\} w)=\mathrm\{ Im\} (\mathscr\{A\} v,w). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上式用 $\displaystyle \mathrm\{ i\} v$ 代替 $\displaystyle v$ 得
\begin\{aligned\} &\mathrm\{ Im\}(\mathrm\{ i\} v,\mathscr\{A\} w)=\mathrm\{ Im\}\left\[\mathrm\{ i\} (v,\mathscr\{A\} w)\right\]=\mathrm\{ Re\} (v,\mathscr\{A\} w)\\\\ =&\mathrm\{ Im\} \left(\mathscr\{A\} (iv),w\right) =\mathrm\{ Im\} \left\[\mathrm\{ i\} (\mathscr\{A\} v,w)\right\]=\mathrm\{ Re\}(\mathscr\{A\} v,w). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
至此, 我们证明了 $\displaystyle (v,\mathscr\{A\} w), (\mathscr\{A\} v,w)$ 的实部和虚部各自相等, 而 $\displaystyle (v,\mathscr\{A\} w)=(\mathscr\{A\} v,w)$.
\begin\{aligned\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1554、 7、 在 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 中, $\displaystyle \gamma$ 是非零向量, 定义 $\displaystyle V$ 上的线性变换
\begin\{aligned\} \mathscr\{A\} \alpha=\alpha-\frac\{2(\alpha,\gamma)\}\{(\gamma,\gamma)\}\gamma, \forall\ \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\}
再设 $\displaystyle W\_0=\left\\{\alpha\in V; (\alpha,\gamma)=0\right\\}$. (1)、 证明: $\displaystyle W\_0$ 是 $\displaystyle \mathscr\{A\}$ 的不变子空间, 并求 $\displaystyle W\_0$ 的维数; (2)、 若 $\displaystyle W$ 是 $\displaystyle \mathscr\{A\}$ 的不变子空间, 证明: $\displaystyle \gamma\in W$ 或 $\displaystyle W\subset W\_0$. (华南理工大学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)、 设 $\displaystyle \eta\_1=\frac\{\gamma\}\{|\gamma|\}$, 则将其扩充为 $\displaystyle V$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_n$. 由 $\displaystyle \mathscr\{A\}$ 的定义容易知道 $\displaystyle \mathscr\{A\}\eta\_i=\left\\{\begin\{array\}\{llllllllllll\}-\eta\_1,&i=1,\\\\ \eta\_i,&2\leq i\leq n.\end\{array\}\right.$ 于是
\begin\{aligned\} \alpha=\sum\_\{i=1\}^n x\_i\eta\_i\in W\_0&\Leftrightarrow 0=(\alpha,\gamma) \Leftrightarrow 0=(\alpha,\eta\_1)=x\_1\Leftrightarrow \alpha=\sum\_\{i=2\}^n x\_i\eta\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle W\_0=L(\eta\_2,\cdots,\eta\_n)$. 由 $\displaystyle 2\leq i\leq n\Rightarrow \mathscr\{A\}\eta\_i=\eta\_i$ 知 $\displaystyle W\_0$ 是 $\displaystyle \mathscr\{A\}$ 的不变子空间, 且 $\displaystyle \dim W\_0=n-1$. (2)、 若 $\displaystyle W\subset W\_0$, 则已证. 若 $\displaystyle W\not\subset W\_0$, 则
\begin\{aligned\} \exists\ \alpha=\sum\_i x\_i\eta\_i\in W,\mathrm\{ s.t.\} \alpha\not\in W\_0=L(\eta\_2,\cdots,\eta\_n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle x\_1\neq 0$,
\begin\{aligned\} &x\_1\eta\_1+\cdots+x\_n\eta\_n\in W,\\\\ &-x\_1\eta\_1+x\_2\eta\_2+\cdots+x\_n\eta\_n=\mathscr\{A\}(x\_1\eta\_1+\cdots+x\_n\eta\_n)\in W. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相减即得 $\displaystyle 2x\_1\eta\_1\in W\stackrel\{x\_1\neq 0\}\{\Rightarrow\}\eta\_1\in W\Rightarrow \gamma\in W$.跟锦数学微信公众号. [在线资料](
