## 张祖锦2023年数学专业真题分类70天之第62天
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1404、 7、 设线性空间 $\displaystyle V\_1$ 的一组基为
\begin\{aligned\} \alpha\_1=(1,2,1)^\mathrm\{T\}, \alpha\_2=(1,1,-1)^\mathrm\{T\}, \alpha\_3=(1,3,3)^\mathrm\{T\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
线性空间 $\displaystyle V\_2$ 的一组基为
\begin\{aligned\} \beta\_1=(2,3,-1)^\mathrm\{T\}, \beta\_2=(1,2,2)^\mathrm\{T\}, \beta\_3=(1,1,-3)^\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 $\displaystyle V\_1+V\_2$ 与 $\displaystyle V\_1\cap V\_2$ 的一组基与维数. (河北工业大学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\} A=&(\alpha\_1,\alpha\_2,\alpha\_3,\beta\_1,\beta\_2,\beta\_3) =\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1&2&1&1\\\\ 2&1&3&3&2&1\\\\ 1&-1&3&-1&2&-3\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&2&0&2&-2\\\\ 0&1&-1&0&1&-1\\\\ 0&0&0&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\}
知 $\displaystyle \alpha\_1,\alpha\_2,\beta\_1$ 线性无关, 是 $\displaystyle V\_1+V\_2$ 的一组基, $\displaystyle \dim(V\_1+V\_2)=3$, 且 [张祖锦注: 初等行变换不改变列向量组的秩及极大无关组所在的位置, 并且可以一下得到其余向量用极大无关组的表示法, 这是张祖锦独创的, 具体证明见张祖锦编著的《樊启斌参考书》中张祖锦常用的结论.]
\begin\{aligned\} \alpha\_3=2\alpha\_1-\alpha\_2, \beta\_2=2\alpha\_1+\alpha\_2-\beta\_1, \beta\_3=-2\alpha\_1-\alpha\_2+2\beta\_1. \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 V\_1\cap V\_2\Leftrightarrow \alpha=\sum\_i x\_i\alpha\_i=-\sum\_i x\_i\beta\_i\\\\ \Leftrightarrow&\alpha=\sum\_i x\_i\alpha\_i, Ax=0\\\\ \stackrel\{(I)\}\{\Leftrightarrow\}&\alpha=\sum\_i x\_i\alpha\_i, x\in L\left(\begin\{array\}\{cccccccccccccccccccc\}(-2,1,1,0,0,0)^\mathrm\{T\}, (-2,-1,0,1,1,0)^\mathrm\{T\},\\\\ (2,1,0,-2,0,1)^\mathrm\{T\}\end\{array\}\right)\\\\ \Leftrightarrow&\alpha\in L(-2\alpha\_1+\alpha\_2+\alpha\_3, -2\alpha\_1-\alpha\_2, 2\alpha\_1+\alpha\_2)\\\\ &=L(2\alpha\_1+\alpha\_2)=L((3,5,1)^\mathrm\{T\}) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle (3,5,1)^\mathrm\{T\}$ 是 $\displaystyle V\_1\cap V\_2$ 的一组基, $\displaystyle \dim(V\_1\cap V\_2)=1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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1405、 (2)、 二维线性空间 $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{A\}$ 在 $\displaystyle V$ 的基 $\displaystyle \varepsilon\_1,\varepsilon\_2$ 下的矩阵为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}2&4\\\\ 4&1\end\{array\}\right)$, 则 $\displaystyle \mathscr\{A\}$ 在基 $\displaystyle \varepsilon\_1+\varepsilon\_2,\varepsilon\_2$ 下的矩阵为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (黑龙江大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设,
\begin\{aligned\} \mathscr\{A\}(\varepsilon\_1,\varepsilon\_2)=&(\varepsilon\_1,\varepsilon\_2)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&4\\\\ 4&1\end\{array\}\right);\\\\ (\varepsilon\_1+\varepsilon\_1,\varepsilon\_2)=&(\varepsilon\_1,\varepsilon\_2)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ 1&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\mathscr\{A\}(\varepsilon\_1+\varepsilon\_1,\varepsilon\_2)=\mathscr\{A\}(\varepsilon\_1,\varepsilon\_2)T =(\varepsilon\_1,\varepsilon\_2)AT =(\varepsilon\_1+\varepsilon\_1,\varepsilon\_2)T^\{-1\}AT. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathscr\{A\}$ 在基 $\displaystyle \varepsilon\_1+\varepsilon\_2,\varepsilon\_2$ 下的矩阵为 $\displaystyle T^\{-1\}AT=\left(\begin\{array\}\{cccccccccccccccccccc\}6&4\\\\-1&-3\end\{array\}\right)$.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)/
---
1406、 (2)、 设 $\displaystyle \mathbb\{P\}[x]\_4$ 为数域 $\displaystyle \mathbb\{P\}$ 上的次数小于 $\displaystyle 4$ 的多项式与零多项式构成的线性空间, $\displaystyle \mathscr\{A\}$ 是 $\displaystyle \mathbb\{P\}[x]\_4$ 上的线性变换, 满足
\begin\{aligned\} \mathscr\{A\} f(x)=f'(x)-2f(x),\quad \forall\ f(x)\in \mathbb\{P\}[x]\_4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求 $\displaystyle \mathscr\{A\}$ 在基 $\displaystyle 1,x,x^2,x^3$ 下的矩阵; (2)、 求 $\displaystyle \mathscr\{A\}$ 的核与值域. (黑龙江大学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\} \mathscr\{A\} 1=-2, \mathscr\{A\} x=1-2x, \mathscr\{A\} x^2=2x-2x^2, \mathscr\{A\} x^3=3x^2-2x^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\} \mathscr\{A\}(1,x,x^2,x^3)=(1,x,x^2,x^3)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&1&0&0\\\\ 0&-2&2&0\\\\ 0&0&-2&3\\\\ 0&0&0&-2\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle A$ 可逆知 $\displaystyle \mathscr\{A\}$ 可逆, 而 $\displaystyle \ker\mathscr\{A\}=\left\\{0\right\\}, \mathrm\{im\}\mathscr\{A\}=\mathbb\{P\}[x]\_4$.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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1407、 (4)、 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性无关, $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ 线性相关, $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_5$ 线性无关. 证明: (1)、 $\displaystyle \alpha\_4$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表示; (2)、 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4-\alpha\_5$ 线性无关. (黑龙江大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ 线性相关知存在不全为 $\displaystyle 0$ 的数 $\displaystyle k\_1,\cdots,k\_4$, 使得
\begin\{aligned\} k\_1\alpha\_1+k\_2\alpha\_2+k\_3\alpha\_3+k\_4\alpha\_4=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
往用反证法证明 $\displaystyle k\_4\neq 0$. 若不然, $\displaystyle k\_4=0$, 则 $\displaystyle k\_1,k\_2,k\_3$ 不全为 $\displaystyle 0$, 且 $\displaystyle k\_1\alpha\_1+k\_2\alpha\_2+k\_3\alpha\_3=0$. 而 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性相关. 这与题设矛盾. 故 $\displaystyle k\_4\neq 0$,
\begin\{aligned\} \alpha\_4=-\frac\{1\}\{k\_4\}(k\_1\alpha\_1+k\_2\alpha\_2+k\_3\alpha\_3). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \alpha\_4$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表示. (2)、 由
\begin\{aligned\} &(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4-\alpha\_5)\\\\ =&\left(\alpha\_1,\alpha\_2,\alpha\_3,-\frac\{1\}\{k\_4\}(k\_1\alpha\_1+k\_2\alpha\_2+k\_3\alpha\_3)-\alpha\_5\right)\\\\ =&(\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_5)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-k\_1/k\_4\\\\ 0&1&0&-k\_2/k\_4\\\\ 0&0&1&-k\_3/k\_4\\\\ 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\}
及 $\displaystyle A$ 可逆知 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4-\alpha\_5$ 线性无关.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)/
---
1408、 3、 设数域 $\displaystyle \mathbb\{K\}$ 上的多项式 $\displaystyle f\_1(x), \cdots, f\_s(x)$ 的最大公因式为 $\displaystyle f\_0(x)$, 且 $\displaystyle f\_0(x)$ 的次数 $\displaystyle \deg f(x)=d < n$. 设
