## 张祖锦2023年数学专业真题分类70天之第67天
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1519、 6、 设 $\displaystyle A$ 是 $\displaystyle m\times n$ 矩阵, $\displaystyle B$ 是 $\displaystyle n\times s$ 矩阵. 证明:
\begin\{aligned\} \dim N(AB)=\dim N(A)+\dim N(B) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的充分必要条件是 $\displaystyle N(A)\subset R(B)$, 其中
\begin\{aligned\} N(A)=&\left\\{X\in\mathbb\{R\}^n; AX=0\right\\},\\\\ R(B)=&\left\\{BX\in\mathbb\{R\}^n; X\in \mathbb\{R\}^s\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国人民大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 写出
\begin\{aligned\} \mathbb\{R\}^s\xrightarrow\{B\}\mathbb\{R\}^n\xrightarrow\{A\}\mathbb\{R\}^m. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
易知 $\displaystyle N(B)\subset N(AB)$. 设 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$ 是 $\displaystyle N(B)$ 的一组基, 将其扩充为 $\displaystyle N(AB)$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r, \eta\_1,\cdots,\eta\_s$. 由
\begin\{aligned\} &\sum\_i x\_i B\eta\_i=0\Rightarrow B\left(\sum\_i x\_i\eta\_i\right)=0 \Rightarrow \sum\_i x\_i\eta\_i\in N(B)\\\\ \Rightarrow&\sum\_i x\_i\eta\_i=-\sum\_j y\_j\varepsilon\_j \Rightarrow x\_i=0, y\_j=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle B\eta\_1,\cdots, B\eta\_s$ 线性无关, 且
\begin\{aligned\} \eta\_i\in N(AB)\Rightarrow AB\eta\_i=0\Rightarrow B\eta\_i\in N(A). \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\} \dim N(A)\geq s=(r+s)-r=\dim N(AB)-\dim N(B), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且等号成立
\begin\{aligned\} \Leftrightarrow \dim N(A)=s\Leftrightarrow N(A)=L(B\eta\_1,\cdots,B\eta\_s) \Rightarrow N(A)\subset R(B). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
反之, 若 $\displaystyle N(A)\subset R(B)$, 则取定 $\displaystyle N(A)\subset R(B)$ 的一组基 $\displaystyle B\eta\_1,\cdots, B\eta\_s$. 再取 $\displaystyle N(B)$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$, 则由
\begin\{aligned\} &\sum\_i x\_i\varepsilon\_i+\sum\_j y\_j\eta\_j=0 \stackrel\{B\cdot\}\{\Rightarrow\} \sum\_j y\_jB\eta\_j=0\Rightarrow y\_j=0\Rightarrow x\_i=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r, \eta\_1,\cdots,\eta\_s$ 线性无关, 都是 $\displaystyle N(AB)$ 中的向量. 又由
\begin\{aligned\} &\alpha\in N(AB)\Rightarrow AB\alpha=0\Rightarrow B\alpha\in N(A)=L(B\eta\_1,\cdots, B\eta\_s)\\\\ \Rightarrow& B\alpha=\sum\_i x\_iB\eta\_i \Rightarrow B\left(\alpha-\sum\_i x\_i\eta\_i\right)=0\\\\ \Rightarrow& \alpha-\sum\_i x\_i\eta\_i\in N(B)=L(\varepsilon\_1,\cdots,\varepsilon\_r) \Rightarrow \alpha-\sum\_i x\_i\eta\_i=\sum\_j y\_j\varepsilon\_j \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &N(AB)=L(\varepsilon\_1,\cdots,\varepsilon\_r,\eta\_1,\cdots,\eta\_s)\\\\ \Rightarrow&\dim N(AB)=r+s=\dim N(B)+\dim N(A). \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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1520、 7、 (16 分) 设 $\displaystyle \sigma$ 是 $\displaystyle n$ 维线性空间 $\displaystyle V$ 上的可逆线性变换, $\displaystyle V$ 的子空间 $\displaystyle W$ 是 $\displaystyle \sigma$ 的不变子空间. 证明: $\displaystyle W$ 是 $\displaystyle \sigma^\{-1\}$ 的不变子空间. (中南大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$ 是 $\displaystyle W$ 的一组基, 将其扩充为 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 则由 $\displaystyle W$ 是 $\displaystyle \sigma$ 不变的知
\begin\{aligned\} \sigma(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}A&\star\\\\ &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\}
由 $\displaystyle \sigma$ 可逆知
\begin\{aligned\} 0\neq \left|\begin\{array\}\{cccccccccc\}A&\star\\\\ &B\end\{array\}\right|=|A|\cdot |B|, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle A,B$ 可逆, 且
\begin\{aligned\} &\sigma^\{-1\}(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}A^\{-1\}&\star\\\\ &B^\{-1\}\end\{array\}\right)\\\\ \Rightarrow&\sigma^\{-1\}(\varepsilon\_1,\cdots,\varepsilon\_r)=(\varepsilon\_1,\cdots,\varepsilon\_r)A^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这即表明 $\displaystyle W$ 是 $\displaystyle \sigma^\{-1\}$ 的不变子空间.跟锦数学微信公众号. [在线资料](
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1521、 5、 (15 分) 设 $\displaystyle V$ 是 $\displaystyle n$ 维复线性空间, $\displaystyle \mathscr\{A\},\mathscr\{B\}$ 是 $\displaystyle V$ 上的线性变换, 且
\begin\{aligned\} \mathscr\{A\}\mathscr\{B\}-\mathscr\{B\}\mathscr\{A\}=2023\mathscr\{B\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 存在 $\displaystyle \mathscr\{A\}$ 的特征向量 $\displaystyle u$, 使得 $\displaystyle \mathscr\{B\} u=0$. (中山大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 先证明如下结论. 设 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 是 $\displaystyle n$ 个数, 且满足
\begin\{aligned\} \lambda\_1+\cdots+\lambda\_n=\lambda\_1^2+\cdots+\lambda\_n^2=\cdots =\lambda\_1^n+\cdots+\lambda\_n^n=0.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \lambda\_1=\lambda\_2=\cdots=\lambda\_n=0$. 用反证法. 若存在某个 $\displaystyle \lambda\_i\neq 0$, 则设 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 中非零且互异的为 $\displaystyle \mu\_1,\cdots,\mu\_s$, $\displaystyle \mu\_i$ 出现了 $\displaystyle n\_i\geq 1$ 次, 则 $\displaystyle \sum\_\{i=1\}^n n\_i=n$, 且
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}\mu\_1&\cdots&\mu\_s\\\\ \vdots&&\vdots\\\\ \mu\_1^s&\cdots&\mu\_s^s\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}n\_1\\\\\vdots\\\\nabla \_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 $\displaystyle n\_i\geq 1$ 知上述关于 $\displaystyle n\_1,\cdots,n\_s$ 的方程组有非零解, 而系数矩阵行列式