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---
1555、 (6)、 欧氏空间 $\displaystyle \mathbb\{R\}^4$ 中, 已知 $\displaystyle \alpha=(1,2,2,3), \beta=(3,1,5,1)$, 则 $\displaystyle \alpha$ 和 $\displaystyle \beta$ 之间的夹角为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华南师范大学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) / 夹角 $\displaystyle \theta$ 满足
\begin\{aligned\} \cos\theta=\frac\{\alpha\cdot \beta\}\{|\alpha|\cdot |\beta|\}=\frac\{18\}\{3\sqrt\{2\}\cdot 6\}=\frac\{1\}\{\sqrt\{2\}\}\Rightarrow \theta=\frac\{\pi\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1556、 4、 设 $\displaystyle W=L(\alpha\_1,\alpha\_2,\alpha\_3)$ 是欧氏空间 $\displaystyle \mathbb\{R\}^5$ 的一个子空间, 其中
\begin\{aligned\} \alpha\_1=(0,3,-6,6,4)^\mathrm\{T\}, \alpha\_2=(3,-7,8,-5,8)^\mathrm\{T\}, \alpha\_3=(1,-3,4,-3,2)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle W$ 的正交补空间 $\displaystyle W^\perp$. (华南师范大学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\} x\in W^\perp\Leftrightarrow \alpha\_i^\perp x=0, i=1,2,3 \Leftrightarrow Ax=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\\alpha\_2^\mathrm\{T\}\\\\\alpha\_3^\mathrm\{T\}\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0&3&-6&6&4\\\\ 3&-7&8&-5&8\\\\ 1&-3&4&-3&2\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-2&3&0\\\\ 0&1&-2&2&0\\\\ 0&0&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\}
取 $\displaystyle x\_3,x\_4$ 为自由变量知 $\displaystyle Ax=0$ 的基础解系为
\begin\{aligned\} \eta\_1=(2,2,1,0,0)^\mathrm\{T\}, \eta\_2=(-3,-2,0,1,0)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最终, $\displaystyle W^\perp=L(\eta\_1,\eta\_2)$.跟锦数学微信公众号. [在线资料](
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---
1557、 (5)、 设 $\displaystyle \mathbb\{R\}$ 为实数域, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&-2\\\\ 2&5&-4\\\\ -2&-4&5\end\{array\}\right)$, 在三维列向量空间 $\displaystyle \mathbb\{R\}^3$ 上定义双线性型
\begin\{aligned\} f\_A(X,Y)=X^\mathrm\{T\} AY, \forall\ X,Y\in \mathbb\{R\}^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里, $\displaystyle X^\mathrm\{T\}$ 表示 $\displaystyle X$ 的转置. (5-1)、 证明: $\displaystyle \left(\mathbb\{R\}^3, f\_A(\cdot,\cdot)\right)$ 是一个欧氏空间; (5-2)、 求上述欧氏空间的一组标准正交基. (华中师范大学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) / (5-1)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 10,1,1$, 而 $\displaystyle A$ 正定. 我们现给出欧氏空间的定义. 若线性空间 $\displaystyle V$ 上有一个满足如下四条性质的二元函数 $\displaystyle (\cdot,\cdot)$, 则称 $\displaystyle V$ 是一个欧氏空间, $\displaystyle (\cdot,\cdot)$ 成为 $\displaystyle V$ 上的内积.
