## 张祖锦2023年数学专业真题分类70天之第63天
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1427、 2、 (20 分) 设 $\displaystyle V$ 是数域 $\displaystyle \varOmega$ 上的 $\displaystyle n$ 维向量空间, $\displaystyle n\geq 2$, $\displaystyle \sigma$ 是 $\displaystyle V$ 上的线性变换. 若 $\displaystyle 3\sigma^3+2\sigma^2+\sigma=\mathscr\{O\}$, 证明:
\begin\{aligned\} V=\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\}
(吉林大学2023年高等代数与解析几何考研试题) [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\} &\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 \alpha=\sigma(\beta)=-(3\sigma^3+2\sigma^2)(\beta)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \ker \sigma\cap \mathrm\{im\}\sigma=\left\\{0\right\\}$,
\begin\{aligned\} \dim\left(\ker\sigma+\mathrm\{im\} \sigma\right)=&\dim \ker \sigma+\dim \mathrm\{im\}\sigma -\dim\left(\ker \sigma\cap \mathrm\{im\}\sigma\right)\\\\ =&n-0=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \ker \sigma+\mathrm\{im\}\sigma=V$. 联合 $\displaystyle \ker \sigma\cap \mathrm\{im\}\sigma=\left\\{0\right\\}$ 知 $\displaystyle V=\ker \sigma\oplus \mathrm\{im\} \sigma$.跟锦数学微信公众号. [在线资料](
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1428、 4、 已知
\begin\{aligned\} \alpha\_1=(a,2,10)^\mathrm\{T\}, \alpha\_2=(-2,1,5)^\mathrm\{T\}, \alpha\_3=(-1,1,4)^\mathrm\{T\}, \beta=(1,b,c)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle a,b,c$ 满足什么条件时, 下面结论分别成立? (1)、 $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出, 且表示唯一; (2)、 $\displaystyle \beta$ 不可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出; (3)、 $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出, 且表示不唯一, 并写出表示形式. (暨南大学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) /
\begin\{aligned\} &(A,\beta)=(\alpha\_1,\alpha\_2,\alpha\_3,\beta)=\left(\begin\{array\}\{cccccccccccccccccccc\}a&-2&-1&1\\\\ 2&1&1&b\\\\ 10&5&4&c\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}a+4&0&1&1+2b\\\\ 2&1&1&b\\\\ 0&0&-1&c-5b\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}4+a&0&0&1-3b+c\\\\ 2&1&0&c-4b\\\\ 0&0&1&5b-c\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)、 若 $\displaystyle a\neq -4$, 则 $\displaystyle |A|\neq 0$, $\displaystyle Ax=\beta$ 有唯一解, $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出, 且表示唯一. (2)、 若 $\displaystyle a=-4, 3b-c\neq 1$, 则 $\displaystyle Ax=\beta$ 无解, $\displaystyle \beta$ 不可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出. (3)、 若 $\displaystyle a=-4, 3b-c=1$, 则
\begin\{aligned\} (A,\beta)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&\frac\{1\}\{2\}&0&-\frac\{1+b\}\{2\}\\\\ 0&0&1&1+2b\\\\ 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\}
故 $\displaystyle Ax=\beta$ 的通解为 $\displaystyle k\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\2\\\\0\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1+b\}\{2\}\\\\ 0\\\\1+2b\end\{array\}\right), \forall\ k$. 故 $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出, 且表示不唯一, 并