\begin\{aligned\} V\_n=\left\\{f(x)=\sum\_\{i=1\}^s u\_i(x)f\_i(x); u\_i(x)\in\mathbb\{K\}[x], \deg f(x) < n\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 证明: $\displaystyle V\_n$ 对于多项式加法及数乘运算构成 $\displaystyle \mathbb\{K\}$ 上的线性空间; (2)、 计算线性空间 $\displaystyle V\_n$ 的维数, 并写出 $\displaystyle V\_n$ 的一组基. (湖南大学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) / (1)、 对 $\displaystyle \forall\ k,l\in\mathbb\{K\}$,
\begin\{aligned\} f(x)=\sum\_\{i=1\}^s u\_i(x)f\_i(x), g(x)=\sum\_\{i=1\}^s v\_i(x)f\_i(x)\in V\_n, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
有
\begin\{aligned\} kf(x)+lg(x)=\sum\_\{i=1\}^s [ku\_i(x)+lv\_i(x)]f\_i(x)\in V\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle V\_n$ 是 $\displaystyle \mathbb\{P\}[x]\_n$ 的线性子空间, 而是 $\displaystyle \mathbb\{K\}$ 上的线性空间. (2)、 为证第 2 问, 我们先给出一个结果. 设 $\displaystyle p\_1(x),\cdots,p\_s(x)$ 为 $\displaystyle \mathbb\{R\}[x]$ 中的非零多项式, 其次数各不相同, 分别为 $\displaystyle n\_1,\cdots,n\_s$. 则 $\displaystyle p\_1(x),\cdots, p\_s(x)$ 在实数域上线性无关. 事实上, 不妨设 $\displaystyle n\_1 < \cdots < n\_s$, $\displaystyle p\_i(x)$ 的最高次项系数为 $\displaystyle a\_i$. 再设
\begin\{aligned\} c\_1p\_1(x)+\cdots+c\_sp\_s(x)=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\}
若 $\displaystyle c\_s\neq 0$, 则由
\begin\{aligned\} \deg\left(c\_1p\_1(x)+\cdots+c\_\{s-1\}p\_\{s-1\}(x)\right)\leq n\_\{s-1\} < n\_s \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle (I)$ 的左端是 $\displaystyle n\_s$ 次多项式, 不等于 $\displaystyle 0$. 这与 $\displaystyle (I)$ 矛盾. 故 $\displaystyle c\_s=0$. 同理可不断推得 $\displaystyle c\_\{s-1\}=0, \cdots,c\_1=0$. 故有结论. (3)、 由 $\displaystyle (f\_1,\cdots,f\_n)=f\_0$ 知
\begin\{aligned\} \exists\ g\_i\in \mathbb\{K\}[x],\mathrm\{ s.t.\} f\_i=g\_if\_0; \exists\ w\_i\in\mathbb\{K\}[x],\mathrm\{ s.t.\} f\_0=\sum\_\{i=1\}^s w\_if\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle f(x)=\sum\_\{i=1\}^s u\_i(x)f\_i(x)\in V\_n$,
\begin\{aligned\} f(x)=\sum\_\{i=1\}^s u\_i(x)\cdot g\_i(x)f\_0(x)\Rightarrow f\_0\mid f. \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 $\displaystyle f\_0\mid f$, 则
\begin\{aligned\} \exists\ g\in\mathbb\{K\}[x],\mathrm\{ s.t.\} f=gf\_0=g\sum\_\{i=1\}^s w\_if\_i =\sum\_\{i=1\}^s (gw\_i)f\_i\in V\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了
\begin\{aligned\} f\in V\_n\Leftrightarrow&\exists\ g\in \mathbb\{K\}[x],\mathrm\{ s.t.\} f=gf\_0, \deg f < n\\\\ &\Leftrightarrow \exists g(x)=\sum\_\{i=0\}^\{n-1-d\}a\_ix^i, \mathrm\{ s.t.\} f(x)=\sum\_\{i=0\}^\{n-1-d\}a\_ix^if\_0(x)\\\\ &\Leftrightarrow f\in L\left(f\_0(x), xf\_0(x),\cdots, x^\{n-1-d\}f\_0(x)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 2 步知 $\displaystyle f\_0(x), xf\_0(x),\cdots, x^\{n-1-d\}f\_0(x)$ 线性无关, 而是 $\displaystyle V\_n$ 的一组基, $\displaystyle \dim V\_n=n-d$.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)/
---
1409、 5、 设
\begin\{aligned\} \mathbb\{K\}[x]\_n=\left\\{f(x)\in\mathbb\{K\}[x]; \deg f(x) < n\right\\}\cup\left\\{0\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为数域 $\displaystyle \mathbb\{K\}$ 上次数小于 $\displaystyle n$ 的多项式全体构成的线性空间, 定义 $\displaystyle \mathbb\{K\}[x]\_n$ 上的线性变换 $\displaystyle \mathscr\{A\}$ 如下:
\begin\{aligned\} \mathscr\{A\} f(x)=f(x)+f'(x)+\frac\{f‘(x)\}\{2\}+\cdots+\frac\{f^\{(n-1)\}(x)\}\{(n-1)!\}, \forall\ f(x)\in \mathbb\{K\}[x]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 写出线性变换 $\displaystyle \mathscr\{A\}$ 在 $\displaystyle \mathbb\{K\}[x]\_n$ 的基
\begin\{aligned\} \varepsilon\_0=1, \varepsilon\_1=x, \varepsilon\_2=x^2, \cdots, \varepsilon\_\{n-1\}=x^\{n-1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
下的矩阵 $\displaystyle A$; (2)、 证明:
\begin\{aligned\} \eta\_0=1, \eta\_1=x-1, \eta\_2=(x-1)^2, \cdots, \eta\_\{n-1\}=(x-1)^\{n-1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle \mathbb\{K\}[x]\_n$ 的一组基; (3)、 证明: 线性变换 $\displaystyle \mathscr\{A\}$ 是双射, 且
\begin\{aligned\} \mathscr\{A\}^\{-1\} f(x)=f(x)-f'(x)+\frac\{f''(x)\}\{2\}+\cdots+(-1)^\{n-1\}\frac\{f^\{(n-1)\}(x)\}\{(n-1)!\}; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 判断线性变换 $\displaystyle \mathscr\{A\}$ 是否可对角化, 并给出理由. (湖南大学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) / (1)、 由
\begin\{aligned\} \mathscr\{A\} x^k=&x^k+kx^\{k-1\}+\frac\{k(k-1)x^\{k-2\}\}\{2\} +\frac\{k(k-1)(k-2)x^\{k-3\}\}\{3!\}+\cdots\\\\ =&\sum\_\{i=0\}^k C\_k^ix^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\} \mathscr\{A\}(\varepsilon\_0,\cdots,\varepsilon\_\{n-1\})=(\varepsilon\_0,\cdots,\varepsilon\_\{n-1\})A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1&\cdots&1\\\\ &1&2&\cdots&C\_\{n-1\}^2\\\\ &&1&\cdots&C\_\{n-1\}^3\\\\ &&&\ddots&\vdots\\\\ &&&&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)、 (2-1)、 我们先给出一个结果. 设 $\displaystyle p\_1(x),\cdots,p\_s(x)$ 为 $\displaystyle \mathbb\{R\}[x]$ 中的非零多项式, 其次数各不相同, 分别为 $\displaystyle n\_1,\cdots,n\_s$. 则 $\displaystyle p\_1(x),\cdots, p\_s(x)$ 在实数域上线性无关. 事实上, 不妨设 $\displaystyle n\_1 < \cdots < n\_s$, $\displaystyle p\_i(x)$ 的最高次项系数为 $\displaystyle a\_i$. 再设
\begin\{aligned\} c\_1p\_1(x)+\cdots+c\_sp\_s(x)=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\}
若 $\displaystyle c\_s\neq 0$, 则由
\begin\{aligned\} \deg\left(c\_1p\_1(x)+\cdots+c\_\{s-1\}p\_\{s-1\}(x)\right)\leq n\_\{s-1\} < n\_s \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle (I)$ 的左端是 $\displaystyle n\_s$ 次多项式, 不等于 $\displaystyle 0$. 这与 $\displaystyle (I)$ 矛盾. 故 $\displaystyle c\_s=0$. 同理可不断推得 $\displaystyle c\_\{s-1\}=0, \cdots,c\_1=0$. 故有结论. (2-2)、 由 $\displaystyle \deg \eta\_i=i, 0\leq i\leq n-1$ 及第 i 步知 $\displaystyle \eta\_0,\cdots,\eta\_\{n-1\}$ 线性无关, 是 $\displaystyle n$ 维线性空间 $\displaystyle \mathbb\{K\}[x]\_n$ 的一组基. (3)、 由 $\displaystyle |A|=1\neq 0$ 知 $\displaystyle \mathscr\{A\}$ 是可逆的, 是双射. 再者, 考虑 $\displaystyle \mathbb\{K\}[x]\_n$ 的两组基
\begin\{aligned\} \varepsilon\_i: 1,x,x^2,\cdots,x^\{n-1\};\quad \eta\_i: 1,x-1,(x-1)^2,\cdots,(x-1)^\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} &\left(1,x-1,(x-1)^2,\cdots,(x-1)^\{n-1\}\right)\\\\ =&(1,x,x^2,\cdots,x^\{n-1\})\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1&\cdots&(-1)^\{n-1\}\\\\ 0&1&-2&\cdots&(-1)^\{n-2\}C\_\{n-1\}^1\\\\ 0&0&1&\cdots&(-1)^\{n-3\}C\_\{n-1\}^2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&0&0&\cdots&1\end\{array\}\right)\\\\ =&(1,x,x^2,\cdots,x^\{n-1\})T. \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\} (1,x,x^2,\cdots,x^\{n-1\})=&\left(1,x-1,(x-1)^2,\cdots,(x-1)^\{n-1\}\right)T^\{-1\}.\qquad(1) \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