\begin\{aligned\} (\alpha,\beta)=&(\beta,\alpha);\\\\ (k\alpha,\beta)=&k(\alpha,\beta);\\\\ (\alpha+\beta,\gamma)=&(\alpha,\gamma)+(\beta,\gamma);\\\\ (\alpha,\alpha)\geq& 0, \alpha=0\Leftrightarrow(\alpha,\alpha)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里, $\displaystyle \alpha,\beta,\gamma$ 是 $\displaystyle V$ 的任意向量, $\displaystyle k$ 是任意实数. 为此, 我们只要验证最后一条. 由 $\displaystyle A$ 正定知存在可逆矩阵 $\displaystyle B$ 使得 $\displaystyle A=B^\mathrm\{T\} B$, 而
\begin\{aligned\} 0\neq X\in\mathbb\{R\}^3\Rightarrow f\_A(X,X)=X^\mathrm\{T\} AX=X^\mathrm\{T\} B^\mathrm\{T\} BX\stackrel\{Y=BX\neq 0\}\{=\}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\}
(5-2)、 由
\begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_1^\mathrm\{T\} AP\_1=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}2&0&0\\\\ 0&3&-2\\\\ 0&-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\}
\begin\{aligned\} P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&\frac\{2\}\{3\}\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2=A\_2=\mathrm\{diag\}\left(2,3,\frac\{5\}\{3\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} P\_3=\mathrm\{diag\}\left(\frac\{1\}\{\sqrt\{2\}\},\frac\{1\}\{\sqrt\{3\}\},\frac\{3\}\{\sqrt\{15\}\}\right)\Rightarrow P\_3^\mathrm\{T\} A\_2P\_3=E\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle P=P\_1P\_2P\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{3\}\}&\frac\{1\}\{\sqrt\{15\}\}\\\\ 0&\frac\{1\}\{\sqrt\{3\}\}&\frac\{2\}\{\sqrt\{15\}\}\\\\ 0&0&\frac\{3\}\{\sqrt\{15\}\}\end\{array\}\right)\equiv (\xi\_1,\xi\_2,\xi\_3)$, 则 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 就是 $\displaystyle V$ 的一组标准正交基. 事实上,
\begin\{aligned\} E=P^\mathrm\{T\} AP\Rightarrow \delta\_\{ij\}=\xi\_i^\mathrm\{T\} A\xi\_j=f\_A(\xi\_i,\xi\_j). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1558、 4、 (20 分) 设 $\displaystyle V$ 是 $\displaystyle n$ 维欧氏空间, $\displaystyle n\geq 2$, $\displaystyle \sigma$ 是 $\displaystyle V$ 上的线性变换, 若 $\displaystyle \tau=\sigma'\sigma$ 是正交变换, 证明 $\displaystyle \sigma$ 是正交变换, 其中 $\displaystyle \sigma'$ 是 $\displaystyle \sigma$ 的伴随变换. (吉林大学2023年高等代数与解析几何考研试题) [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 $\displaystyle V$ 的一组标准正交基后, 设 $\displaystyle \sigma,\tau$ 在该基下的矩阵分别为 $\displaystyle A,B$, 则只要证明
\begin\{aligned\} B=A^\mathrm\{T\} A, B^\mathrm\{T\} B=E\_n\Rightarrow A^\mathrm\{T\} A=E\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 由
\begin\{aligned\} x^\mathrm\{T\} Bx=x^\mathrm\{T\} A^\mathrm\{T\} Ax=(Ax)^\mathrm\{T\} (Ax)\geq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle B$ 半正定, 而存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} &P^\mathrm\{T\} BP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\geq 0\\\\ \Rightarrow&E\_n=P^\mathrm\{T\} E\_nP =P^\mathrm\{T\} B^\mathrm\{T\} P\cdot P^\mathrm\{T\} BP=(P^\mathrm\{T\} BP)^2=\mathrm\{diag\}(\lambda\_1^2,\cdots,\lambda\_n^2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lambda\_i=1\Rightarrow P^\mathrm\{T\} BP=E\_n\Rightarrow E\_n=B=A^\mathrm\{T\} A$.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](
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---