\begin\{aligned\} \beta=\left(-k-\frac\{1+b\}\{2\}\right)\alpha\_1+2k\alpha\_2+(1+2b)\alpha\_3, \forall\ k. \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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1429、 7、 线性变换的最小多项式是否一定整除特征多项式, 请给出证明或举出反例. (暨南大学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) / 是. 设 $\displaystyle \mathscr\{A\}$ 的最小多项式和特征多项式分别为 $\displaystyle m(x), f(x)$, 则由 Hamilton-Cayley 定理, $\displaystyle f(\mathscr\{A\})=\mathscr\{O\}$. 又由带余除法,
\begin\{aligned\} f(x)=q(x)m(x)+r(x), r(x)=0\mbox\{或\} \deg r < \deg m. \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 $\displaystyle r(x)=0\Rightarrow m(x)\mid f(x)$. 若不然, $\displaystyle \deg r < \deg m$, 则由上式知 $\displaystyle r(x)$ 是使得 $\displaystyle r(\mathscr\{A\})=\mathscr\{O\}$ 的, 次数比 $\displaystyle m(x0$ 更低的多项式. 这与最小多项式的定义矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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1430、 7、 设 $\displaystyle \mathbb\{P\}$ 是数域, $\displaystyle \sigma$ 是 $\displaystyle \mathbb\{P\}$ 上 $\displaystyle n$ 维线性空间 $\displaystyle V$ 的线性变换, $\displaystyle m(x)$ 是 $\displaystyle \sigma$ 的最小多项式. 证明: 对任意 $\displaystyle f(x)\in\mathbb\{P\}[x]$, 只要 $\displaystyle \left(f(x),m(x)\right)=d(x)$, 则
\begin\{aligned\} \mathrm\{rank\} f(\sigma)=\mathrm\{rank\} d(\sigma). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南昌大学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 $\displaystyle (f,m)=d$ 知
\begin\{aligned\} &\exists\ u,v,\mathrm\{ s.t.\} uf+vm=d\\\\ \Rightarrow&d(\sigma)=u(\sigma)f(\sigma)+v(\sigma)m(\sigma)=u(\sigma)f(\sigma)\\\\ \Rightarrow&\mathrm\{im\} d(\sigma)\subset \mathrm\{im\} f(\sigma). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle d\mid f$ 知
\begin\{aligned\} \exists\ q, \mathrm\{ s.t.\} f=dq \Rightarrow f(\sigma)=d(\sigma)q(\sigma) \Rightarrow \mathrm\{im\} f(\sigma)\subset \mathrm\{im\} d(\sigma). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{im\} f(\sigma)=\mathrm\{im\} d(\sigma)$,
\begin\{aligned\} \mathrm\{rank\} f(\sigma)=\dim \mathrm\{im\} f(\sigma)=\dim \mathrm\{im\} d(\sigma)=\mathrm\{rank\} d(\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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1431、 8、 设 $\displaystyle f(x),g(x)$ 为数域 $\displaystyle \mathbb\{P\}$ 上的互素多项式, $\displaystyle A\in\mathbb\{P\}^\{n\times n\}$. 证明:
\begin\{aligned\} \ker\left(f(A)g(A)\right)=\ker f(A)\oplus \ker g(A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南昌大学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) / 易知 $\displaystyle \ker f(A)\subset \ker\left(f(A)g(A)\right), \ker g(A)\subset \ker\left(f(A)g(A)\right)$. 由
\begin\{aligned\} (f,g)=1\Rightarrow& \exists\ u,v,\mathrm\{ s.t.\} uf+vg=1\\\\ \Rightarrow&E\_n=u(A)f(A)+v(A)g(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\} \alpha\in \mathbb\{P\}^n\Rightarrow&\alpha=E\_n\alpha=v(A)g(A)\alpha+u(A)f(A)\alpha\\\\ &\in \ker f(A)+\ker g(A),\\\\ \alpha\in \ker f(A)\cap \ker g(A)\Rightarrow&f(A)\alpha=0, g(A)\alpha=0\\\\ \Rightarrow&\alpha=E\_n\alpha=v(A)g(A)\alpha+u(A)f(A)\alpha=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle \ker\left(f(A)g(A)\right)=\ker f(A)\oplus \ker g(A)$.跟锦数学微信公众号. [在线资料](
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---
1432、 5、 $\displaystyle W=\mathbb\{R\}^\{n\times 2\}$ 是 $\displaystyle n\times 2$ 实矩阵生成的线性空间, 映射