\begin\{aligned\} \mu\_1\cdots\mu\_s\prod\_\{1\leq i < j\leq s\}(\mu\_j-\mu\_i)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle \mu\_i$ 非零且互异矛盾. 故有结论. (2)、 回到题目. 由
\begin\{aligned\} &\mathrm\{tr\}(2023\mathscr\{B\}^k)=\mathrm\{tr\}\left(\mathscr\{B\}^\{k-1\}(\mathscr\{A\}\mathscr\{B\}-\mathscr\{B\}\mathscr\{A\})\right)\\\\ =&\mathrm\{tr\}\left(\mathscr\{B\}^\{k-1\}\mathscr\{A\}\cdot \mathscr\{B\}-\mathscr\{B\}\cdot \mathscr\{B\}^\{k-1\}\mathscr\{A\}\right)=0, \forall\ k\geq 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \mathscr\{B\}$ 的特征值 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 满足 $\displaystyle (I)$. 由第 1 步知 $\displaystyle \lambda\_1=\cdots=\lambda\_n=0$. 令
\begin\{aligned\} V\_0=\left\\{\alpha\in V; \mathscr\{B\} \alpha=0\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由
\begin\{aligned\} \mathscr\{B\}\mathscr\{A\}\alpha=(\mathscr\{A\}\mathscr\{B\}-2023\mathscr\{B\})\alpha=0\Rightarrow \mathscr\{A\}\alpha\in V\_0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_0$ 是 $\displaystyle \mathscr\{A\}$ 不变的. 设 $\displaystyle u\in V\_0$ 是 $\displaystyle \mathscr\{A\}|\_\{V\_0\}$ 的特征向量, 则 $\displaystyle \mathscr\{B\} u=0$.跟锦数学微信公众号. [在线资料](
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---
1522、 9、 (15 分) 设 $\displaystyle \left < \cdot,\cdot\right > $ 为 $\displaystyle \mathbb\{R\}^2$ 上的标准内积, 记
\begin\{aligned\} S: =\left\\{\left(\cos\frac\{k\pi\}\{3\},\sin\frac\{k\pi\}\{3\}\right); k\in\mathbb\{Z\}\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle f$ 是 $\displaystyle \mathbb\{R\}^2\to\mathbb\{R\}^2$ 的线性变换, 且满足 (1)、 $\displaystyle \forall\ u\in S, f(u)\in S$; (2)、 存在 $\displaystyle S$ 中的向量 $\displaystyle a$ 与 $\displaystyle \mathbb\{R\}^2$ 中非零向量 $\displaystyle b$, 使得 $\displaystyle f(a)=-a, f(b)=b$. 证明: 对任意的 $\displaystyle v\in\mathbb\{R\}^2$, 有 $\displaystyle f(v)=v-2(v,a)a$. (中山大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \alpha\_k=\left(\cos\frac\{k\pi\}\{3\},\sin\frac\{k\pi\}\{3\}\right)$, 则
\begin\{aligned\} &\alpha\_0+\alpha\_2=\alpha\_2\Rightarrow f(\alpha\_0)+f(\alpha\_2)=f(\alpha\_1)\\\\ \Rightarrow& \left(f(\alpha\_0)+f(\alpha\_2), f(\alpha\_0)+f(\alpha\_2)\right)=\left(f(\alpha\_1),f(\alpha\_1)\right)\\\\ \stackrel\{f(\alpha\_i)\in S\}\{\Rightarrow\}&2+2\left(f(\alpha\_0),f(\alpha\_2)\right)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle \forall\ \alpha\in \mathbb\{R\}^2, \alpha=s\alpha\_0+t\alpha\_2$, 我们有
\begin\{aligned\} &|f(\alpha)|^2=\left(f(\alpha),f(\alpha)\right) =\left(sf(\alpha\_0)+tf(\alpha\_2),sf(\alpha\_0)+tf(\alpha\_2)\right)\\\\ =&s^2+2st \left(f(\alpha\_0),f(\alpha\_2)\right)+t^2 =s^2-2st+t^2\\\\ =&s^2|\alpha\_0|^2+2st(\alpha\_0,\alpha\_2)+t^2|\alpha\_2^2|=|s\alpha\_0+t\alpha\_2|^2=|\alpha|^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle f$ 是正交变换. 进而
\begin\{aligned\} (a,b)=\left(f(a),f(b)\right)=\left(-a,b\right)\Rightarrow (a,b)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
用 $\displaystyle \frac\{b\}\{|b|\}$ 代替 $\displaystyle b$ 后不妨设 $\displaystyle b$ 是单位向量, 而 $\displaystyle a,b$ 是 $\displaystyle \mathbb\{R\}^2$ 的标准正交基. 对 $\displaystyle \forall\ v=ka+lb$, 我们有 $\displaystyle k=(v,a), l=(v,b)$, 而
\begin\{aligned\} f(v)=&kf(a)+lf(b)=-ka+lb=(ka+lb)-2ka=v-2(v,a)a. \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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---
1523、 8、 (15 分) 用 $\displaystyle \mathbb\{R\}[x]\_n$ 表示次数不超过 $\displaystyle n$的实系数多项式全体. (1)、 证明: $\displaystyle \mathbb\{R\}[x]\_n$ 按照多项式的加法, 实数与多项式的数量乘法构成一个 $\displaystyle n+1$ 维向量空间. (2)、 证明: 差分算子
\begin\{aligned\} \mathscr\{D\}\_n: f(x)\mapsto f(x+1)-f(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是定义在 $\displaystyle \mathbb\{R\}[x]\_n$ 上的一个线性变换. (重庆大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 易知 $\displaystyle 1,x,\cdots,x^n$ 是 $\displaystyle \mathbb\{R\}[x]\_n$ 的一组基, 而 $\displaystyle \mathbb\{R\}[x]\_n$ 确为 $\displaystyle n+1$ 维向量空间. (2)、 首先 $\displaystyle f(x)\in \mathbb\{R\}[x]\_n\Rightarrow \mathscr\{D\}\_n\left\[f(x)\right\]\in \mathbb\{R\}[x]\_n$. 对 $\displaystyle \forall\ k,l\in\mathbb\{R\}, f(x),g(x)\in \mathbb\{R\}[x]\_n$,
\begin\{aligned\} &\mathscr\{D\}\_n\left\[kf(x)+lg(x)\right\]=kf(x+1)+lg(x+1)-[kf(x)+lg(x)]\\\\ =&k[f(x+1)-f(x)]+l[g(x+1)-g(x)]=k\mathscr\{D\}\_n \left\[f(x)\right\]+l \mathscr\{D\}\_n \left\[g(x)\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathscr\{D\}\_n$ 确为 $\displaystyle \mathbb\{R\}[x]\_n$ 上的一个线性变换.跟锦数学微信公众号. [在线资料](
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---
1524、 9、 (15 分) 设 $\displaystyle \mathscr\{A\}$ 是数域 $\displaystyle \mathbb\{K\}$ 上 $\displaystyle n$ 维线性空间 $\displaystyle V$ 上的一个线性变换, $\displaystyle \lambda$ 是 $\displaystyle \mathscr\{A\}$ 的一个特征值, $\displaystyle \mathscr\{A\}$ 的属于特征值 $\displaystyle \lambda$ 的特征子空间 $\displaystyle V\_\lambda$ 称为特征值 $\displaystyle \lambda$ 的几何重数; $\displaystyle \lambda$ 作为 $\displaystyle \mathscr\{A\}$ 的特征多项式的根的重数称为特征值 $\displaystyle \lambda$ 的代数重数. (1)、 证明: $\displaystyle \lambda$ 的几何重数不超过它的代数重数; (2)、 举例说明几何重数的确可以小于代数重数. (重庆大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle \dim V\_\lambda=r$, 则可设 $\displaystyle V\_\lambda$ 的一组基为 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$. 将其扩充为 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 则
\begin\{aligned\} \mathscr\{A\}(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E\_r&A\\\\ &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\}