1559、 9、 欧氏空间 $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{A\}$ 与其共轭变换可交换, $\displaystyle W$ 是 $\displaystyle \mathscr\{A\}$ 的不变子空间. 问 $\displaystyle W$ 的正交补空间 $\displaystyle N$ 是否也是 $\displaystyle \mathscr\{A\}$ 的不变子空间? (暨南大学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) / 一言以蔽之, 正规变换的不变子空间的正交补也是不变的. (1)、 为此只要证明 $\displaystyle W$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的即可. 事实上, 一旦证得, 我们知对 $\displaystyle \beta\in N=W^\perp$,
\begin\{aligned\} \forall\ \alpha\in W, (\mathscr\{A\}\beta,\alpha)=(\beta,\mathscr\{A\}^\star\alpha)\stackrel\{\beta\in W^\perp, \mathscr\{A\}^\star\alpha\in W\}\{=\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathscr\{A\}\beta\in W^\perp=N$. 这就证明了 $\displaystyle N$ 是 $\displaystyle \mathscr\{A\}$ 的不变子空间. (2)、 往对 $\displaystyle \dim W=k$ 作数学归纳法证明 $\displaystyle W$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的. 当 $\displaystyle k=1$ 时, 由 $\displaystyle W=L(\alpha)$ 是 $\displaystyle \mathscr\{A\}$ 不变的知 $\displaystyle \exists\ \lambda\in \mathbb\{R\},\mathrm\{ s.t.\} \mathscr\{A\}\alpha=\lambda \alpha$. 从而
\begin\{aligned\} &(\mathscr\{A\}^\star\alpha-\lambda\alpha,\mathscr\{A\}^\star\alpha-\lambda \alpha) =(\mathscr\{A\}^\star\alpha,\mathscr\{A\}^\star\alpha)-2\lambda(\mathscr\{A\}^\star\alpha,\alpha)+\lambda^2(\alpha,\alpha)\\\\ =&(\alpha,\mathscr\{A\}\mathscr\{A\}^\star\alpha)-2\lambda(\alpha,\mathscr\{A\}\alpha)+\lambda^2(\alpha,\alpha) \xlongequal\{\tiny\mbox\{题设\}\}(\alpha,\mathscr\{A\}^\star\mathscr\{A\}\alpha)-\lambda^2(\alpha,\alpha)\\\\ =&(\mathscr\{A\}\alpha,\mathscr\{A\}\alpha)-\lambda^2(\alpha,\alpha)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathscr\{A\}^\star\alpha=\lambda\alpha\in W$. 这就证明了 $\displaystyle W$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的. 假设结论对维数小于 $\displaystyle i$ 的 $\displaystyle \mathscr\{A\}$ 不变子空间成立, 即: 只要 $\displaystyle \mathscr\{A\}$ 的不变子空间的维数 $\displaystyle < i$, 那么它就是 $\displaystyle \mathscr\{A\}^\star$ 不变的, 则对维数 $\displaystyle =i$ 的 $\displaystyle \mathscr\{A\}$ 不变子空间 $\displaystyle W$. 考虑 $\displaystyle \mathscr\{A\}|\_W$, 由 $\displaystyle (\mathscr\{A\}|\_W)^\star\mathscr\{A\}|\_W=\mathscr\{A\}|\_W(\mathscr\{A\}|\_W)^\star$ 知 $\displaystyle \mathscr\{A\}|\_W$ 在 $\displaystyle W$ 的某组基下的矩阵是正规矩阵, 特征值为实数. 设 $\displaystyle \lambda$ 是 $\displaystyle \mathscr\{A\}$ 的一个特征值, $\displaystyle 0\neq \alpha\in V$ 为对应的特征向量, 则 $\displaystyle U=L(\alpha)$ 是 $\displaystyle \mathscr\{A\}$ 不变的, 从而 $\displaystyle U^\perp$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的. 再由已证的 $\displaystyle k=1$ 时的结果知 $\displaystyle U$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的, 从而 $\displaystyle U^\perp$ 是 $\displaystyle (\mathscr\{A\}^\star)^\star=\mathscr\{A\}$ 不变的. 这就证明了 $\displaystyle U^\perp$ 是 $\displaystyle \mathscr\{A\}$, 也是 $\displaystyle \mathscr\{A\}^\star$ 不变的. 考虑 $\displaystyle W\_1=W\cap U^\perp$, 由 $\displaystyle \alpha\in W, \alpha\not\in U^\perp\Rightarrow \alpha\not\in W\_1$ 知 $\displaystyle W\_1$ 是 $\displaystyle W$ 的真子空间, $\displaystyle \dim W\_1 < i$. 由归纳假设知 $\displaystyle W\_1$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的. 最后由 $\displaystyle W=W\_1\oplus U$ 知 $\displaystyle W$ 是 $\displaystyle \mathscr\{A\}^\star$ 不变的. 结论得证.跟锦数学微信公众号. [在线资料](