\begin\{aligned\} \mathscr\{T\}: W\to W, \qquad \mathscr\{T\} X=AX, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle A$ 是秩为 $\displaystyle r$ 的 $\displaystyle n$ 阶实方阵. (1)、 验证 $\displaystyle \mathscr\{T\}$ 为线性变换; (2)、 求 $\displaystyle \dim \ker\mathscr\{T\}, \dim \mathrm\{im\}\mathscr\{T\}$. (南方科技大学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)、 对 $\displaystyle \forall\ k,l\in\mathbb\{R\}, X,Y\in W$,
\begin\{aligned\} \mathscr\{T\}(kX+lY)=A(kX+lY)=kAX+lAY=k\mathscr\{T\} X+l\mathscr\{T\} Y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathscr\{T\}$ 是线性变换. (2)、 由 $\displaystyle \mathrm\{rank\} A=r$ 知 $\displaystyle Ax=0$ 有 $\displaystyle n-r$ 个线性无关的基础解系 $\displaystyle \eta\_1,\cdots,\eta\_\{n-r\}$. 于是
\begin\{aligned\} &X\in \ker\mathscr\{T\}\Leftrightarrow AX=0, X=(\alpha\_1,\alpha\_2)\\\\ \Leftrightarrow& X=(\alpha\_1,\alpha\_2), \alpha\_i=\sum\_\{j=1\}^n x\_\{ij\}\eta\_j\\\\ \Leftrightarrow& X=(\alpha\_1,0)+(0,\alpha\_2) =\sum\_\{j=1\}^n x\_\{1j\}(\eta\_j,0)+\sum\_\{j=1\}^n x\_\{2j\}(0,\eta\_j). \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 $\displaystyle \ker\mathscr\{T\}$ 有一组基
\begin\{aligned\} (\eta\_j,0), (0,\eta\_j), j=1,\cdots, n-r, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \dim \ker\mathscr\{T\}=2(n-r), \dim \mathrm\{im\}\mathscr\{T\}=\dim W-\dim \ker \mathscr\{T\}=2r$.跟锦数学微信公众号. [在线资料](
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---
1433、 8、 设 $\displaystyle \varSigma$ 是 $\displaystyle n$ 阶实方阵构成的线性空间, $\displaystyle U$ 是 $\displaystyle \varSigma$ 的子空间且 $\displaystyle U$ 中所有非零元素都是可逆矩阵,
\begin\{aligned\} \varphi(n)=\max\_\{U\in\varSigma\}\dim U. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: (1)、 $\displaystyle n$ 为奇数时, $\displaystyle \varphi(n)=1$; (2)、 $\displaystyle n$ 为偶数时, $\displaystyle \varphi(n)\leq n$. (南方科技大学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)、 用反证法. 若 $\displaystyle \varphi(n)\geq 2$, 则存在 $\displaystyle \varSigma$ 的一个维数 $\displaystyle \geq 2$ 的子空间 $\displaystyle U$, 使得 $\displaystyle U$ 中任意非零矩阵都可逆. 设 $\displaystyle A,B$ 是 $\displaystyle U$ 中两个线性无关的矩阵. 则对 $\displaystyle \forall\ k\in\mathbb\{R\}$, $\displaystyle 1,k$ 不全为 $\displaystyle 0$, $\displaystyle A+kB\neq 0$, 而 $\displaystyle A+kB$ 可逆, $\displaystyle |A+kB|\neq 0$. 这与奇数次多项式 $\displaystyle |A+kB|$ 至少有一个实根矛盾. 故有结论. (2)、 用反证法. 若 $\displaystyle \dim V\geq n+1$, 则存在 $\displaystyle n+1$ 个线性无关的 $\displaystyle n$ 阶方阵 $\displaystyle A\_1,\cdots,A\_\{n+1\}$. 考虑关于 $\displaystyle x\_1,\cdots,x\_\{n+1\}$ 的方程
\begin\{aligned\} |x\_1A\_1+\cdots+x\_\{n+1\}A\_\{n+1\}|=0. \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\} A\_i=(\alpha\_1^i,\cdots,\alpha\_\{n\}^i)\left(1\leq i\leq n+1\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
考虑线性方程组
\begin\{aligned\} x\_1\alpha^1\_1+\cdots+x\_\{n+1\}\alpha^\{n+1\}\_1=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由于未知量的个数大于方程的个数, 而该线性方程组有非零解 $\displaystyle (c\_1,\cdots,c\_\{n+1\})$. 也即 $\displaystyle c\_1A\_1+\cdots+c\_\{n+1\}A\_\{n+1\}$ 的第一列为 $\displaystyle 0$. 这表明 $\displaystyle |c\_1A\_1+\cdots+c\_\{n+1\}A\_\{n+1\}|=0$. 这就证明了 $\displaystyle V$ 中存在非零矩阵