\begin\{aligned\} x^\{i-1\}=&\left\[(x-1)+1\right\]^\{i-1\} =\sum\_\{j=0\}^\{i-1\}C\_\{i-1\}^j (x-1)^j\\\\ =&\sum\_\{j=1\}^i C\_\{i-1\}^\{j-1\} (x-1)^\{j-1\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们知
\begin\{aligned\} &(1,x,x^2,\cdots,x^\{n-1\})\\\\ =&\left(1,x-1,(x-1)^2,\cdots,(x-1)^\{n-1\}\right) \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1&\cdots&C\_\{n-1\}^1\\\\ 0&1&2&\cdots&C\_\{n-1\}^1\\\\ 0&0&1&\cdots&C\_\{n-1\}^2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&0&0&\cdots&1\end\{array\}\right).\qquad(2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
比较 $\displaystyle (1),(2)$ 即知
\begin\{aligned\} &T^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1&\cdots&C\_\{n-1\}^1\\\\ 0&1&2&\cdots&C\_\{n-1\}^1\\\\ 0&0&1&\cdots&C\_\{n-1\}^2\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ 0&0&0&\cdots&1\end\{array\}\right)=A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \mathscr\{A\}^\{-1\}(1,x,x^2,\cdots,x^\{n-1\})=&(1,x,x^2,\cdots,x^\{n-1\})A^\{-1\}\\\\ =&(1,x,x^2,\cdots,x^\{n-1\})T. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此即
\begin\{aligned\} \mathscr\{A\}^\{-1\} x^k=&x^k-C\_k^1 x^\{k-1\}+C\_k^2 x^\{k-2\}+\cdots+(-1)^k\\\\ =&x^k-(x^k)'+\frac\{(x^k)''\}\{2!\}-\cdots+(-1)^k \frac\{(x^k)^\{(k)\}\}\{k!\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} \mathscr\{A\}^\{-1\} f(x)=f(x)-f'(x)+\frac\{f''(x)\}\{2\}+\cdots+(-1)^\{n-1\}\frac\{f^\{(n-1)\}(x)\}\{(n-1)!\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1$. 又由 $\displaystyle A-E$ 的右上角有一个 $\displaystyle n-1$ 阶子式 $\displaystyle =(n-1)!\neq 0$ 知
\begin\{aligned\} \mathrm\{rank\}(A-E)=n-1\Rightarrow A\mbox\{的 Jordan 标准形为 \}\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &\ddots&\ddots&\\\\ &&\ddots&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\}
于是 $\displaystyle \mathscr\{A\}$ 在 $\displaystyle \mathbb\{C\}$ 上不可对角化, 进而在 $\displaystyle \mathbb\{K\}$ 中不可对角化. 当然, 你可以由 $\displaystyle \mathscr\{A\}$ 的属于特征值 $\displaystyle 1$ 的特征向量只有一个知 $\displaystyle \mathscr\{A\}$ 不可对角化.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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---
1410、 (3)、 考虑欧氏空间 $\displaystyle \mathbb\{R\}^4$ 中的向量
\begin\{aligned\} &\alpha\_1=(1,3,1,-1), \alpha\_2=(2,3,2,1),\\\\ &\beta\_1=(3,-1,-3,-5), \beta\_2=(2,-1,0,1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle W\_1$ 是由 $\displaystyle \alpha\_1,\alpha\_2$ 生成的子空间, $\displaystyle W\_2$ 是由 $\displaystyle \beta\_1,\beta\_2$ 生成的子空间, 则 $\displaystyle W\_1\cap W\_2$ 的维数是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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\} A&=(\alpha\_1^\mathrm\{T\}, \alpha\_2^\mathrm\{T\}, \beta\_1^\mathrm\{T\}, \beta\_2^\mathrm\{T\}) =\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&3&2\\\\ 3&3&-1&-1\\\\ 1&2&-3&0\\\\ -1&1&-5&1\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-\frac\{13\}\{9\}\\\\ 0&1&0&\frac\{11\}\{9\}\\\\ 0&0&1&\frac\{1\}\{3\}\\\\ 0&0&0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle AX=0$ 的基础解系为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}13\\\\-11\\\\-3\\\\9\end\{array\}\right)$. 故
\begin\{aligned\} \alpha\in W\_1\cap W\_2\Leftrightarrow&\alpha=x\_1\alpha\_1+x\_2\alpha\_2=-x\_3\beta\_1-x\_4\beta\_2\\\\ \Leftrightarrow&\alpha=x\_1\alpha\_1+x\_2\alpha\_2, Ax=0\\\\ \Leftrightarrow&\alpha=k(13\alpha\_1-11\alpha\_2)=3k(-3,2,-3,-8). \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 $\displaystyle W\_1\cap W\_2=L\left((-3,2,-3,-8)\right), \dim (W\_1\cap W\_2)=1$.跟锦数学微信公众号. [在线资料](
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---
1411、 2、 解答题 (每题 20 分, 共 100 分). (1)、 考虑由所有二阶复系数方阵构成的集合
\begin\{aligned\} M\_2(\mathbb\{C\})=\left\\{\left(\begin\{array\}\{cccccccccccccccccccc\}a&b\\\\ c&d\end\{array\}\right); a,b,c,d\in\mathbb\{C\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
已知 $\displaystyle M\_2(\mathbb\{C\})$ 是以 $\displaystyle E\_\{11\},E\_\{12\},E\_\{21\},E\_\{22\}$ 为基的复线性空间, 这里 $\displaystyle E\_\{ij\}$ 是指出第 $\displaystyle i$ 行第 $\displaystyle j$ 列元素为 $\displaystyle 1$ 外, 其余元素均为 $\displaystyle 0$ 的二阶方阵. 设
\begin\{aligned\} B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ 1&1\end\{array\}\right)=E\_\{11\}+E\_\{12\}+E\_\{21\}+E\_\{22\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-1)、 证明: 如下映射为线性映射,
\begin\{aligned\} \begin\{array\}\{cccc\} \varphi\_B: &M\_2(\mathbb\{C\})&\to&M\_2(\mathbb\{C\}),\\\\ &X&\mapsto&\varphi\_B(X)=BX. \end\{array\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 求 $\displaystyle \varphi\_B$ 在上述基下的表示矩阵. (1-3)、 分别求核空间 $\displaystyle \ker\varphi\_B$ 和像空间 $\displaystyle \mathrm\{im\} \varphi\_B$ 的维数与基. (1-4)、 求 $\displaystyle \varphi\_B$ 的若尔当典范形. (华东师范大学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) / (1-1)、 对 $\displaystyle \forall\ k,l\in\mathbb\{C\}, X,Y\in M\_2(\mathbb\{C\})$,
\begin\{aligned\} \varphi\_B(kX+lY)=&B(kX+lY) =kBX+lBY\\\\ =&k\varphi\_B(X)+l\varphi\_B(Y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \varphi\_B$ 是线性映射. (1-2)、 由
\begin\{aligned\} \varphi\_B E\_\{11\}=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\1&0\end\{array\}\right), \varphi\_B E\_\{12\}=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\0&1\end\{array\}\right),\\\\ \varphi\_B E\_\{21\}=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\1&0\end\{array\}\right), \varphi\_B E\_\{22\}=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\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\}
知
\begin\{aligned\} \varphi\_B(E\_\{11\},E\_\{12\},E\_\{21\},E\_\{22\}) =&(E\_\{11\},E\_\{12\},E\_\{21\},E\_\{22\})A,\\\\ A=&\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&1&0\\\\ 0&1&0&1\\\\ 1&0&1&0\\\\ 0&1&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\}
(1-3)、 由
\begin\{aligned\} A\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1&0\\\\ 0&1&0&1\\\\ 0&0&0&0\\\\ 0&0&0&0\end\{array\}\right)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle Ax=0$ 的基础解系为 $\displaystyle (-1,0,1,0)^\mathrm\{T\}, (0,-1,0,1)^\mathrm\{T\}$, 而
\begin\{aligned\} \ker \varphi\_B=L\left(\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0\\\\1&0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}0&-1\\\\0&1\end\{array\}\right)\right)\Rightarrow \dim \ker \varphi\_B=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此外, $\displaystyle A$ 的列向量组的一个极大无关组为 $\displaystyle Ae\_1,Ae\_2$, 而