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---
1560、 9、 $\displaystyle \left(V,(\cdot,\cdot)\right)$ 是 $\displaystyle n$ 维实内积空间, 向量组
\begin\{aligned\} S=\left\\{\alpha\_1,\cdots,\alpha\_k\right\\}\subset V, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle (\alpha\_i,\alpha\_j) < 0, 1\leq i\neq j\leq n$. 求证: $\displaystyle k\leq n+1$. (南方科技大学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) / 用反证法. 若 $\displaystyle k\geq n+2$, 则
\begin\{aligned\} (\alpha\_i,\alpha\_j) < 0, \forall\ 1\leq i\neq j\leq k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们可以证明其中任意 $\displaystyle n+1$ 个向量线性无关. 比如我们证明 $\displaystyle \alpha\_1,\cdots,\alpha\_\{n+1\}$ 线性无关. 用反证法. 若存在不全为 $\displaystyle 0$ 的 $\displaystyle k\_i\in\mathbb\{R\}$, 使得
\begin\{aligned\} \sum\_\{i=1\}^\{n+1\}k\_i\alpha\_i=0\Rightarrow \sum\_\{i=1\}^\{n+1\}k\_i(\alpha\_i,\alpha\_\{n+2\})=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到 $\displaystyle (\alpha\_i,\alpha\_\{n+2\}) < 0$, 我们知 $\displaystyle k\_1,\cdots,k\_\{n+1\}$ 有正有负,
\begin\{aligned\} &\sum\_\{k\_i < 0\}k\_i\alpha\_i+\sum\_\{k\_i > 0\}k\_i\alpha\_i=0\\\\ \Rightarrow&\alpha=\sum\_\{k\_i > 0\}k\_i\alpha\_i=-\sum\_\{k\_i < 0\}k\_i\alpha\_i\\\\ \Rightarrow&(\alpha,\alpha)=\left(\sum\_\{k\_i > 0\}k\_i\alpha\_i, -\sum\_\{k\_j < 0\}k\_j\alpha\_j\right) =-\sum\_\{k\_i > 0\}\sum\_\{k\_j < 0\}k\_ik\_j(\alpha\_i,\alpha\_j) < 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故 $\displaystyle \alpha\_1,\cdots,\alpha\_\{n+1\}$ 线性无关, $\displaystyle \dim V\geq n+1$, 这与题设 $\displaystyle \dim V=n$ 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](
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---
1561、 9、 (20 分) 设 $\displaystyle e\_1,\cdots,e\_n$ 为欧氏空间 $\displaystyle V=\mathbb\{R\}^n$ 的一组标准正交基, $\displaystyle v\_1,\cdots,v\_n\in V$ 满足对任意的 $\displaystyle 1\leq i\leq n$ 都有 $\displaystyle \left\Vert e\_i-v\_i\right\Vert < \frac\{1\}\{\sqrt\{n\}\}$. 求证: $\displaystyle v\_1,\cdots,v\_n$ 是 $\displaystyle V$ 的一组基. (南京大学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 $\displaystyle W=L(v\_1,\cdots,v\_n)$, 则只要证明
\begin\{aligned\} \dim W^\perp=0&\Rightarrow \dim W=n \Rightarrow W=V\\\\ & \Rightarrow \mbox\{ $\displaystyle v\_1,\cdots,v\_n$ 线性无关, 而是 $\displaystyle V$ 的一组基\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
用反证法. 若
\begin\{aligned\} \dim W^\perp > 0\Rightarrow \exists\ 0\neq \alpha\in V,\mathrm\{ s.t.\} \alpha\in W^\perp \Rightarrow \alpha\neq 0, \left < \alpha,v\_i\right > =0, \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\} \left\Vert \alpha\right\Vert ^2&=\sum\_\{i=1\}^n |\left < \alpha,e\_i\right > |^2\left(\mbox\{勾股定理: $\displaystyle \alpha=\sum\_\{i=1\}^n \left < \alpha,e\_i\right > e\_i$\}\right)\\\\ &=\sum\_\{i=1\}^n |\left < \alpha,v\_i-e\_i\right > |^2 \leq \sum\_\{i=1\}^n \left\Vert \alpha\right\Vert ^2 \cdot \left\Vert v\_i-e\_i\right\Vert ^2\\\\ & < \left\Vert \alpha\right\Vert ^2\sum\_\{i=1\}^n \left(\frac\{1\}\{\sqrt\{n\}\}\right)^2 =\left\Vert \alpha\right\Vert ^2. \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](