\begin\{aligned\} c\_1A\_1+\cdots+c\_\{n+1\}A\_\{n+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不可逆, 与题设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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1434、 2、 (15 分) 设 $\displaystyle V$ 是实线性空间 $\displaystyle M\_n(\mathbb\{R\})$ 的线性空间, 假设 $\displaystyle V$ 中任意非零矩阵都是可逆阵, 证明: $\displaystyle V$ 的维数不超过 $\displaystyle n$. (南京大学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) / 用反证法. 若 $\displaystyle \dim V\geq n+1$, 则存在 $\displaystyle n+1$ 个线性无关的 $\displaystyle n$ 阶方阵 $\displaystyle A\_1,\cdots,A\_\{n+1\}$. 考虑关于 $\displaystyle x\_1,\cdots,x\_\{n+1\}$ 的方程
\begin\{aligned\} |x\_1A\_1+\cdots+x\_\{n+1\}A\_\{n+1\}|=0. \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\} A\_i=(\alpha\_1^i,\cdots,\alpha\_\{n\}^i)\left(1\leq i\leq n+1\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
考虑线性方程组
\begin\{aligned\} x\_1\alpha^1\_1+\cdots+x\_\{n+1\}\alpha^\{n+1\}\_1=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由于未知量的个数大于方程的个数, 而该线性方程组有非零解 $\displaystyle (c\_1,\cdots,c\_\{n+1\})$. 也即 $\displaystyle c\_1A\_1+\cdots+c\_\{n+1\}A\_\{n+1\}$ 的第一列为 $\displaystyle 0$. 这表明 $\displaystyle |c\_1A\_1+\cdots+c\_\{n+1\}A\_\{n+1\}|=0$. 这就证明了 $\displaystyle V$ 中存在非零矩阵
\begin\{aligned\} c\_1A\_1+\cdots+c\_\{n+1\}A\_\{n+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不可逆. 这与题设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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1435、 8、 (20 分) 设 $\displaystyle n\geq 2, A\in M\_n(\mathbb\{R\})$ 且 $\displaystyle \mathrm\{rank\} A=n-1$. 证明: 映射
\begin\{aligned\} \sigma: M\_n(\mathbb\{R\})\to M\_n(\mathbb\{R\}),\quad X\mapsto AX+XA \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不是满射. (南京大学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 $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}(A^\mathrm\{T\})=n-1$ 知 $\displaystyle Ax=0, A^\mathrm\{T\} x=0$ 的基础解系各有 $\displaystyle 1$ 个线性无关的解向量 $\displaystyle \alpha,\beta$. 设 $\displaystyle B=\alpha\beta^\mathrm\{T\}\neq 0$, 则
\begin\{aligned\} \sigma(B)=AB+BA=A\alpha \beta^\mathrm\{T\}+\alpha\beta^\mathrm\{T\} A =0+\alpha(A^\mathrm\{T\} \beta)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \ker \sigma\neq \left\\{0\right\\}$. 由
\begin\{aligned\} \dim \ker \sigma+\dim \mathrm\{im\}\sigma=\dim \mathbb\{R\}^n=n\Rightarrow \dim \mathrm\{im\} \sigma\leq n-1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sigma$ 不是满射.跟锦数学微信公众号. [在线资料](
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---
1436、 3、 设 $\displaystyle V\_1$ 是由向量
\begin\{aligned\} \alpha\_1=(1,1,\alpha)^\mathrm\{T\}, \alpha\_2=(-2,\alpha,4)^\mathrm\{T\}, \alpha\_3=(-2,\alpha,-2)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
生成的 $\displaystyle \mathbb\{R\}^3$ 的子空间, $\displaystyle V\_2$ 是由
\begin\{aligned\} \beta\_1=(1,1,\alpha)^\mathrm\{T\}, \beta\_2=(1,\alpha,1)^\mathrm\{T\}, \beta\_3=(\alpha,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\}