这表明 $\displaystyle \mathscr\{A\}$ 的特征多项式为 $\displaystyle f(x)=(x-\lambda)^r |x E\_\{n-r\}-B|$, $\displaystyle \lambda$ 的代数重数, 就是作为 $\displaystyle f(x)$ 的根的重数 $\displaystyle \geq r$. (2)、 设 $\displaystyle \mathscr\{A\}$ 在 $\displaystyle V=\mathbb\{K\}^2$ 的基 $\displaystyle e\_1,e\_2$ 下的矩阵为 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\ 0&0\end\{array\}\right)$, 则 $\displaystyle 0$ 的代数重数为 $\displaystyle 2$, 而几何重数为 $\displaystyle 1$.跟锦数学微信公众号. [在线资料](
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1525、 10、 (16 分) 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{K\}$ 上的线性空间, $\displaystyle f$ 是 $\displaystyle V$ 上的双线性函数. (1)、 证明: $\displaystyle V$ 不可能是它的两个真子空间的并; (2)、 给定向量 $\displaystyle \alpha\in V$. 证明:
\begin\{aligned\} V\_\alpha=\left\\{\beta\in V; f(\alpha,\beta)=f(\beta,\alpha)\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为 $\displaystyle V$ 的一个子空间; (3)、 若已知对任意向量 $\displaystyle \alpha,\beta\in V$, 有
\begin\{aligned\} f(\alpha,\beta)=f(\beta,\alpha)\mbox\{或\} f(\alpha,\beta)=-f(\beta,\alpha) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
成立. 证明: $\displaystyle f$ 是对称的或反对称的. (重庆大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 用反证法. 若 $\displaystyle V=V\_1\cup V\_2$, 其中 $\displaystyle V\_i$ 是 $\displaystyle V$ 的真子空间. 于是 $\displaystyle V\_1\not\subset V\_2, V\_2\not\subset V\_1$. 而
\begin\{aligned\} \exists\ \alpha\_1\in V\_1, \mathrm\{ s.t.\} \alpha\_1\not\in V\_2; \exists\ \alpha\_2\in V\_2,\mathrm\{ s.t.\} \alpha\_2\not\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\}
考查 $\displaystyle \alpha=\alpha\_1+\alpha\_2\in V=V\_1\cup V\_2$. (1-1)、 若 $\displaystyle \alpha\in V\_1$, 则 $\displaystyle \alpha\_2=\alpha-\alpha\_1\in V\_1$. 矛盾. (1-2)、 若 $\displaystyle \alpha\in V\_2$, 则 $\displaystyle \alpha\_1=\alpha-\alpha\_2\in V\_2$. 矛盾. 故有结论. (2)、 对 $\displaystyle \forall\ k,l\in\mathbb\{K\}, \beta,\gamma\in V\_\alpha$,
\begin\{aligned\} &f(\alpha, k\beta+l\gamma) =kf(\alpha,\beta)+lf(\alpha,\gamma)\\\\ =&kf(\beta,\alpha)+lf(\gamma,\alpha) =f(k\beta+l\gamma,\alpha). \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 $\displaystyle k\beta+l\gamma\in V\_\alpha$. 而 $\displaystyle V\_\alpha$ 是 $\displaystyle V$ 的子空间. (3)、 对 $\displaystyle \forall\ \alpha\in V$, 设
\begin\{aligned\} V\_\alpha=&\left\\{\beta\in V; f(\alpha,\beta)=f(\beta,\alpha)\right\\},\\\\ W\_\alpha=&\left\\{\beta\in V; f(\alpha,\beta)=-f(\beta,\alpha)\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由题设, $\displaystyle V=V\_\alpha\cup W\_\alpha$. 由第 2 步知 $\displaystyle V$ 是它的两个子空间的并. 再由第 1 步知
\begin\{aligned\} V\_\alpha=V\mbox\{或\} W\_\alpha=V.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再设
\begin\{aligned\} U\_1=\left\\{\alpha\in V; V\_\alpha=V\right\\}, U\_2=\left\\{\alpha\in V; W\_\alpha=V\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则对 $\displaystyle \forall\ k,l\in \mathbb\{K\}, \alpha,\beta\in U\_1$,
\begin\{aligned\} &V\_\alpha=V, V\_\beta=V\\\\ \Rightarrow&\forall\ \gamma\in V, f(\alpha,\gamma)=f(\gamma,\alpha), f(\beta,\gamma)=f(\gamma,\beta)\\\\ \Rightarrow&\forall\ \gamma\in V, f(k\alpha+l\beta,\gamma)=f(\gamma,k\alpha+l\beta)\\\\ \Rightarrow&k\alpha+l\beta\in U\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle U\_1$ 是 $\displaystyle V$ 的线性子空间. 同理, $\displaystyle U\_2$ 也是 $\displaystyle V$ 的线性子空间. 由 $\displaystyle (I)$ 知 $\displaystyle V=U\_1\cup U\_2$. 再由第 1 步知 (3-1)、 $\displaystyle U\_1=V$, 此时 $\displaystyle f$ 是对称变换. (3-2)、 或 $\displaystyle U\_2=V$. 此时 $\displaystyle f$ 是反对称变换.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](
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---
1526、 4、 (12 分) 设 $\displaystyle t\_1,\cdots,t\_n$ 互不相同, $\displaystyle \alpha\_i= (1,t\_i,t\_i^2,\cdots,t\_i^\{n-1\})$, $\displaystyle i=1,2,\cdots,n$. 试证: $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 线性无关. (重庆师范大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} &k\_1\alpha\_1+\cdots +k\_n\alpha\_n=0\\\\ \Leftrightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\cdots&1\\\\ t\_1&t\_2&\cdots&t\_n\\\\ \vdots&\vdots&&\vdots\\\\ t\_1^\{n-1\}&t\_2^\{n-1\}&\cdots&t\_n^\{n-1\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}k\_1\\\\k\_2\\\\\vdots\\\\k\_n\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\\vdots\\\\0\end\{array\}\right)\\\\ \Leftrightarrow&k\_1=\cdots=k\_n=0\left(\mbox\{系数矩阵的行列式 $\displaystyle =\prod\_\{1\leq i < j\leq n\}(t\_j-t\_i)\neq 0$\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 线性无关.跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1527、 8、 设 $\displaystyle V$ 为 $\displaystyle [0,1]$ 上全体实函数构成的集合, 对 $\displaystyle f\_1,f\_2\in V, k\in\mathbb\{R\}$, 定义
\begin\{aligned\} (f\_1+f\_2)(x)=f\_1(x)+f\_2(x), \quad (kf)(x)=k\cdot f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 试证: $\displaystyle V$ 在上述运算下构成一个线性空间, 并指出零元以及 $\displaystyle f$ 的负元. (2)、 证明: $\displaystyle V$ 不是有限维线性空间. (重庆师范大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由
\begin\{aligned\} &f(x)+g(x)=g(x)+f(x),\\\\ &[f(x)+g(x)]+h(x)=f(x)+[g(x)+h(x)],\\\\ &0+f(x)=f(x), \left\[-f(x)\right\]+f(x)=0,\\\\ &1\cdot f(x)=f(x), (kl)f(x)=k\left\[lf(x)\right\],\\\\ &(k+l)f(x)=kf(x)+lf(x), k\left\[f(x)+g(x)\right\]=kf(x)+kg(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V$ 是一个线性空间, 且零函数 $\displaystyle 0$ 是零元, $\displaystyle -f(x)$ 是 $\displaystyle f(x)$ 的负元. (2)、 用反证法. 若 $\displaystyle V$ 是有限维的, 则 $\displaystyle \dim V=m$. 但 $\displaystyle 1,x,\cdots,x^m$ 线性无关, 而