\begin\{aligned\} \mathrm\{im\}\varphi\_B=L\left(\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\1&0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\0&1\end\{array\}\right)\right)\Rightarrow \dim \mathrm\{im\}\varphi\_B=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-4)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,0,0$. 由
\begin\{aligned\} 2E-A\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1&0\\\\ 0&1&0&-1\\\\ 0&0&0&0\\\\ 0&0&0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 $\displaystyle (I)$ 知属于特征值 $\displaystyle 2$ 与 $\displaystyle 0$ 的特征向量都有 $\displaystyle 2$ 个. 故 $\displaystyle \varphi\_B$ 可对角化, 其 Jordan 标准形为 $\displaystyle \mathrm\{diag\}(2,2,0,0)$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1412、 (3)、 已知 $\displaystyle D: \mathbb\{R\}[x]\to \mathbb\{R\}[x]$ 是实系数多项式空间上的线性映射, 满足: (3-1)、 $\displaystyle D(fg)=D(f)g+fD(g), \forall\ f,g\in \mathbb\{R\}[x]$; (3-2)、 $\displaystyle D(x)=1$. (3-1)、 证明: $\displaystyle D(f)=f'$ 是 $\displaystyle f$ 的形式导数; (3-2)、 求 $\displaystyle D$ 限制在 $\displaystyle \mathbb\{R\}[x]\_n$ 上的所有不变子空间, 其中 $\displaystyle \mathbb\{R\}[x]\_n$ 是次数不超过 $\displaystyle n$ 的实多项式空间. [张祖锦注: 如果没有条件’线性‘, 则 $\displaystyle D$ 未必是导数! 反例见参考解答.] (华东师范大学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) / (3-1)、 若 $\displaystyle D(x^k)=x^\{k-1\}$, 则
\begin\{aligned\} D(x^\{k+1\})=&D(x^k\cdot x) =D(x^k)\cdot x+x^kD(x)\\\\ =&kx^\{k-1\}\cdot x+x^k\cdot 1=(k+1)x^k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故由数学归纳法知 $\displaystyle D(x^k)=kx^\{k-1\}, k\geq 1$. 由 $\displaystyle D$ 的线性性知 $\displaystyle D(f)=f'$. 注意. 如果没有要求 $\displaystyle D$ 是线性的, 则得不到结果. 反例如下. 对任意 $\displaystyle \mathbb\{R\}$ 上的不可约多项式 $\displaystyle f$, 定义 $\displaystyle D(f)=1$. 对任一非零多项式 $\displaystyle f(x)=c\prod\_k f\_k$, 其中 $\displaystyle f\_k$ 不可约, 可能相同, 定义
\begin\{aligned\} Df=f\sum\_k \frac\{D(f\_k)\}\{f\_k\}=f\sum\_k \frac\{1\}\{f\_k\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle D$ 是良定义的, 且由
\begin\{aligned\} &f=c \prod\_k f\_k, g=d \prod\_l g\_l\\\\ \Rightarrow&fg=cd \prod\_k f\_k \prod\_l g\_l, Df=f\sum\_k \frac\{1\}\{f\_k\}, Dg=g\sum\_l \frac\{1\}\{g\_l\}\\\\ \Rightarrow&D(fg)=fg\left(\sum\_k \frac\{1\}\{f\_k\}+\sum\_l \frac\{1\}\{g\_l\}\right) =gD(f)+fD(g) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle D$ 确实满足
\begin\{aligned\} D(fg)=D(f)g+fD(g), \forall\ f,g\in \mathbb\{R\}[x];\quad D(x)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但由 $\displaystyle D(x^2+1)=1$ 知 $\displaystyle D$ 不是求导算子. (3-2)、 设 $\displaystyle W$ 是 $\displaystyle D$ 的非零不变子空间, 则考虑 $\displaystyle W$ 中次数最大的多项式
\begin\{aligned\} f(x)=a\_mx^m+\cdots+a\_0\in W, a\_m\neq 0, m\geq 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} D^mf(x)=m!a\_m\in W\Rightarrow 1\in 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\} D^\{m-1\}f(x)=a\_m m!x+a\_\{m-1\}(m-1)!\in W\Rightarrow x\in 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\} &D^\{m-2\}f(x)=m(m-1)\cdots 3x^2 +(m-1)!x+(m-2)!\in W\\\\ \Rightarrow& x^2\in W, \cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最终 $\displaystyle x^m\in W$. 于是 $\displaystyle D$ 的所有不变子空间为
\begin\{aligned\} \left\\{0\right\\}, L(1), L(1,x), L(1,x,x^2), \cdots, L(1,x,\cdots,x^\{n-1\}), V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1413、 (5)、 设 $\displaystyle U$ 和 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{K\}$ 上的有限维线性空间, 如果 $\displaystyle \mathbb\{K\}$ 上的线性空间 $\displaystyle T$ 和双线性映射 $\displaystyle \sigma: U\times V\to T$ 满足: 对 $\displaystyle \mathbb\{K\}$ 上的任一线性空间 $\displaystyle W$ 以及任意双线性映射 $\displaystyle \theta: U\times V\to W$ 都存在唯一的线性映射 $\displaystyle \varphi: T\to W$, 使得 $\displaystyle \theta=\varphi\circ\sigma$, 则称 $\displaystyle T$ 是 $\displaystyle U$ 和 $\displaystyle V$ 的张量积. 证明: 张量积在同构意义下是唯一的. (华东师范大学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) / (5-1)、 先证明 $\displaystyle \sigma: U\times V\to T$ 是满射. 用反证法. 设 $\displaystyle \mathrm\{im\}\sigma$ 的一组基为 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$, 将其看扩充为 $\displaystyle T$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ ($r < n$). 则
\begin\{aligned\} \varphi: T\to T, \varphi(\varepsilon\_i)=\left\\{\begin\{array\}\{llllllllllll\}\varepsilon\_i, &1\leq i\leq r,\\\\ 0,&r+1\leq i\leq n\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
满足 $\displaystyle \varphi\neq 1\_T$, 但 $\displaystyle \varphi\circ \sigma=1\_T \circ \sigma=\sigma: T\to T$. 这与题设中的’唯一性‘矛盾. 故有结论. (5-2)、 设 $\displaystyle T,T\_1$ 都是 $\displaystyle U,V$ 的张量积, 则存在
\begin\{aligned\} \sigma: U\times V\to T,\quad \sigma\_1: U\times V\to T\_1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
使得题中的那些结论成立. 特别地,
\begin\{aligned\} &\exists\ \varphi: T\to T\_1,\mathrm\{ s.t.\} \sigma\_1=\varphi\circ \sigma,\\\\ &\exists\ \varphi\_1: T\_1\to T,\mathrm\{ s.t.\} \sigma=\varphi\_1\circ \sigma\_1\\\\ \Rightarrow&\sigma=\varphi\_1\circ \varphi\circ \sigma, \sigma\_1=\varphi\circ \varphi\_1\circ \sigma\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \sigma: U\times V\to T, \sigma\_1: U\times V\to T\_1$ 都是满射知
\begin\{aligned\} &t=\varphi\_1\circ \varphi(t),\quad \forall\ t\in T;\qquad(I) &t\_1=\varphi\circ \varphi\_1(t\_1),\quad \forall\ t\_1\in T\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle (I)$ 蕴含
\begin\{aligned\} \varphi(t)=\varphi(t')\Rightarrow t=\varphi\_1\circ \varphi(t) =\varphi\_1\circ \varphi(t')=t', \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle \varphi$ 是单射. $\displaystyle (II)$ 蕴含 $\displaystyle \varphi$ 是满射. 故 $\displaystyle \varphi: T: \to T\_1$ 是可逆的, 是同构映射. 这就证明了张量积在同构意义下是唯一的.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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---
1414、 5、 $\displaystyle V$ 是有限维实线性空间, $\displaystyle \mathscr\{A\}$ 是 $\displaystyle V$ 上的线性变换, $\displaystyle f(x)$ 是 $\displaystyle \mathscr\{A\}$ 的特征多项式, 且存在复数 $\displaystyle a+b\mathrm\{ i\}\ (b\neq 0)$ 使得 $\displaystyle f(a+b\mathrm\{ i\})=0$. 证明: 存在 $\displaystyle V$ 的二维子空间 $\displaystyle W$, 使得 $\displaystyle \mathscr\{A\} W\subset W$. (华南理工大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 取定 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 设