生成的 $\displaystyle \mathbb\{R\}^3$ 的子空间. (1)、 若 $\displaystyle V\_2$ 的维数为 $\displaystyle 1$, 求 $\displaystyle \alpha$ 的值; (2)、 若 $\displaystyle V\_1=V\_2$, 求 $\displaystyle \alpha$ 的值; (3)、 求 $\displaystyle V\_1+V\_2$ 维数的取值范围. (南京航空航天大学大学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 $\displaystyle A=(\alpha\_1,\alpha\_2,\alpha\_3), B=(\beta\_1,\beta\_2,\beta\_3)$. (1)、 由
\begin\{aligned\} B=&\left(\beta\_1,\beta\_2,\beta\_3\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\alpha\\\\ 1&\alpha&1\\\\ \alpha&1&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\alpha\\\\ 0&\alpha-1&1-\alpha\\\\ 0&1-\alpha&1-\alpha^2\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\alpha\\\\ 0&\alpha-1&1-\alpha\\\\ 0&0&-(\alpha-1)(\alpha+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\}
及 $\displaystyle 1=\dim V\_2=\mathrm\{rank\} B$ 知 $\displaystyle \alpha=1$. (2)、 由 $\displaystyle V\_1=V\_2$ 知 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 与 $\displaystyle \beta\_1,\beta\_2,\beta\_3$ 等价, 即它们可以相互线性表出: $\displaystyle AX=B, BX=A$ 都有解,
\begin\{aligned\} \mathrm\{rank\} A=\mathrm\{rank\}(A,B)=\mathrm\{rank\} B . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} &(A,B)= (\alpha\_1,\alpha\_2,\alpha\_3,\beta\_1,\beta\_2,\beta\_3)\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&-2&1&1&\alpha\\\\ 0&\alpha+2&\alpha+2&0&\alpha-1&1-\alpha\\\\ 0&2(\alpha+2)&2(\alpha-1)&0&1-\alpha&1-\alpha^2\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&-2&1&1&\alpha\\\\ 0&\alpha+2&\alpha+2&0&\alpha-1&1-\alpha\\\\ 0&0&-6&0&3(1-\alpha)&-(1-\alpha)^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\}
知当 $\displaystyle \alpha=-2$ 时, $\displaystyle \mathrm\{rank\} A=2 < 3=\mathrm\{rank\}(A,B)$; 当 $\displaystyle \alpha=1$ 时, $\displaystyle \mathrm\{rank\} B=1 < 3=\mathrm\{rank\}(A,B)$. 故 $\displaystyle \alpha\neq -2$ 且 $\displaystyle \alpha\neq 1$. (3)、 (3-1)、 当 $\displaystyle \alpha=-2$ 时,
\begin\{aligned\} (A,B)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&0&1&0&-1\\\\ 0&0&1&0&0&0\\\\ 0&0&0&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 $\displaystyle \alpha\_1,\alpha\_2,\beta\_2$ 线性无关, 是 $\displaystyle V\_1+V\_2$ 的一组基, 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\_2=-2\alpha\_1, \beta\_1=\alpha\_1, \beta\_3=-\alpha\_1-\beta\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此时, $\displaystyle \dim (V\_1+V\_2)=3$. (3-2)、 若 $\displaystyle \alpha=1$, 则
\begin\{aligned\} (A,B)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1&1&1\\\\ 0&1&0&0&0&0\\\\ 0&0&1&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\}
故 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性无关, 是 $\displaystyle V\_1+V\_2$ 的一组基, $\displaystyle \dim (V\_1+V\_2)=3$. (3-3)、 若 $\displaystyle \alpha\neq -2$ 且 $\displaystyle \alpha\neq 1$, 则 $\displaystyle |A|\neq 0, |B|\neq 0$, 而 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性无关, $\displaystyle \beta\_1,\beta\_2,\beta\_3$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出. 故 $\displaystyle \dim(V\_1+V\_2)=3$. 综上即知 $\displaystyle \dim(V\_1+V\_2)=3$.跟锦数学微信公众号. [在线资料](
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---
1437、 4、 设 $\displaystyle \sigma$ 为 $\displaystyle \mathbb\{R\}^3$ 上的线性变换, $\displaystyle \varepsilon\_1=(1,1,0)^\mathrm\{T\}, \varepsilon\_2=(0,1,1)^\mathrm\{T\}, \varepsilon\_3=(1,1,1)^\mathrm\{T\}$, 且