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---
1562、 (9)、 $\displaystyle \alpha\_1,\cdots,\alpha\_\{n-1\}$ 是 $\displaystyle n$ 维欧氏空间的非零正交向量组,
\begin\{aligned\} (\beta\_j,\alpha\_i)=0, i=1,2, j=1,\cdots,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 $\displaystyle \beta\_1,\beta\_2$ 必定 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (选填: 线性无关, 线性相关) (厦门大学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) / 线性相关. 事实上,
\begin\{aligned\} \beta\_j\in L(\alpha\_1,\cdots,\alpha\_\{n-1\})^\perp, j=1,2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \dim L(\alpha\_1,\cdots,\alpha\_\{n-1\})=n-1$ 知 $\displaystyle \dim L(\alpha\_1,\cdots,\alpha\_\{n-1\})^\perp=1$, 而 $\displaystyle \beta\_1,\beta\_2$ 线性相关.跟锦数学微信公众号. [在线资料](
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---
1563、 6、 设欧几里得空间 $\displaystyle V=\mathbb\{R\}^4$ 中的三个向量为
\begin\{aligned\} \alpha\_1=(1,-1,-1,1), \alpha\_2=(1,0,-1,1), \alpha\_3=(0,1,-1,1), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
子空间 $\displaystyle W=L(\alpha\_1,\alpha\_2,\alpha\_3)$, 求向量 $\displaystyle \beta=(2,1,4,2)$ 在 $\displaystyle W$ 上的正交投影. (山东大学2023年高等代数与常微分方程考研试题) [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) / 先把题中向量看成列向量. 设 $\displaystyle \beta$ 在 $\displaystyle W$ 上的正交投影为 $\displaystyle \gamma=x\_1\alpha\_1+x\_2\alpha\_2+x\_3\alpha\_3$, 则 ($X=\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1\\\\x\_2\\\\x\_3\end\{array\}\right)$)
\begin\{aligned\} &(\beta-\gamma)\perp\alpha\_i=0, i=1,2,3\Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\\alpha\_2^\mathrm\{T\}\\\\\alpha\_3^\mathrm\{T\}\end\{array\}\right)\left\[\beta-(\alpha\_1,\alpha\_2,\alpha\_3)X\right\]=0\\\\ \Leftrightarrow&A^\mathrm\{T\} AX=A^\mathrm\{T\} \beta, A=(\alpha\_1,\alpha\_2,\alpha\_3). \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^\mathrm\{T\} A,A^\mathrm\{T\} \beta)=\left(\begin\{array\}\{cccccccccccccccccccc\}4&3&1&-1\\\\ 3&3&2&0\\\\ 1&2&3&-1\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-4\\\\ 0&1&0&6\\\\ 0&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\}
知 $\displaystyle \beta$ 在 $\displaystyle W$ 上的正交投影为
\begin\{aligned\} \gamma=x\_1\alpha\_1+x\_2\alpha\_2+x\_3\alpha\_3=-4\alpha\_1+6\alpha\_2-3\alpha\_3=(2,1,1,-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)/
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1564、 9、 (10 分) 设 $\displaystyle V\_1,V\_2$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 的子空间, 且 $\displaystyle V\_1$ 的维数小于 $\displaystyle V\_2$ 的维数, 证明: $\displaystyle V\_2$ 中必有非零向量正交于 $\displaystyle V\_1$ 中的一切向量. (陕西师范大学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) / 由
\begin\{aligned\} \dim (V\_2\cap V\_1^\perp)&=\dim V\_2+\dim V\_1^\perp-\dim (V\_2\cap V\_1^\perp)\\\\ &=\dim V\_2+(n-\dim V\_1)-\dim (V\_2\cap V\_1^\perp)\\\\ &=(\dim V\_2-\dim V\_1)+\left\[n-\dim (V\_2\cap V\_1^\perp)\right\]\\\\ &\geq \dim V\_2-\dim V\_1 > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \exists\ 0\neq \alpha\in V\_2\cap V\_1^\perp$. 此 $\displaystyle \alpha$ 即满足题意.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 13:25:49