\begin\{aligned\} \mathscr\{A\}(\varepsilon\_1,\cdots,\varepsilon\_n)=(\varepsilon\_1,\cdots,\varepsilon\_n)A, A\in\mathbb\{R\}^\{n\times n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle A$ 的特征多项式为 $\displaystyle f(x)\in\mathbb\{R\}[x]$. 由题设, $\displaystyle a+b\mathrm\{ i\}$ 是 $\displaystyle A$ 的特征值, 设 $\displaystyle 0\neq \xi+\mathrm\{ i\} \eta\ (\xi,\eta\in\mathbb\{R\}^n)$ 为对应的特征向量, 则
\begin\{aligned\} A(\xi+\mathrm\{ i\} \eta)=(a+b\mathrm\{ i\})(\xi+\mathrm\{ i\} \eta) \Rightarrow A\xi=a\xi-b\eta, A\eta=b\xi+a\eta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 往证 $\displaystyle \xi,\eta$ 线性无关. 设 $\displaystyle k\xi+l\eta=0, k,l\in\mathbb\{R\}$, 则
\begin\{aligned\} 0=&kA\xi+lA\eta=k(a\xi-b\eta)+l(b\xi+a\eta)\\\\ =&a(k\xi+l\eta)+b(-k\eta+l\xi) =b(-k\eta+l\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle b\neq 0$ 知 $\displaystyle -k\eta+l\xi=0$. 于是
\begin\{aligned\} 0=&(k\xi+l\eta)^\mathrm\{T\} (k\xi+l\eta)=k^2\xi^\mathrm\{T\}\xi+2kl\xi^\mathrm\{T\} \eta+l^2\eta^\mathrm\{T\}\eta,\\\\ 0=&(-k\eta+l\xi)^\mathrm\{T\} (-k\eta+l\xi)=l^2\xi^\mathrm\{T\} \xi-2kl\xi^\mathrm\{T\} \eta+k^2\eta^\mathrm\{T\} \eta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相加即得
\begin\{aligned\} (k^2+l^2)(\xi^\mathrm\{T\} \xi+\eta^\mathrm\{T\} \eta)=0\stackrel\{\xi+\mathrm\{ i\}\eta\neq 0\}\{\Rightarrow\}k^2+l^2=0\Rightarrow k=l=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 设
\begin\{aligned\} \alpha=(\varepsilon\_1,\cdots,\varepsilon\_n)\xi, \beta=(\varepsilon\_1,\cdots,\varepsilon\_n)\eta\in V, W=L(\alpha,\beta), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \dim W=2, \mathscr\{A\} W\subset W$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1415、 6、 设 $\displaystyle f(x)$ 是数域 $\displaystyle \mathbb\{P\}$ 上的二次多项式, $\displaystyle x\_1,x\_2\in\mathbb\{P\}$ 是它的两个不同根. 再设 $\displaystyle \mathscr\{A\}$ 是数域 $\displaystyle \mathbb\{P\}$ 上的线性空间 $\displaystyle V$ 上的非数乘线性变换, 满足 $\displaystyle f(\mathscr\{A\})=\mathscr\{O\}$. (1)、 证明: $\displaystyle x\_1,x\_2$ 是 $\displaystyle \mathscr\{A\}$ 的特征值; (2)、 证明: $\displaystyle V=V\_\{x\_1\}\oplus V\_\{x\_2\}$, 其中 $\displaystyle V\_\{x\_i\}$ 是 $\displaystyle \mathscr\{A\}$ 的属于特征值 $\displaystyle x\_i$ 的特征子空间. (华南理工大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由题设, $\displaystyle f(x)=(x-x\_1)(x-x\_2)$, 且
\begin\{aligned\} (\mathscr\{A\}-x\_1\mathscr\{E\})(\mathscr\{A\}-x\_2\mathscr\{E\})=\mathscr\{O\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
($\mathscr\{E\}, \mathscr\{O\}$ 分别是单位变换和零变换). 往用反证法证明 $\displaystyle \mathscr\{A\}-x\_1\mathscr\{E\}$ 不是可逆的, 若不然, 由上式知 $\displaystyle \mathscr\{A\}-x\_2\mathscr\{E\}=\mathscr\{O\}$, 与题设矛盾. 故有结论. 从而
\begin\{aligned\} \exists\ 0\neq\alpha\in V,\mathrm\{ s.t.\} (\mathscr\{A\}-x\_1\mathscr\{E\})\alpha=0\Rightarrow \mathscr\{A\}\alpha=x\_1\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 $\displaystyle x\_1$ 是 $\displaystyle \mathscr\{A\}$ 的特征值. 同理, $\displaystyle x\_2$ 也是 $\displaystyle \mathscr\{A\}$ 的特征值. (2)、 对 $\displaystyle \alpha\in V$, 若 $\displaystyle \alpha=\alpha\_1+\alpha\_2, \alpha\_i\in V\_\{x\_i\}$, 则 $\displaystyle \mathscr\{A\}\alpha=x\_1\alpha\_1+x\_2\alpha\_2$. 从而求得 $\displaystyle \alpha\_1=\frac\{\mathscr\{A\}\alpha-x\_2\alpha\}\{x\_1-x\_2\}, \alpha\_2=\frac\{-\mathscr\{A\}\alpha+x\_1\alpha\}\{x\_1-x\_2\}$. 这就证明了 $\displaystyle V=V\_\{x\_1\}+V\_\{x\_2\}$. 又由
\begin\{aligned\} \alpha\in V\_\{x\_1\}\cap V\_\{x\_2\}\Rightarrow x\_1\alpha=\mathscr\{A\}\alpha=x\_2\alpha\stackrel\{x\_1\neq 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\}
知 $\displaystyle V=V\_\{x\_1\}\oplus V\_\{x\_2\}$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1416、 (4)、 向量空间 $\displaystyle \mathbb\{F\}^3$ 中定义线性变换
\begin\{aligned\} \sigma(x\_1,x\_2,x\_3)=(x\_3-x\_2, 2x\_2, 2x\_1-x\_2), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则线性变换关于基 $\displaystyle \alpha\_1=(1,0,1), \alpha\_2=(0,1,0), \alpha\_3=(0,0,1)$ 的矩阵是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华南师范大学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\} \sigma \alpha\_1=(1,0,2), \sigma \alpha\_2=(-1,2,-1), \sigma \alpha\_3=(1,0,0) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \sigma(\alpha\_1,\alpha\_2,\alpha\_3)=(e\_1,e\_2,e\_3)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&2&0\\\\ 2&-1&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} (\alpha\_1,\alpha\_2,\alpha\_3)=(e\_1,e\_2,e\_3)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&0\\\\ 1&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\}
知
\begin\{aligned\} \sigma(\alpha\_1,\alpha\_2,\alpha\_3)=(\alpha\_1,\alpha\_2,\alpha\_3)T^\{-1\}A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故线性变换 $\displaystyle \sigma$ 关于基 $\displaystyle \alpha\_1=(1,0,1), \alpha\_2=(0,1,0), \alpha\_3=(0,0,1)$ 的矩阵是 $\displaystyle T^\{-1\}A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&1\\\\ 0&2&0\\\\ 1&0&-1\end\{array\}\right)$.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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---
1417、 (5)、 向量空间 $\displaystyle \mathbb\{F\}[x]\_3$ 中多项式 $\displaystyle f(x)=x^3-3x^2+2x+1$ 关于基
\begin\{aligned\} \left\\{x-1,1-x^2,x^2+2x-2, x^3\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的坐标是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华南师范大学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\} \left(x-1,1-x^2,x^2+2x-2, x^3\right)=(1,x,x^2,x^3)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&1&-2&0\\\\ 1&0&2&0\\\\ 0&-1&1&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\}
知
\begin\{aligned\} f(x)=&(1,x,x^2,x^3)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\-3\\\\1\end\{array\}\right)\\\\ =&\left(x-1,1-x^2,x^2+2x-2, x^3\right)T^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\-3\\\\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 $\displaystyle f$ 在题中基下的坐标为 $\displaystyle T^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\-3\\\\1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\3\\\\0\\\\1\end\{array\}\right)$.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)/
---
1418、 6、 设欧氏空间 $\displaystyle \mathbb\{R\}^4$ 上的一个线性变换 $\displaystyle \sigma$ 关于基 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ 的矩阵为
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1&-1\\\\ 1&0&-1&1\\\\ 1&-1&0&1\\\\ -1&1&1&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求正交矩阵 $\displaystyle P$ 和对角矩阵 $\displaystyle \varLambda$, 使得 $\displaystyle P^\mathrm\{T\} AP=\varLambda$; (2)、 判断线性变换 $\displaystyle \sigma$ 是否可逆? (3)、 证明: 向量组