\begin\{aligned\} \sigma \varepsilon\_1=(0,-1,1)^\mathrm\{T\}, \sigma \varepsilon\_2=(1,1+a,0)^\mathrm\{T\}, \sigma \varepsilon\_3=(1,a-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\}
(1)、 求 $\displaystyle \sigma$ 在基
\begin\{aligned\} \eta\_1=(1,0,0)^\mathrm\{T\}, \eta\_2=(0,1,0)^\mathrm\{T\}, \eta\_3=(0,0,1)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
下的矩阵 $\displaystyle A$; (2)、 若 $\displaystyle \sigma$ 可对角化, 求 $\displaystyle a$ 的值; (3)、 当 $\displaystyle a=2$ 时, 求一多项式 $\displaystyle g(x)$, 使得 $\displaystyle g(A)=A^\{-1\}$. (南京航空航天大学大学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,
\begin\{aligned\} (\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)=&(\eta\_1,\eta\_2,\eta\_3)T, T=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 1&1&1\\\\ 0&1&1\end\{array\}\right),\\\\ \sigma(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)=&(\eta\_1,\eta\_2,\eta\_3)B, B=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1\\\\ -1&1+a&a-1\\\\ 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\}
故
\begin\{aligned\} \sigma(\eta\_1,\eta\_2,\eta\_3)=\sigma(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)T^\{-1\} =(\eta\_1,\eta\_2,\eta\_3)BT^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle A=BT^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&1\\\\-2&1&a\\\\1&0&0\end\{array\}\right)$. (2)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,-1$. 由 $\displaystyle \sigma$ 可对角化知 $\displaystyle A$ 可对角化, $\displaystyle \sim \mathrm\{diag\}(1,1,-1)$. 从而
\begin\{aligned\} \mathrm\{rank\}(A-E)=1, A-E=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0&1\\\\ -2&0&a\\\\ 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\}
故 $\displaystyle a=2$. (3)、 当 $\displaystyle a=2$ 时, 由第 2 步知 $\displaystyle A$ 可对角化, 且 $\displaystyle A$ 的最小多项式为 $\displaystyle (x-1)(x+1)=x^2-1$. 故 $\displaystyle A^2=E\Rightarrow A^\{-1\}=A$. 取 $\displaystyle g(x)=x$ 即可.跟锦数学微信公众号. [在线资料](
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---
1438、 8、 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n$ 维线性空间, $\displaystyle U,W$ 是 $\displaystyle V$ 的两个子空间, 并且 $\displaystyle V=U\oplus W$. 任给 $\displaystyle \alpha=\alpha\_1+\alpha\_2\in V$, 其中 $\displaystyle \alpha\_1\in U, \alpha\_2\in W$. 令 $\displaystyle \sigma\alpha=\alpha\_1$. 记
\begin\{aligned\} \ker\sigma=&\left\\{\alpha\in V; \sigma\alpha=0\right\\},\\\\ \mathrm\{im\}\sigma=&\left\\{\sigma\alpha; \alpha\in V\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: (1)、 $\displaystyle \sigma$ 是 $\displaystyle V$ 上的线性变换, 且 $\displaystyle \sigma^2=\sigma$; (2)、 $\displaystyle \ker\sigma=W, \mathrm\{im\}\sigma=U$; (3)、 $\displaystyle \sigma$ 可对角化. (南京航空航天大学大学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 $\displaystyle \forall\ k,l\in\mathbb\{P\}, \alpha,\beta\in V$, 设 $\displaystyle \sigma\alpha=\alpha\_1,\sigma\beta=\beta\_1$, 则
\begin\{aligned\} &\alpha=\alpha\_1+\alpha\_2, \beta=\beta\_1+\beta\_2, \alpha\_1,\beta\_1\in U, \alpha\_2,\beta\_2\in W\\\\ \Rightarrow&k\alpha+l\beta=(k\alpha\_1+l\beta\_1)+(k\alpha\_2+l\beta\_2), k\alpha\_1+l\beta\_1\in U, k\alpha\_2+l\beta\_2\in W. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由直和分解的唯一性知