\begin\{aligned\} m=\dim V\geq \dim L(1,x,\cdots,x^m)=m+1 > m. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1528、 9、 (12 分) 设 $\displaystyle \sigma\_1,\cdots,\sigma\_s$ 是线性空间 $\displaystyle V$ 上的 $\displaystyle n$ 个两两不同的线性变换. 试证: 存在 $\displaystyle \alpha\in V$, 使得 $\displaystyle \sigma\_1(\alpha),\cdots,\sigma\_s(\alpha)$ 两两不同. (重庆师范大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 先证明一个结论: 非完全覆盖原理 (线性空间不能被其任意有限个真子空间覆盖). 设 $\displaystyle V$ 是 $\displaystyle n$ 维线性空间, $\displaystyle V\_1,\cdots,V\_m$ 是 $\displaystyle V$ 的 $\displaystyle m$ 个真子空间, 则
\begin\{aligned\} \exists\ \mbox\{无限多个\}\alpha\in\left\\{\beta\_j\right\\}\_\{j=1\}^\infty\subset V,\mathrm\{ s.t.\} \alpha\not\in V\_1\cup \cdots \cup V\_m. \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, 取定 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 令
\begin\{aligned\} \alpha\_i=\varepsilon\_1+i \varepsilon\_2+\cdots+i^\{n-1\}\varepsilon\_n, i=1,2,\cdots, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \alpha\_1,\alpha\_2,\cdots$ 中的任意 $\displaystyle n$ 个
\begin\{aligned\} (\alpha\_\{i\_1\},\cdots,\alpha\_\{i\_n\})=(\varepsilon\_1,\cdots,\varepsilon\_n)\left(\begin\{array\}\{cccccccccccccccccccc\} 1&1&\cdots&1\\\\ i\_1&i\_2&\cdots&i\_n\\\\ \vdots&\vdots&&\vdots\\\\ i\_1^\{n-1\}&i\_2^\{n-1\}&\cdots&i\_n^\{n-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\}
由范德蒙行列式知 $\displaystyle \alpha\_\{i\_1\},\cdots,\alpha\_\{i\_n\}$ 是 $\displaystyle V$ 的一组基. 由于 $\displaystyle V\_i$ 是 $\displaystyle V$ 的真子空间, 而每个 $\displaystyle V\_i$ 至多包含 $\displaystyle \alpha\_1,\alpha\_2,\cdots$ 中 的 $\displaystyle n-1$ 个, $\displaystyle \bigcup\_\{i=1\}^m V\_i$ 至多包含 $\displaystyle \alpha\_1,\alpha\_2,\cdots$ 中的 $\displaystyle m(n-1)$ 个. 于是
\begin\{aligned\} \exists\ \mbox\{无限多个\} k,\mathrm\{ s.t.\} \alpha\_k\not\in \bigcup\_\{i=1\}^mV\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle \sigma\_1,\sigma\_2,\cdots,\sigma\_s$ 是线性空间 $\displaystyle V$ 中两两不同的线性变换知 $\displaystyle \sigma\_1-\sigma\_2,\sigma\_1-\sigma\_3,\cdots,\sigma\_\{s-1\}-\sigma\_s$ 都不是零变换, 而
\begin\{aligned\} \ker(\sigma\_i-\sigma\_j)\subsetneq V, \quad \forall\ 1\leq i < j\leq s. \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
\begin\{aligned\} \exists\ \alpha\in V,\mathrm\{ s.t.\} \alpha\not\in\bigcup\_\{1\leq i < j\leq s\}\ker(\sigma\_i-\sigma\_j). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
换句话说,
\begin\{aligned\} \forall\ 1\leq i < j\leq s, \sigma\_i(\alpha)\neq \sigma\_j(\alpha). \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)/
---
1529、 10、 (12 分) 设 $\displaystyle V$ 是线性空间, $\displaystyle W\_1,W\_2$ 均为 $\displaystyle V$ 的子空间, 且 $\displaystyle W\_1\cup W\_2$ 也是 $\displaystyle V$ 的子空间. 试证: $\displaystyle W\_1\subset W\_2$ 或 $\displaystyle W\_2\subset W\_1$, 且 $\displaystyle W\_1+W\_2=W\_1\cup W\_2$. (重庆师范大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 用反证法证明证明 $\displaystyle W\_1\subset W\_2$ 或 $\displaystyle W\_2\subset W\_1$. 若 $\displaystyle W\_1\not\subset W\_2$ 且 $\displaystyle W\_2\not\subset W\_1$, 则
\begin\{aligned\} \exists\ \alpha\_1\in W\_1,\mathrm\{ s.t.\} \alpha\_1\not \in W\_2; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \exists\ \alpha\_2\in W\_2,\mathrm\{ s.t.\} \alpha\_2\not\in W\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle \alpha=\alpha\_1+\alpha\_2$, 则 $\displaystyle \alpha\not\in W\_1\cup W\_2$. 事实上,
\begin\{aligned\} &\quad \alpha\in W\_i\quad (i=1,2)\\\\ &\Rightarrow \alpha\_j=\alpha-\alpha\_i\in W\_i\quad (j\neq i). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故有结论. (2)、 若 $\displaystyle W\_1\subset W\_2$, 则 $\displaystyle W\_1+W\_2=W\_2=W\_1\cup W\_2$; 若 $\displaystyle W\_2\subset W\_1$, 则 $\displaystyle W\_1+W\_2=W\_1=W\_1\cup W\_2$.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)/
---
1530、 (3)、 设 $\displaystyle V$ 为欧氏空间, $\displaystyle X$ 为 $\displaystyle V$ 的子集,
\begin\{aligned\} X^\perp=\left\\{\alpha\in V; \forall\ \beta\in X, (\alpha,\beta)=0\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle X^\perp$ 为 $\displaystyle V$ 的子空间的充要条件为 $\displaystyle X$ 为 $\displaystyle V$ 的子空间. (安徽大学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) / $\displaystyle \times$. 由
\begin\{aligned\} \alpha\_1,\alpha\_2\in X^\perp&\Rightarrow \forall\ \beta\in X, (k\_1\alpha\_1+k\_2\alpha\_2,\beta)=\sum\_\{i=1\}^2 k\_i(\alpha\_i,\beta)=0\\\\ &\Rightarrow k\_1\alpha\_1+k\_2\alpha\_2\in X^\perp \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知不管 $\displaystyle X$ 是否为 $\displaystyle V$ 的子空间, $\displaystyle X^\perp$ 始终为 $\displaystyle V$ 的子空间. 比如 $\displaystyle V=\mathbb\{R\}^3$, $\displaystyle X=\left\\{e\_1\right\\}$ 不是子空间, 但$X^\perp=L(e\_2,e\_3)$ 是子空间.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)/
---
1531、 (5)、 在带有标准内积的欧氏空间 $\displaystyle \mathbb\{R\}^3$ 中,
\begin\{aligned\} \alpha\_1=(1,1,0)^\mathrm\{T\}, \alpha\_2=(1,0,1)^\mathrm\{T\}, \alpha\_3=(-1,1,1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(5-1)、 求 $\displaystyle \mathbb\{R\}^3$ 的一组标准正交基 $\displaystyle \xi\_1,\xi\_2,\xi\_3$, 满足
\begin\{aligned\} L(\xi\_1)=L(\alpha\_1), L(\xi\_1,\xi\_2)=L(\alpha\_1,\alpha\_2), L(\xi\_1,\xi\_2,\xi\_3)=L(\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\}
(5-2)、 设矩阵 $\displaystyle A=(\alpha\_1,\alpha\_2,\alpha\_3)$, 求正交矩阵 $\displaystyle Q$ 和上三角阵 $\displaystyle R$, 使得 $\displaystyle A=QR$. (安徽大学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)、 对 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 施行 Gram-Schmidt 标准正交化过程得