\begin\{aligned\} \eta\_1=\alpha\_1+\alpha\_2, \eta\_2=\alpha\_2, \eta\_3=\alpha\_3+\alpha\_4, \eta\_4=\alpha\_4 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
仍是 $\displaystyle \mathbb\{R\}^4$ 的一个基, 并求 $\displaystyle \sigma$ 关于这组基的矩阵; (4)、 向量 $\displaystyle \alpha$ 关于基 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ 的坐标为 $\displaystyle (1,2,1,2)$, 求 $\displaystyle \alpha$ 关于基 $\displaystyle \eta\_1,\eta\_2,\eta\_3,\eta\_4$ 的坐标. (华南师范大学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)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,1,-3$. 由
\begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1&1\\\\ 0&0&0&0\\\\ 0&0&0&0\\\\ 0&0&0&0\end\{array\}\right), -3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&0&-1\\\\ 0&1&0&1\\\\ 0&0&1&1\\\\ 0&0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,-3$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\1\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\0\\\\1 \end\{array\}\right), \xi\_4=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\-1\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle \xi\_1,\xi\_2,\xi\_3,\xi\_4$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3,\eta\_4$. 令
\begin\{aligned\} P=(\eta\_1,\eta\_2,\eta\_3,\eta\_4)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&-\frac\{1\}\{2\sqrt\{3\}\}&\frac\{1\}\{2\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{2\sqrt\{3\}\}&-\frac\{1\}\{2\}\\\\ 0&\frac\{2\}\{\sqrt\{6\}\}&\frac\{1\}\{2\sqrt\{3\}\}&-\frac\{1\}\{2\}\\\\ 0&0&\frac\{\sqrt\{3\}\}\{2\}&\frac\{1\}\{2\} \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle P$ 正交, 且
\begin\{aligned\} P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}\left(1,1,1,-3\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle A$ 可逆知 $\displaystyle \sigma$ 可逆. (3)、 由
\begin\{aligned\} (\eta\_1,\cdots,\eta\_4)=(\alpha\_1,\cdots,\alpha\_4)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ 1&1&&\\\\ &&1&\\\\ &&1&1\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle |T|=1\neq 0$ 知 $\displaystyle \eta\_1,\cdots,\eta\_4$ 是 $\displaystyle \mathbb\{R\}^4$ 的一组基, 且
\begin\{aligned\} &\sigma(\eta\_1,\cdots,\eta\_4)=\sigma(\alpha\_1,\cdots,\alpha\_4)T\\\\ =&(\alpha\_1,\cdots,\alpha\_4)AT =(\eta\_1,\cdots,\eta\_4)T^\{-1\}AT. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sigma$ 在基 $\displaystyle \eta\_1,\cdots,\eta\_4$ 下的矩阵 $\displaystyle T^\{-1\}AT=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&-1\\\\ 0&-1&0&2\\\\ 0&-1&1&1\\\\ 0&2&0&-1\end\{array\}\right)$. (4)、
\begin\{aligned\} \alpha=(\alpha\_1,\cdots,\alpha\_4)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\1\\\\2\end\{array\}\right) =(\eta\_1,\cdots,\eta\_4)T^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\1\\\\2\end\{array\}\right)=(\eta\_1,\cdots,\eta\_4)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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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---
1419、 7、 在 $\displaystyle \mathbb\{R\}[x]$ 中, $\displaystyle V$ 是由多项式 $\displaystyle 1,x,x^2,\cdots,x^n$ 所生成的子空间, 即
\begin\{aligned\} V=\left\\{a\_0+a\_1x+a\_2x^2+\cdots+a\_nx^n; a\_i\in\mathbb\{R\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle V$ 上线性变换 $\displaystyle \sigma$ 定义为
\begin\{aligned\} \sigma\left\[f(x)\right\]=xf'(x)-f(x), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle f'(x)$ 表示 $\displaystyle f(x)$ 的导数. (1)、 求 $\displaystyle \ker\sigma$ 和 $\displaystyle \mathrm\{im\}\sigma$; (2)、 证明: $\displaystyle V=\ker \sigma\oplus \mathrm\{im\} \sigma$. (华南师范大学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)、
\begin\{aligned\} \sigma(kf(x)+lg(x))&=x[kf(x)+lg(x)]'-[kf(x)+lg(x)]\\\\ &=k[xf'(x)-f(x)] +l[xg'(x)-g(x)]\\\\ &=k\sigma f(x)+l \sigma g(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由
\begin\{aligned\} f\in \ker\sigma&\Leftrightarrow xf'(x)=f(x)\\\\ &\Leftrightarrow f\equiv 0\mbox\{或\}\frac\{\mathrm\{ d\} f\}\{f\}=\frac\{1\}\{x\}\\\\ &\Leftrightarrow f(x)\equiv cx \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \ker\sigma=L(x). \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\} \sigma x^k&=x\cdot kx^\{k-1\}-x^k\\\\ &=(k-1)x^k\left(0\leq k\leq n-1\right)\\\\ &=\left\\{\begin\{array\}\{llllllllllll\} (k-1)x^k,&k\neq 1\\\\ 0,&k=1 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mathrm\{im\}\sigma&=L(\sigma (1),\sigma (x),\cdots,\sigma (x^n))\\\\ &=L(-1,0,x^2,\cdots,nx^n)\\\\ &=L(1,x^2,\cdots,x^n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、
\begin\{aligned\} V&=L(1,x,x^2,\cdots,x^n)\\\\ &=L(x)\oplus L(1,x^2,\cdots,x^n)\\\\ &=\ker\sigma\oplus \mathrm\{im\}\sigma. \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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---
1420、 8、 设 $\displaystyle \sigma,\tau$ 是数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 维向量空间 $\displaystyle V$ 的两个线性变换, 且满足 $\displaystyle \sigma\tau=\tau\sigma$. 证明: (1)、 $\displaystyle \ker\sigma+\ker\tau$ 是 $\displaystyle V$ 的子空间; (2)、 $\displaystyle \ker\sigma+\ker\tau$ 在 $\displaystyle \sigma\tau$ 之下不变. (华南师范大学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)、 $\displaystyle \ker\sigma, \ker\tau$ 是 $\displaystyle V$ 的子空间, 而 $\displaystyle \ker \sigma+\ker \tau$ 作为它们的直和, 是 $\displaystyle V$ 的子空间. (2)、 对 $\displaystyle \forall\ \alpha\in \ker \sigma$, $\displaystyle \sigma\tau(\alpha)=\tau\sigma(\alpha)=\tau(0)=0\in \ker \sigma$. 对 $\displaystyle \forall\ \alpha\in \ker \tau$, $\displaystyle \sigma\tau(\alpha)=\sigma(0)=0\in \ker \tau$. 故
\begin\{aligned\} &\alpha\in \ker \sigma+\ker\tau\Rightarrow \alpha=\beta+\gamma, \beta\in \ker \sigma, \gamma\in \ker \tau\\\\ \Rightarrow& \sigma\tau(\alpha)=0+0=0\in \ker \sigma+\ker \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \ker\sigma+\ker\tau$ 在 $\displaystyle \sigma\tau$ 之下不变.跟锦数学微信公众号. [在线资料](
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---
1421、 5、 (20 分) 已知 $\displaystyle \sigma$ 为线性空间 $\displaystyle V$ 的一个线性变换, 回答如下问题: (1)、 $\displaystyle \ker\sigma=\ker\sigma^2$ 的充要条件是 $\displaystyle \ker\sigma\cap \mathrm\{im\}\sigma=\left\\{0\right\\}$; (2)、 $\displaystyle \mathrm\{im\}\sigma=\mathrm\{im\}\sigma^2$ 的充要条件是 $\displaystyle V=\ker \sigma+\mathrm\{im\}\sigma$. (华中科技大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 (1-1)、 $\displaystyle \Rightarrow$:
\begin\{aligned\} &\alpha\in \ker\sigma\cap \mathrm\{im\}\sigma\Rightarrow \sigma\alpha=0; \exists\ \beta\in V,\mathrm\{ s.t.\} \alpha=\sigma\beta\\\\ \Rightarrow&0=\sigma\alpha=\sigma^2\beta\Rightarrow \beta\in\ker\sigma^2=\ker\sigma \Rightarrow \alpha=\sigma\beta=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)、 $\displaystyle \Leftarrow$: 显然 $\displaystyle \ker\sigma\subset \ker\sigma^2$. 反之, 由
\begin\{aligned\} \alpha\in \ker\sigma^2\Rightarrow& \sigma^2\alpha=0\Rightarrow \sigma\alpha\in\mathrm\{im\} \sigma\cap \ker\sigma=\left\\{0\right\\}\\\\ \Rightarrow& \sigma\alpha=0\Rightarrow \alpha\in\ker\sigma \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \ker\sigma^2\subset \ker\sigma$. 故 $\displaystyle \ker\sigma=\ker\sigma^2$. (2)、 写出
\begin\{aligned\} \dim(\ker\sigma+\mathrm\{im\}\sigma)=&\dim \ker\sigma+\dim \mathrm\{im\}\sigma-\dim\left(\ker\sigma\cap \mathrm\{im\}\sigma\right)\\\\ =&n-\dim\left(\ker\sigma\cap \mathrm\{im\}\sigma\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\}
我们有
\begin\{aligned\} &\mathrm\{im\}\sigma=\mathrm\{im\}\sigma^2\stackrel\{\mathrm\{im\} \sigma^2\subset \mathrm\{im\}\sigma\}\{\Leftrightarrow\} \dim\mathrm\{im\}\sigma=\dim\mathrm\{im\}\sigma^2\\\\ \stackrel\{\mbox\{维数公式\}\}\{\Leftrightarrow\}&\dim\ker \sigma=\dim\ker\sigma^2 \stackrel\{\ker\sigma\subset \ker\sigma^2\}\{\Leftrightarrow\}\ker \sigma=\ker\sigma^2\\\\ \stackrel\{\mbox\{第 1 步\}\}\{\Leftrightarrow\}&\ker\sigma\cap \mathrm\{im\}\sigma=\left\\{0\right\\} \stackrel\{\mbox\{(I)\}\}\{\Leftrightarrow\}\dim(\ker\sigma+\mathrm\{im\}\sigma)=n\\\\ \Leftrightarrow& \ker\sigma+\mathrm\{im\}\sigma=V. \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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---
1422、 6、 (20 分) 设 $\displaystyle \sigma$ 为 $\displaystyle \mathbb\{R\}^3$ 上的一个线性变换, 且 $\displaystyle \sigma^2=\mathscr\{O\}$. 证明: 存在一个线性映射 $\displaystyle f: \mathbb\{R\}^3\to \mathbb\{R\}$, 存在 $\displaystyle y\in\mathbb\{R\}^3$, 使得
\begin\{aligned\} \sigma(x)=f(x)y,\quad \forall\ x\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\}
(华中科技大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 若 $\displaystyle \sigma=\mathscr\{O\}$, 则取 $\displaystyle y=0$, 任取线性映射 $\displaystyle f$ 即可. 若 $\displaystyle \sigma\neq \mathscr\{O\}$, 则 $\displaystyle \dim \mathrm\{im\}\sigma\geq 1$. 又由
\begin\{aligned\} &\sigma^2=\mathscr\{O\}\Rightarrow \mathrm\{im\}\sigma\subset \ker\sigma\\\\ \Rightarrow&3=\dim\mathrm\{im\}\sigma+\dim\ker\sigma \geq 2\dim \mathrm\{im\} \sigma\Rightarrow \dim\mathrm\{im\}\sigma\leq\frac\{3\}\{2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \dim \mathrm\{im\}\sigma=1$. 设 $\displaystyle 0\neq y$ 是 $\displaystyle \mathrm\{im\}\sigma$ 的一组基, 则对 $\displaystyle \forall\ x\in\mathbb\{R\}^3$,
\begin\{aligned\} &\sigma x\in \mathrm\{im\} \sigma\Rightarrow \exists\ f(x)\in\mathbb\{R\},\mathrm\{ s.t.\} \sigma x=f(x)y. \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\} \sigma(kx+k'x')=\left\\{\begin\{array\}\{llllllllllll\}f(kx+k'x')y\\\\ k\sigma x+k'\sigma x'=\left\[kf(x)+k'f(x')\right\]y\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 $\displaystyle f(kx+k'x')=kf(x)+k'f(x')$. 故 $\displaystyle f$ 是线性函数, 且 $\displaystyle \sigma x=f(x)y$.跟锦数学微信公众号. [在线资料](
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---
1423、 8、 (20 分) 若 $\displaystyle n$ 维线性空间 $\displaystyle V$ 中共存在一组至少有一个不为零变换的线性变换 $\displaystyle \sigma\_\{ij\}, 1\leq i,j\leq n$ 满足以下条件:
\begin\{aligned\} \sigma\_\{iji\}\sigma\_\{hk\}=\left\\{\begin\{array\}\{llllllllllll\}\sigma\_\{ik\},&j=h,\\\\ 0,&\mbox\{其它\}.\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\}
证明: 在 $\displaystyle V$ 中存在一组基 $\displaystyle v\_1,\cdots,v\_n$, 使得
\begin\{aligned\} \sigma\_\{ij\}(v\_k)=\left\\{\begin\{array\}\{llllllllllll\}v\_i,&j=k,\\\\ 0,&\mbox\{其它\}.\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\}
(华中科技大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由
\begin\{aligned\} \sigma\_\{ik\}\sigma\_\{kh\}\sigma\_\{hj\}=\sigma\_\{ih\}\sigma\_\{hj\}=\sigma\_\{ij\}\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 [从矩阵的观点看: $\displaystyle A=BCD\Rightarrow \mathrm\{rank\} A\leq \mathrm\{rank\} C$] $\displaystyle \mathrm\{rank\} \sigma\_\{ij\}\leq \mathrm\{rank\} \sigma\_\{kh\}$. 同理,
\begin\{aligned\} \sigma\_\{kh\}=\sigma\_\{ki\}\sigma\_\{ij\}\sigma\_\{jh\}\Rightarrow \mathrm\{rank\}\sigma\_\{kh\}\leq \mathrm\{rank\} \sigma\_\{ij\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\} \sigma\_\{kh\}=\mathrm\{rank\}\sigma\_\{ij\}, \forall\ i,j,k,l$. (2)、 由 $\displaystyle (I)$ 知若某个 $\displaystyle \sigma\_\{kh\}=\mathscr\{O\}$, 则 $\displaystyle \forall\ i,j, \sigma\_\{ij\}=\mathscr\{O\}$. 这与题设矛盾. 故
\begin\{aligned\} \forall\ i,j, \sigma\_\{ij\}\neq\mathscr\{O\}\Rightarrow \mathrm\{rank\} \sigma\_\{ij\}\geq 1.\qquad(II) \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\} &\sigma\_\{ii\}=\sigma\_\{ii\}\sigma\_\{ii\}=\sigma\_\{ii\}^2,\qquad(III)\\\\ i\neq j\Rightarrow&\sigma\_\{ii\}\sigma\_\{jj\}=\mathscr\{O\}\qquad(IV) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知若
\begin\{aligned\} 0=\sigma\_\{11\}\alpha\_1+\cdots+\sigma\_\{nn\}\alpha\_n, \alpha\_i\in V\Rightarrow \sigma\_\{ii\}\alpha\_i\in \mathrm\{im\} \sigma\_\{ii\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则用 $\displaystyle \sigma\_\{ii\}$ 作用后得
\begin\{aligned\} 0\stackrel\{(III)\}\{=\}\sigma\_\{ii\}^2\alpha\_i\stackrel\{(IV)\}\{=\}\sigma\_\{ii\}\alpha\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \mathrm\{im\} \sigma\_\{11\}+\cdots+\mathrm\{im\}\sigma\_\{nn\}$ 中零向量分解的唯一性, $\displaystyle \mathrm\{im\} \sigma\_\{11\}\oplus \cdots\oplus \mathrm\{im\} \sigma\_\{nn\}$ 是直和,
\begin\{aligned\} n\geq&\dim \left(\mathrm\{im\} \sigma\_\{11\}\oplus \cdots\oplus\mathrm\{im\}\sigma\_\{nn\}\right) =\sum\_\{i=1\}^n \dim \mathrm\{im\} \sigma\_\{ii\} \overset\{\tiny\mbox\{第1步\}\}\{=\} n\mathrm\{rank\} \sigma\_\{11\}\stackrel\{(II)\}\{\geq\}n. \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\} \mathrm\{rank\} \sigma\_\{11\}=1\overset\{\tiny\mbox\{第1步\}\}\{\Longrightarrow\} \mathrm\{rank\} \sigma\_\{ij\}=1, \forall\ i,j, \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\} \mathrm\{im\} \sigma\_\{11\}\oplus \cdots\oplus \mathrm\{im\} \sigma\_\{nn\}=V.\qquad(IV) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 设 $\displaystyle 0\neq u\_1$ 是 $\displaystyle \mathrm\{im\} \sigma\_\{11\}$ 的一组基, 则