\begin\{aligned\} \sigma(k\alpha+l\beta)=k\alpha\_1+l\beta\_1=k\sigma\alpha+l\sigma\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sigma$ 是线性变换, 且由 $\displaystyle \alpha\_1=\alpha\_1+0\in U+W$ 知
\begin\{aligned\} \sigma^2\alpha=\sigma\alpha\_1=\alpha\_1=\sigma\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sigma^2=\sigma$. (2)、 (2-1)、 设 $\displaystyle \alpha\in \ker\sigma$, 则 $\displaystyle 0=\sigma\alpha=\alpha\_1$,
\begin\{aligned\} \alpha=\alpha\_1+\alpha\_2=\alpha\_2\in W. \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 $\displaystyle \alpha\in W$, 则
\begin\{aligned\} \alpha=0+\alpha\in U+W\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\}
故有 $\displaystyle \ker\sigma=W$. (2-2)、 设 $\displaystyle \alpha\in \mathrm\{im\}\sigma$, 则 $\displaystyle \exists\ \beta\in V,\mathrm\{ s.t.\} \alpha=\sigma\beta\in U$. 反之, 若 $\displaystyle \alpha\in U$, 则
\begin\{aligned\} \alpha=\alpha+0\in U+W\Rightarrow \alpha=\sigma\alpha\in \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\}
故有 $\displaystyle \mathrm\{im\}\sigma=U$. (3)、 取定 $\displaystyle U$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$, $\displaystyle W$ 的一组基 $\displaystyle \varepsilon\_\{r+1\},\cdots,\varepsilon\_n$, 则由 $\displaystyle U+W=V$ 知 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组基, 且
\begin\{aligned\} 1\leq i\leq r\Rightarrow& \varepsilon\_i=\varepsilon\_i+0\in U+W\Rightarrow \sigma\varepsilon\_i=\varepsilon\_i;\\\\ r+1\leq i\leq n\Rightarrow&\varepsilon\_i\in W\Rightarrow \sigma\varepsilon\_i=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sigma$ 在 $\displaystyle V$ 的基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 下的矩阵是对角阵 $\displaystyle \mathrm\{diag\}(E\_r,0)$. 这就证明了 $\displaystyle \sigma$ 可对角化.跟锦数学微信公众号. [在线资料](
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---
1439、 (4)、 已知 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 到
\begin\{aligned\} \beta\_1=\alpha\_1+\alpha\_2, \beta\_2=\alpha\_2+\alpha\_3, \beta\_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\}
的过渡矩阵为 $\displaystyle A$, 则 $\displaystyle A^4=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (南京理工大学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) /
\begin\{aligned\} (\beta\_1,\beta\_2,\beta\_3)=(\alpha\_1,\alpha\_2,\alpha\_3)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 1&1&0\\\\ 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\}
设 $\displaystyle N=A-E$, 则
\begin\{aligned\} A^4=(E+N)^4=E+C\_4^1 N+C\_4^2 N^2 =\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 4&1&0\\\\ 6&4&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1440、 4、 (12 分) 在线性恐艾金你 $\displaystyle \mathbb\{R\}^\{2\times 2\}$ 中定义
\begin\{aligned\} \mathscr\{A\}: \mathbb\{R\}^\{2\times 2\}\to\mathbb\{R\}^\{2\times 2\}, X\mapsto \left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ 1&1\end\{array\}\right)X-X\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\}