\begin\{aligned\} \xi\_1=&\frac\{\alpha\_1\}\{|\alpha\_1|\} =\frac\{\alpha\_1\}\{\sqrt\{2\}\}=\left(\frac\{1\}\{\sqrt\{2\}\},\frac\{1\}\{\sqrt\{2\}\},0\right)^\mathrm\{T\},\\\\ \xi\_2=&\frac\{\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1\}\{|\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1|\} =\frac\{\alpha\_2-\frac\{1\}\{\sqrt\{2\}\}\alpha\_1\}\{\frac\{3\}\{\sqrt\{6\}\}\}=\left(\frac\{1\}\{\sqrt\{6\}\},-\frac\{1\}\{\sqrt\{6\}\},\frac\{2\}\{\sqrt\{6\}\}\right)^\mathrm\{T\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \xi\_3=&\frac\{\alpha\_3-(\alpha\_3,\beta\_1)\beta\_1-(\alpha\_3,\beta\_2)\beta\_2\}\{|\alpha\_3-(\alpha\_3,\beta\_1)\beta\_1-(\alpha\_3,\beta\_2)\beta\_2|\} =\frac\{\alpha\_3\}\{\sqrt\{3\}\}=\left(-\frac\{1\}\{\sqrt\{3\}\},\frac\{1\}\{\sqrt\{3\}\},\frac\{1\}\{\sqrt\{3\}\}\right)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这样得到的 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 就满题意. (5-2)、 设 $\displaystyle Q=(\xi\_1,\xi\_2,\xi\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&-\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\}\\\\ 0&\frac\{2\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{3\}\}\end\{array\}\right)$, 则由第 1 步的结果
\begin\{aligned\} \alpha\_1=\sqrt\{2\}\xi\_1, \alpha\_2=\frac\{1\}\{\sqrt\{2\}\}\xi\_1+\frac\{3\}\{\sqrt\{6\}\}\xi\_2, \alpha\_3=\sqrt\{3\}\xi\_3 \Rightarrow R=\left(\begin\{array\}\{cccccccccccccccccccc\}\sqrt\{2\}&\frac\{1\}\{\sqrt\{2\}\}&0\\\\ 0&\frac\{3\}\{\sqrt\{6\}\}&0\\\\ 0&0&\sqrt\{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\}
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)/
---
1532、 5、 (20 分) 设 $\displaystyle V$ 是 $\displaystyle n$ 维欧氏空间, $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组标准正交基, $\displaystyle \mathrm\{End\}(V)$ 表示 $\displaystyle V$ 的全体线性变换构成的线性空间. 定义
\begin\{aligned\} \left < \sigma,\tau\right > =\sum\_\{i=1\}^n \left(\sigma(\varepsilon\_i),\tau(\varepsilon\_i)\right), \forall\ \sigma,\tau\in \mathrm\{End\}(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 $\displaystyle (-,-)$ 为 $\displaystyle V$ 上的内积. (1)、 证明 $\displaystyle \left < -,-\right > $ 为 $\displaystyle \mathrm\{End\}(V)$ 上的内积; (2)、 求 $\displaystyle \mathrm\{End\}(V)$ 的一组标准正交基. (北京工业大学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)、 (1-1)、 由 $\displaystyle \left < \sigma,\sigma\right > =\sum\_\{i=1\}^n \left(\sigma(\varepsilon\_i),\sigma(\varepsilon\_i)\right)\geq 0$,
\begin\{aligned\} &\left < \sigma,\sigma\right > =0\Leftrightarrow \forall\ 1\leq i\leq n, \left(\sigma(\varepsilon\_i),\sigma(\varepsilon\_i)\right)=0\\\\ \Leftrightarrow&\forall\ 1\leq i\leq n, \sigma(\varepsilon\_i)=0\Leftrightarrow \sigma=0; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、
\begin\{aligned\} &\left < k\sigma+l\rho,\tau\right > =\sum\_\{i=1\}^n \left((k\sigma+l\rho)(\varepsilon\_i),\tau(\varepsilon\_i)\right)\\\\ =&k\sum\_\{i=1\}^n \left(\sigma(\varepsilon\_i),\tau(\varepsilon\_i)\right) +l\sum\_\{i=1\}^n \left(\rho(\varepsilon\_i),\tau(\varepsilon\_i)\right)\\\\ =&k\left < \sigma,\tau\right > +l\left < \rho,\tau\right > ; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-3)、
\begin\{aligned\} &\left < \sigma,\tau\right > =\sum\_\{i=1\}^n \left(\sigma(\varepsilon\_i),\tau(\varepsilon\_i)\right) =\sum\_\{i=1\}^n \left(\tau(\varepsilon\_i),\sigma(\varepsilon\_i)\right)=\left < \tau,\sigma\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 \left < -,-\right > $ 为 $\displaystyle \mathrm\{End\}(V)$ 上的内积; (2)、 取 $\displaystyle \sigma\_i\in \mathrm\{End\}(V)$ 使得
\begin\{aligned\} \sigma\_i(\varepsilon\_k)=\delta\_\{ik\}\varepsilon\_i=\left\\{\begin\{array\}\{llllllllllll\}\varepsilon\_i,&k=i,\\\\ 0,&k\neq i,\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} &\left < \sigma\_i,\sigma\_j\right > =\sum\_\{k=1\}^n \left(\sigma\_i(\varepsilon\_k),\sigma\_j(\varepsilon\_k)\right) =\sum\_\{k=1\}^n \left(\delta\_\{ik\}\varepsilon\_i,\delta\_\{jk\}\varepsilon\_j\right)\\\\ =&\sum\_\{k=1\}^n \delta\_\{ik\}\delta\_\{jk\}\delta\_\{ij\} =\delta\_\{ij\}=\left\\{\begin\{array\}\{llllllllllll\}1,&i=j,\\\\ 0,&i\neq j.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sigma\_1,\cdots,\sigma\_n$ 是 $\displaystyle \mathrm\{End\}(V)$ 的一组标准正交基.跟锦数学微信公众号. [在线资料](
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---
1533、 6、 (20 分) $\displaystyle \sigma$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 上的一个线性反对称变换, 即
\begin\{aligned\} \left(\sigma(\alpha),\beta\right)+\left(\alpha,\sigma(\beta)\right)=0,\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\}
其中 $\displaystyle \left(\cdot,\cdot\right)$ 表示欧氏空间的内积. 证明: 存在 $\displaystyle V$ 的一组基标准正交基, 使得 $\displaystyle \sigma^2$ 在此组基下的矩阵为对角阵. (北京科技大学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) / 设 $\displaystyle \tau=\sigma^2$, 则对 $\displaystyle \forall\ \alpha,\beta\in V$,
\begin\{aligned\} &\left(\tau(\alpha),\beta\right)=\left(\sigma^2(\alpha),\beta\right) =-\left(\sigma(\alpha),\sigma(\beta)\right) =-\left\[-\left(\alpha,\sigma^2(\beta)\right)\right\] =\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\}
取定 $\displaystyle V$ 的一组标准正交基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 设
\begin\{aligned\} \tau(\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\}
则由
\begin\{aligned\} \left(\tau(\varepsilon\_i),\varepsilon\_j\right)=&\left(\sum\_k a\_\{ki\}\varepsilon\_k,\varepsilon\_j\right)=a\_\{ji\},\\\\ \left(\varepsilon\_i,\tau(\varepsilon\_j)\right)=&\left(v\mathrm\{e\}\_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 $\displaystyle a\_\{ij\}=a\_\{ji\}$, 而 $\displaystyle A$ 实对称, 存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\in\mathbb\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle (\eta\_1,\cdots,\eta\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)P$, 则 $\displaystyle \eta\_1,\cdots,\eta\_n$ 是 $\displaystyle V$ 的一组标准正交基, 且