\begin\{aligned\} u\_1\in \mathrm\{im\} \sigma\_\{11\}\Rightarrow& \exists\ \alpha\in V,\mathrm\{ s.t.\} u\_1=\sigma\_\{11\}\alpha\\\\ \Rightarrow&\sigma\_\{11\}u\_1=\sigma\_\{11\}^2\alpha\stackrel\{(III)\}\{=\}\sigma\_\{11\}\alpha=u\_1. \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\_i=\sigma\_\{i1\}u\_1=\sigma\_\{ii\}\sigma\_\{i1\}u\_1\in \mathrm\{im\} \sigma\_\{ii\}, 2\leq i\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
往用反证法证明 $\displaystyle u\_i\neq 0$, 而由 $\displaystyle \mathrm\{rank\} \sigma\_\{ii\}=1$ 知 $\displaystyle u\_i$ 是 $\displaystyle \mathrm\{im\} \sigma\_\{ii\}$ 的一组基. 若不然,
\begin\{aligned\} &\left\\{\begin\{array\}\{llllllllllll\} \sigma\_\{i1\}u\_1=0\Rightarrow \sigma\_\{i1\}|\_\{\mathrm\{im\}\sigma\_\{11\}\}=\mathscr\{O\}\\\\ 2\leq i\leq n\Rightarrow 1\neq i\Rightarrow \sigma\_\{i1\}|\_\{\mathrm\{im\} \sigma\_\{ii\}\}=\mathscr\{O\}\end\{array\}\right. \stackrel\{(IV)\}\{\Rightarrow\}&\sigma\_\{i1\}=\mathscr\{O\},\mbox\{与 $\displaystyle (II)$ 矛盾\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由 $\displaystyle (IV)$ 知 $\displaystyle u\_1,\cdots,u\_n$ 是 $\displaystyle V$ 的一组基, 且
\begin\{aligned\} \sigma\_\{ij\} v\_k=\sigma\_\{ij\}\sigma\_\{k1\}v\_1=\left\\{\begin\{array\}\{llllllllllll\}\sigma\_\{i1\}v\_1=v\_i,&j=k,\\\\ \mathscr\{O\} v\_1=0,&j\neq k.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1424、 (5)、 设 $\displaystyle M\_n(\mathbb\{R\})$ 是所有 $\displaystyle n$ 阶实方阵构成的实向量空间, 那么 它的由所有迹等于 $\displaystyle 0$ 的方阵构成的子空间的维数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华中师范大学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 $\displaystyle a\_\{11\}+\cdots+a\_\{nn\}=0$ 的系数矩阵的秩为 $\displaystyle 1$ 知应填 $\displaystyle n^2-1$.跟锦数学微信公众号. [在线资料](
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---
1425、 (4)、 证明: 向量组 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 线性无关当且仅当存在 $\displaystyle \alpha\in L(\alpha\_1,\cdots,\alpha\_k)$ 是 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 的唯一线性组合. (华中师范大学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) / (4-1)、 $\displaystyle \Rightarrow$: 取 $\displaystyle \alpha\_1\in L(\alpha\_1,\cdots,\alpha\_k)$, 则由 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 线性无关知 $\displaystyle \alpha\_1=1\cdot \alpha\_1+0\cdot \alpha\_2+\cdots+0\cdot \alpha\_k$ 是 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 的唯一线性组合. (4-2)、 $\displaystyle \Leftarrow$: 设存在 $\displaystyle \alpha\in L(\alpha\_1,\cdots,\alpha\_k)$ 是 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 的唯一线性组合, 则 $\displaystyle \alpha=\sum\_\{i=1\}^k x\_i\alpha\_i$, 且表示法唯一. 往用反证法证明 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 线性无关. 设 $\displaystyle \sum\_\{i=1\}^k y\_i\alpha\_i=0$, 则 $\displaystyle \alpha=\sum\_\{i=1\}^k x\_i\alpha\_i+0=\sum\_\{i=1\}^k x\_i\alpha\_i+\sum\_\{i=1\}^k y\_i\alpha\_i =\sum\_\{i=1\}^k (x\_i+y\_i)\alpha\_i$. 由表示法唯一知 $\displaystyle x\_i+y\_i=x\_i\Rightarrow y\_i=0$. 这就证明了 $\displaystyle \alpha\_1,\cdots,\alpha\_k$ 线性无关.跟锦数学微信公众号. [在线资料](
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---
1426、 (6)、 设 $\displaystyle \mathbb\{C\}$ 为复数域, $\displaystyle \mathbb\{C\}^n$ 是所有 $\displaystyle n$ 维列向量构成的复向量空间, $\displaystyle M\_n(\mathbb\{C\})$ 是所有 $\displaystyle n$ 阶复方阵构成的集合. (6-1)、 证明: 对于 $\displaystyle \mathbb\{C\}^n$ 中任意的非零向量 $\displaystyle \alpha$, 由向量集合
\begin\{aligned\} \left\\{A\alpha; A\in M\_n(\mathbb\{C\})\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
生成的子空间等于 $\displaystyle \mathbb\{C\}^n$. (6-2)、 若 $\displaystyle \theta$ 为 $\displaystyle \mathbb\{C\}^n$ 的一个线性变换, 且满足对任意的 $\displaystyle \alpha\in \mathbb\{C\}^n$ 以及任意的 $\displaystyle A\in M\_n(\mathbb\{C\})$, 均有 $\displaystyle \theta(A\alpha)=A\theta(\alpha)$. 证明: 存在一个复数 $\displaystyle a$, 使得 $\displaystyle \theta(\alpha)=a\alpha$ 对任意的 $\displaystyle \alpha\in \mathbb\{C\}^n$ 都成立. [张祖锦注: 第 1 问中可以去掉’生成的子空间‘, 见参考解答.] (华中师范大学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) / (6-1)、 将 $\displaystyle \alpha$ 扩充为 $\displaystyle \mathbb\{C\}^n$ 的一组基 $\displaystyle \alpha,\alpha\_2,\cdots,\alpha\_n$. 对 $\displaystyle \forall\ \beta\in\mathbb\{C\}^n$, 取 $\displaystyle \mathbb\{C\}^n$ 上的线性变换 $\displaystyle \mathscr\{A\}$ 使得
\begin\{aligned\} \mathscr\{A\} \alpha=\beta, \mathscr\{A\}\alpha\_i=0, 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\}
设 $\displaystyle \mathscr\{A\}$ 在 $\displaystyle \mathbb\{C\}^n$ 的标准基 $\displaystyle e\_1,\cdots,e\_n$ 下的矩阵为 $\displaystyle A$, 则
\begin\{aligned\} \beta=\mathscr\{A\}\alpha=\mathscr\{A\}(e\_1,\cdots,e\_n)\alpha =(e\_1,\cdots,e\_n)A\alpha=A\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \left\\{A\alpha; A\in M\_n(\mathbb\{C\})\right\\}=\mathbb\{C\}^n$. (6-2)、 为证第 ii 问, 我们先给出两个结论. 如果 $\displaystyle A$ 与所有对角矩阵可交换, 则 $\displaystyle A$ 也是对角矩阵. 如果 $\displaystyle A$ 与所有矩阵可交换, 则 $\displaystyle A$ 是数量矩阵. 事实上,
\begin\{aligned\} &\mbox\{ $\displaystyle A$ 与所有对角矩阵可交换\}\\\\ \Rightarrow& \forall\ D=\mathrm\{diag\}(d\_1,\cdots,d\_n), AD=DA\\\\ \Rightarrow&\forall\ D, a\_\{ij\}d\_j=d\_ia\_\{ij\}\\\\ \Rightarrow&\mbox\{取 $\displaystyle d\_i$ 互不相同, 则 $\displaystyle a\_\{ij\}(d\_j-d\_i)=0$ 蕴含 $\displaystyle a\_\{ij\}=0, \forall\ i\neq j$\}\\\\ \Rightarrow&\mbox\{ $\displaystyle A$ 是对角矩阵\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle A$ 与所有矩阵可交换, 则 $\displaystyle A$ 与所有对角矩阵可交换, 而由已征知 $\displaystyle A$ 是对角矩阵 $\displaystyle \mathrm\{diag\}(d\_1,\cdots,d\_n)$. 进一步,
\begin\{aligned\} &\mbox\{ $\displaystyle A$ 与所有矩阵可交换\}\\\\ \Rightarrow& d\_ib\_\{ij\}=b\_\{ij\}d\_j, \forall\ B=(b\_\{ij\})\in \mathbb\{F\}^\{n\times n\}\\\\ \Rightarrow&\mbox\{取 $\displaystyle B$ 使得非对角元非零, 则知 $\displaystyle d\_i=d\_j$\}\\\\ \Rightarrow&\mbox\{ $\displaystyle A$ 是数量矩阵 $\displaystyle dI\_n, d=d\_1=\cdots=d\_n$\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(6-3)、 回到第 ii 问的证明. 设 $\displaystyle \theta$ 在 $\displaystyle \mathbb\{C\}^n$ 的标准基 $\displaystyle e\_1,\cdots,e\_n$ 下的矩阵为 $\displaystyle B$, 则对 $\displaystyle \forall\ A\in M\_n(\mathbb\{C\})$,
\begin\{aligned\} \theta(Ae\_i)=&\theta(e\_1,\cdots,e\_n)Ae\_i=(e\_1,\cdots,e\_n)BAe\_i,\\\\ A\theta(e\_i)=&A\theta(e\_1,\cdots,e\_n)e\_i=ABe\_i. \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\} BAe\_i=ABe\_i\Rightarrow BA=BA(e\_1,\cdots,e\_n)=AB(e\_1,\cdots,e\_n)=AB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 ii 步的结果知 $\displaystyle B$ 是数量矩阵 $\displaystyle \alpha I\_n, \alpha\in\mathbb\{C\}$, 此即
\begin\{aligned\} \theta(\alpha)=\theta(e\_1,\cdots,e\_n)\alpha =(e\_1,\cdots,e\_n)aI\_n\alpha=a\alpha,\forall\ \alpha\in \mathbb\{C\}^n. \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)/
发表于 2023-3-5 13:22:58