\begin\{aligned\} \tau^2(\eta\_1,\cdots,\eta\_n)=(\eta\_1,\cdots,\eta\_n)P^\{-1\}AP =(\eta\_1,\cdots,\eta\_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\}
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---
1534、 5、 若对于任意向量 $\displaystyle \alpha=(x\_1,x\_2)^\mathrm\{T\}, \beta=(y\_1,y\_2)^\mathrm\{T\}$, 定义
\begin\{aligned\} f(\alpha,\beta)=x\_1y\_1-x\_2y\_2. \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: $\displaystyle f$ 为双线性函数, 求其在 $\displaystyle \varepsilon\_1=(1,0)^\mathrm\{T\}, \varepsilon\_2=(0,1)^\mathrm\{T\}$ 下的度量矩阵; (2)、 $\displaystyle f$ 是否非退化; (3)、 求一组基, 使得 $\displaystyle f$ 在这组基下的度量矩阵为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\ 1&0\end\{array\}\right)$. (4)、 求所有使得 $\displaystyle f(\alpha,\alpha)=0$ 的非零向量 $\displaystyle \alpha$. (北京理工大学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)、 对 $\displaystyle \forall\ k,k',\alpha,\alpha'$,
\begin\{aligned\} &f(k\alpha+k'\alpha',\beta)=(kx\_1+k'x\_1')y\_1-(kx\_2+k'x\_2')y\_2\\\\ =&k(x\_1y\_1-x\_2y\_2)+k'(x\_1'y\_1-x\_2'y\_2) =kf(\alpha,\beta)+k'f(\alpha',\beta), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle f(\alpha,\beta)=f(\beta,\alpha)$. 故 $\displaystyle f$ 是双线性函数. 进一步, $\displaystyle f$ 在 $\displaystyle \varepsilon\_1,\varepsilon\_2$ 下的度量矩阵为
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}f(\varepsilon\_1,\varepsilon\_1)&f(\varepsilon\_1,\varepsilon\_2)\\\\ f(\varepsilon\_2,\varepsilon\_1)&f(\varepsilon\_1,\varepsilon\_2)\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ 0&-1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle \forall\ \beta$,
\begin\{aligned\} 0=&f(\alpha,\beta)=f(x\_1\varepsilon\_1+x\_2\varepsilon\_2,y\_1\varepsilon\_1+y\_2\varepsilon\_2)\\\\ =&\sum\_\{i,j\}x\_iy\_jf(\varepsilon\_i,\varepsilon\_j) =(x\_1,x\_2)A\left(\begin\{array\}\{cccccccccccccccccccc\}y\_1\\\\y\_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\}
则取 $\displaystyle \beta=\varepsilon\_1,\varepsilon\_2$ 后知
\begin\{aligned\} &0=(x\_1,x\_2)A\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\end\{array\}\right), 0=(x\_1,x\_2)A\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1\end\{array\}\right)\\\\ \Rightarrow&(0,0)=(x\_1,x\_2)A\stackrel\{|A|=-1\}\{\Rightarrow\}(0,0)=(x\_1,x\_2)\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\}
故 $\displaystyle f$ 非退化. 当然你可以直接由 $\displaystyle |A|\neq 0$ 知 $\displaystyle f$ 非退化. (3)、 设在基 $\displaystyle (\eta\_1,\eta\_2)=(\varepsilon\_1,\varepsilon\_2)T$ 下 $\displaystyle f$ 的度量矩阵为 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\ 1&0\end\{array\}\right)$, 则
\begin\{aligned\} b\_\{ij\}=&f(\eta\_i,\eta\_j)=f\left(\sum\_k t\_\{ki\}\varepsilon\_k, \sum\_l t\_\{lj\}\varepsilon\_l\right) =\sum\_\{k,l\}t\_\{ki\}a\_\{kl\}t\_\{lj\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle B=T^\mathrm\{T\} AT$. 由二次型化为规范形的结果知可取 $\displaystyle T=\frac\{1\}\{\sqrt\{2\}\}\left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ -1&1\end\{array\}\right)$,
\begin\{aligned\} \eta\_1=\frac\{(1,-1)^\mathrm\{T\}\}\{\sqrt\{2\}\}, \eta\_2=\frac\{(1,1)^\mathrm\{T\}\}\{\sqrt\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 $\displaystyle 0=f(\alpha,\alpha)=x\_1^2-x\_2^2\Leftrightarrow x\_2=\pm x\_1\Leftrightarrow \alpha=k\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\end\{array\}\right), \forall\ k\in\mathbb\{R\}$. 故所有使得 $\displaystyle f(\alpha,\alpha)=0$ 的非零向量 $\displaystyle \alpha=k\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\\pm 1\end\{array\}\right), \forall\ 0\neq k\in\mathbb\{R\}$.跟锦数学微信公众号. [在线资料](
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---
1535、 6、 设 $\displaystyle U$ 是内积空间 $\displaystyle V$ 的一个有限维子空间, $\displaystyle U^\perp$ 是 $\displaystyle U$ 的正交补. (1)、 证明: $\displaystyle U^\perp$ 是 $\displaystyle V$ 的子空间; (2)、 $\displaystyle V=U\oplus U^\perp$. (北京理工大学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)、 对 $\displaystyle \forall\ k,l\in\mathbb\{R\}, \alpha,\beta\in U^\perp$,
\begin\{aligned\} \forall\ \gamma\in U, (k\alpha+l\beta,\gamma)=k(\alpha,\gamma)+l(\beta,\gamma)=k0+l0=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle k\alpha+l\beta\in U^\perp$. 而 $\displaystyle U^\perp$ 是 $\displaystyle V$ 的子空间. (2)、 取定 $\displaystyle U$ 的一组标准正交基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 则对 $\displaystyle \forall\ \alpha\in V$,
\begin\{aligned\} \beta=\sum\_\{i=1\}^n (\alpha,\varepsilon\_i)\varepsilon\_i\in U, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且
\begin\{aligned\} (\alpha-\beta,\varepsilon\_j)=&(\alpha,\varepsilon\_j)-\left(\sum\_\{i=1\}^n (\alpha,\varepsilon\_i)\varepsilon\_i,\varepsilon\_j\right)\\\\ =&(\alpha,\varepsilon\_j)-(\alpha,\varepsilon\_j)=0, \forall\ 1\leq j\leq 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 $\displaystyle \alpha-\beta\in U^\perp$. 从而 $\displaystyle \alpha=\beta+(\alpha-\beta)\in U+U^\perp$. 又由
\begin\{aligned\} \alpha\in U\cap U^\perp\Rightarrow (\alpha,\alpha)\stackrel\{\alpha\in U, \alpha\in U^\perp\}\{=\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle U\cap U^\perp=\left\\{0\right\\}$. 这就证明了 $\displaystyle V=U\oplus U^\perp$.跟锦数学微信公众号. [在线资料](
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---
1536、 9、 设 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 是 $\displaystyle 3$ 维欧氏空间的一个基, $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 的度量矩阵为
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&1\\\\ -2&3&-1\\\\ 1&-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\}
令 $\displaystyle W=L(\alpha\_1+\alpha\_2,\alpha\_2+\alpha\_3)$. (1)、 求 $\displaystyle W$ 的一个规范正交基; (2)、 求 $\displaystyle W$ 的正交补 $\displaystyle W^\perp$, 并求 $\displaystyle W^\perp$ 的维数及一个规范正交基. (北京邮电大学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)、 设题中矩阵为 $\displaystyle A$, $\displaystyle \xi\_1=(1,1,0)^\mathrm\{T\}, \xi\_2=(0,1,1)^\mathrm\{T\}$,
\begin\{aligned\} \beta\_1=\alpha\_1+\alpha\_2=(\alpha\_1,\alpha\_2,\alpha\_3)\xi\_1, \beta\_2=\alpha\_2+\alpha\_3=(\alpha\_1,\alpha\_2,\alpha\_3\xi\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle |\beta\_1|=\sqrt\{\xi\_1^\mathrm\{T\} A\xi\_1\}=1, (\beta\_1,\beta\_2)=\xi\_1^\mathrm\{T\} A\xi\_2=1$. 将 $\displaystyle \beta\_1,\beta\_2$ 标准正交化为 $\displaystyle \gamma\_1=\frac\{\beta\_1\}\{|\beta\_1|\}=\beta\_1$,
\begin\{aligned\} \gamma\_2=&\frac\{\beta\_2-(\beta\_2,\gamma\_1)\gamma\_1\}\{|\beta\_2-(\beta\_2,\gamma\_1)\gamma\_1|\} =\frac\{\beta\_2-\beta\_1\}\{|\beta\_2-\beta\_1|\}=\frac\{-\alpha\_1+\alpha\_3\}\{\sqrt\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle W$ 的一个规范正交基为 $\displaystyle \alpha\_1+\alpha\_2,\frac\{-\alpha\_1+\alpha\_3\}\{\sqrt\{2\}\}$. (2)、 由
\begin\{aligned\} &\gamma=\sum\_i x\_i\alpha\_i\in W^\perp\Leftrightarrow 0=(\gamma,\beta\_i)=\xi\_i^\mathrm\{T\} Ax, i=1,2\\\\ \Leftrightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\xi\_1^\mathrm\{T\}\\\\ \xi\_2^\mathrm\{T\}\end\{array\}\right)Ax \Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0\\\\ -1&2&1\end\{array\}\right)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0\\\\ -1&2&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&0\end\{array\}\right)$ 知 $\displaystyle x=k(1,0,1)^\mathrm\{T\}, \forall\ k$. 于是 $\displaystyle W=L(\alpha\_1+\alpha\_3)$, $\displaystyle \dim W=1$, 且 $\displaystyle W$ 有一组标准正交基 $\displaystyle \frac\{\alpha\_1+\alpha\_3\}\{\sqrt\{6\}\}$.跟锦数学微信公众号. [在线资料](
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---
1537、 (7)、 设 $\displaystyle \mathscr\{A\}$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$ 上的线性变换, 若 $\displaystyle \mathscr\{A\}$ 不改变向量的距离, 即
\begin\{aligned\} |\mathscr\{A\}\alpha-\mathscr\{A\}\beta|=|\alpha-\beta|,\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: $\displaystyle \mathscr\{A\}$ 是正交变换. (大连理工大学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) / 由 $\displaystyle \mathscr\{A\}$ 是线性变换知 $\displaystyle \mathscr\{A\} 0=\mathscr\{A\}(\alpha-\alpha)=\mathscr\{A\}\alpha-\mathscr\{A\}\alpha=0$, 从而
\begin\{aligned\} |\mathscr\{A\}\alpha|=|\mathscr\{A\}\alpha-\mathscr\{A\} 0|=|\alpha-0|=|\alpha|,\quad \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\}
对 $\displaystyle \forall\ \alpha,\beta\in V$, 写出
\begin\{aligned\} &\left(\mathscr\{A\}(\alpha),\mathscr\{A\}(\alpha)\right)=(\alpha,\alpha), \left(\mathscr\{A\}(\beta),\mathscr\{A\}(\beta)\right)=(\beta,\beta),\\\\ &\left(\mathscr\{A\}(\alpha+\beta),\mathscr\{A\}(\alpha+\beta)\right)=(\alpha+\beta,\alpha+\beta). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后一个等式展开后利用前两个等式知
\begin\{aligned\} \left(\mathscr\{A\}(\alpha),\mathscr\{A\}(\beta)\right)=(\alpha,\beta). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就是说, $\displaystyle \mathscr\{A\}$ 是正交变换.跟锦数学微信公众号. [在线资料](
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---
1538、 (4)、 设 $\displaystyle V$ 是 $\displaystyle n\ ( > 1)$ 维欧氏空间, $\displaystyle \alpha$ 是 $\displaystyle V$ 中的单位向量. (4-1)、 $\displaystyle V$ 上的正交变换 $\displaystyle \mathscr\{B\}$ 称为关于 $\displaystyle \alpha$ 的镜面反射, 如果 $\displaystyle \mathscr\{B\}$ 满足:
\begin\{aligned\} \mbox\{若 $\displaystyle (\alpha,\beta)=0$, 则 $\displaystyle \mathscr\{B\}\beta=\beta; \mathscr\{B\}\alpha=-\alpha$.\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 如果正交变换 $\displaystyle \mathscr\{B\}$ 是关于 $\displaystyle \alpha$ 的镜面反射, 则
\begin\{aligned\} \mathscr\{B\}\gamma=\gamma-2(\alpha,\gamma)\alpha, \forall\ \gamma\in V; \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 $\displaystyle \beta$ 也是 $\displaystyle V$ 中的单位向量, 证明: 存在正交变换 $\displaystyle \mathscr\{A\}$ 使得
\begin\{aligned\} \dim\ker(\mathscr\{A\}-1\_V)=n-1\mbox\{且\} \mathscr\{A\}\alpha=\beta, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle 1\_V$ 表示 $\displaystyle V$ 上的恒等变换. (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) / (4-1)、 将 $\displaystyle \alpha$ 扩充为 $\displaystyle V$ 的一组标准正交基 $\displaystyle \alpha,\varepsilon\_2,\cdots,\varepsilon\_n$, 则由题设,
\begin\{aligned\} \mathscr\{B\}\alpha=-\alpha;\quad \mathscr\{B\}\varepsilon\_i=\varepsilon\_i, 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\}
对 $\displaystyle \forall\ \gamma=x\alpha+\sum\_\{i=2\}^n x\_i\varepsilon\_i\in V$,
\begin\{aligned\} \mathscr\{B\}\gamma=&-x\alpha+\sum\_\{i=2\}^n x\_i\varepsilon\_i =-2x\alpha+\left(x\alpha+\sum\_\{i=2\}^n x\_i\varepsilon\_i\right)\\\\ =&-2(\alpha,\gamma)\alpha+\gamma. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4-2)、 取一 $\displaystyle \beta-\alpha$ 为法向量的镜面反射
\begin\{aligned\} \mathscr\{A\}(\gamma)=\gamma-2(\gamma,\eta)\eta,\quad \forall\ \gamma\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 $\displaystyle \eta=\frac\{\beta-\alpha\}\{|\beta-\alpha|\}$, 则 $\displaystyle \mathscr\{A\}\alpha=\beta$. 由第 i 问知将 $\displaystyle \eta$ 扩充为 $\displaystyle V$ 的标准正交基 $\displaystyle \eta,\eta\_2,\cdots,\eta\_n$ 后, $\displaystyle \mathscr\{A\}$ 在该基下的矩阵就是 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}-1&\\\\ &E\_\{n-1\}\end\{array\}\right)$. 故
\begin\{aligned\} \dim \ker(\mathscr\{A\}-1\_V)=\dim L(\eta\_2,\cdots,\eta\_n)=n-1. \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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---
1539、 5、 设 $\displaystyle \varepsilon\_1,\varepsilon\_2,\varepsilon\_3$ 是 $\displaystyle 3$ 维欧氏空间 $\displaystyle V$ 的一组标准正交基. 已知
\begin\{aligned\} \alpha=3\varepsilon\_1+4\varepsilon\_2, \quad \beta=4\varepsilon\_1-3\varepsilon\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求一正交变换 $\displaystyle \mathscr\{A\}$ 使得 $\displaystyle \mathscr\{A\}\alpha=\beta$. (东北大学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 $\displaystyle \eta=\frac\{\beta-\alpha\}\{|\beta-\alpha|\}=\frac\{\varepsilon\_1-7\varepsilon\_2\}\{5\sqrt\{2\}\}$, 则
\begin\{aligned\} \mathscr\{A\}\gamma=2(\gamma,\eta)\eta, \forall\ \gamma\in V \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
就是满足 $\displaystyle \mathscr\{A\}\alpha=\beta$ 的正交变换. 事实上, 设 $\displaystyle \gamma=x\_1\varepsilon\_1+x\_2\varepsilon\_2+x\_3\varepsilon\_3$, 则由
\begin\{aligned\} (\gamma,\eta)\eta =\frac\{x\_1-7x\_2\}\{5\sqrt\{2\}\} \cdot \frac\{1\}\{5\sqrt\{2\}\}(\varepsilon\_1-7\varepsilon\_2) =\frac\{x\_1-7x\_2\}\{50\}(\varepsilon\_1-7\varepsilon\_2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mathscr\{A\}(x\_1\varepsilon\_1+x\_2\varepsilon\_2+x\_3\varepsilon\_3)=\frac\{24x\_1+7x\_2\}\{25\}\varepsilon\_1 +\frac\{7x\_1-24x\_2\}\{25\}\varepsilon\_2+x\_3\varepsilon\_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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---
1540、 6、 设 $\displaystyle \varepsilon\_1,\varepsilon\_2,\varepsilon\_3$ 是欧氏空间 $\displaystyle V$ 的一组基.已知 $\displaystyle \varepsilon\_1,\varepsilon\_2,\varepsilon\_3$ 的度量矩阵为
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}3&2&-2\\\\ 2&2&-1\\\\ -2&-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\}
令 $\displaystyle W=L(\varepsilon\_1+\varepsilon\_3,\varepsilon\_2+\varepsilon\_3)$. (1)、 求 $\displaystyle W$ 的维数及一组标准正交基; (2)、 求 $\displaystyle W^\perp$ 的维数和一组基. (东北大学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
\begin\{aligned\} \xi\_1=\varepsilon\_1+\varepsilon\_3, \xi\_2=\varepsilon\_2+\varepsilon\_3, \xi\_3=\varepsilon\_3, \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\} (\xi\_1,\xi\_2,\xi\_3)=(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&&1&0\\\\ 1&1&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} (\xi\_i,\xi\_j)=&\left(\begin\{array\}\{cccccccccccccccccccc\}\sum\_k a\_\{ki\}\varepsilon\_k, \sum\_l a\_\{lj\}\varepsilon\_l\end\{array\}\right) =\sum\_\{kl\}a\_\{ki\}g\_\{kl\}a\_\{lj\}, (g\_\{ij\})=G=\left(\begin\{array\}\{cccccccccccccccccccc\}3&2&-2\\\\ 2&2&-1\\\\ -2&-1&2\end\{array\}\right)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 的度量矩阵为
\begin\{aligned\} B=A^\mathrm\{T\} GA=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 1&2&0\\\\ 0&1&2\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&1&0\\\\ 0&0&1\end\{array\}\right)\Rightarrow B\_1=P\_1^\mathrm\{T\} BP\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&1\\\\ 0&1&2\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&-1\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} B\_1P\_2=E\_3 \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\} P=P\_1P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&1&-1\\\\ 0&0&1\end\{array\}\right)\Rightarrow P^\mathrm\{T\} BP=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
\begin\{aligned\} &(\eta\_1,\eta\_2,\eta\_3)=(\xi\_1,\xi\_2,\xi\_3)P\\\\ =&(\xi\_1,-\xi\_1+\xi\_2,\xi\_1-\xi\_2+\xi\_3) =(\varepsilon\_1+\varepsilon\_3,-\varepsilon\_1+\varepsilon\_2,\varepsilon\_1-\varepsilon\_2), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则类似 $\displaystyle (I)$ 的推理知 $\displaystyle \eta\_1,\eta\_2$ 是 $\displaystyle W$ 的一组标准正交基, $\displaystyle \dim W=2$; 而 $\displaystyle \eta\_3$ 是 $\displaystyle W^\perp$ 的一组基, $\displaystyle \dim W^\perp=1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1541、 6、 (15 分) 设 $\displaystyle V$ 是欧氏空间, $\displaystyle W$ 是 $\displaystyle V$ 的子空间, $\displaystyle V$ 中的向量 $\displaystyle \alpha$ 不在 $\displaystyle W$ 中, 问是否存在 $\displaystyle \alpha\_0\in W$, 使得 $\displaystyle \alpha-\alpha\_0$ 与 $\displaystyle W$ 的任意向量都正交? 如果不存在, 举出例子; 如果存在, 说明理由并讨论其唯一性. (东北师范大学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) / 存在且唯一. 设 $\displaystyle \eta\_1,\cdots,\eta\_m$ 是 $\displaystyle W$ 的一组标准正交基, 将其扩充为 $\displaystyle V$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_n$. 而可设 $\displaystyle \alpha=\sum\_\{i=1\}^n x\_i\eta\_i$, 其中 $\displaystyle x\_i=(\alpha,\eta\_i)$. 令 $\displaystyle \alpha\_0=\sum\_\{i=1\}^m x\_i\eta\_i\in W$, 则
\begin\{aligned\} \forall\ \beta=\sum\_\{i=1\}^m y\_i\eta\_i\in W, (\alpha-\alpha\_0,\beta) =\left(\sum\_\{i=m+1\}^n x\_i\eta\_i,\sum\_\{j=1\}^m y\_j\eta\_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\}
故 $\displaystyle \alpha-\alpha\_0\in W^\perp$. 若 $\displaystyle \alpha\_1\in W$ 也满足 $\displaystyle \alpha-\alpha\_1\in W^\perp$, 则
\begin\{aligned\} &\alpha\_0\in W, \alpha\_1\in W\Rightarrow \alpha\_0-\alpha\_1\in W, &\alpha\_0-\alpha\_1=(\alpha-\alpha\_1)-(\alpha-\alpha\_0)\in W^\perp. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \alpha\_0-\alpha\_1\in W\cap W^\perp$,
\begin\{aligned\} (\alpha\_0-\alpha\_1,\alpha\_0-\alpha\_1)\stackrel\{\mbox\{第一个向量在 $\displaystyle W$ 中, 第二个向量在 $\displaystyle W^\perp$ 中\}\}\{=\}0\Rightarrow \alpha\_0=\alpha\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
唯一性得证.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 13:25:12