(1)、 记 $\displaystyle E\_\{ij\}\ (i,j=1,2)$ 是第 $\displaystyle i$ 行第 $\displaystyle j$ 列元素为 $\displaystyle 1$ 其余元素为 $\displaystyle 0$ 的二阶方阵, 求 $\displaystyle \mathscr\{A\}$ 在 $\displaystyle E\_\{11\},E\_\{12\},E\_\{21\},E\_\{22\}$ 下的矩阵; (2)、 求 $\displaystyle \mathscr\{A\}$ 的秩与零度. (南京理工大学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
\begin\{aligned\} &\mathscr\{A\}(E\_\{11\})=\left(\begin\{array\}\{cccccccccccccccccccc\}0&-1\\\\1&0\end\{array\}\right), \mathscr\{A\}(E\_\{12\})=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0\\\\0&1\end\{array\}\right),\\\\ &\mathscr\{A\}(E\_\{21\}=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\0&-1\end\{array\}\right), \mathscr\{A\}(E\_\{22\})=\left(\begin\{array\}\{cccccccccccccccccccc\}0&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\}
知
\begin\{aligned\} \mathscr\{A\}(E\_\{11\},E\_\{12\},E\_\{21\},E\_\{22\})=&(E\_\{11\},E\_\{12\},E\_\{21\},E\_\{22\})A,\\\\ A=&\left(\begin\{array\}\{cccccccccccccccccccc\}0&-1&1&0\\\\ -1&0&0&1\\\\ 1&0&0&-1\\\\ 0&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\}
(2)、 由 $\displaystyle A\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-1\\\\ 0&1&-1&0\\\\ 0&0&0&0\\\\ 0&0&0&0\end\{array\}\right)$ 知 $\displaystyle \mathscr\{A\}$ 的秩 $\displaystyle \mathrm\{rank\} \mathscr\{A\}=\mathrm\{rank\} A=2$, $\displaystyle \mathscr\{A\}$ 的零度
\begin\{aligned\} \dim\ker \mathscr\{A\}=2^2-\dim \mathrm\{im\} \mathscr\{A\}=4-2=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1441、 6、 (12 分) 设 $\displaystyle A$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n$ 阶矩阵. 证明:
\begin\{aligned\} \left\\{g(A); g(x)\in\mathbb\{P\}[x]\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle \mathbb\{P\}^\{n\times n\}$ 的维数不超过 $\displaystyle n$ 的线性子空间. (南京理工大学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) / 设 $\displaystyle W=\left\\{g(A); g(x)\in\mathbb\{P\}[x]\right\\}$, 则对 $\displaystyle \forall\ k,l\in\mathbb\{P\}, g(x), h(x)\in\mathbb\{P\}[x]$,
\begin\{aligned\} kg(A)+lh(A)=(kg+lh)(A)\in W. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle W$ 是 $\displaystyle \mathbb\{P\}^\{n\times n\}$ 的子空间. 又由 Hamilton-Cayley 定理知对 $\displaystyle f(x)=|xE\_n-A|$, 有 $\displaystyle f(A)=0$. 设 $\displaystyle A$ 的最小多项式为 $\displaystyle m(x)=a\_0+a\_1x+\cdots+a\_mx^m$, 则 $\displaystyle m(x)\mid f(x)\Rightarrow m=\deg m(x)\leq n$, 且
\begin\{aligned\} \forall\ g(x)\in \mathbb\{P\}[x], g(x)=q(x)m(x)+r(x), r(x)=0\mbox\{或\} \deg r(x)\leq m-1. \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\} g(A)=q(A)m(A)+r(A)=r(A)\in L(E,A,\cdots,A^\{m-1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此外, 显然有 $\displaystyle L(E,A,\cdots,A^\{m-1\})\subset V$, 而 $\displaystyle V=L(E,A,\cdots,A^\{m-1\})$. 若 $\displaystyle E,A,\cdots,A^\{m-1\}$ 线性相关, 则存在不全为 $\displaystyle 0$ 的 $\displaystyle k\_i$, 使得
\begin\{aligned\} k\_0E+k\_1A+\cdots+k\_\{m-1\}A^\{m-1\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而多项式 $\displaystyle g(x)=k\_0+k\_1x+\cdots+k\_\{m-1\}x^\{m-1\}$ 是 $\displaystyle A$ 的零化多项式. 由最小多项式的定义知
\begin\{aligned\} m(x)\mid g(x)\Rightarrow m=\partial\left(m(x)\right)\leq \deg g(x)\leq m-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故 $\displaystyle E,A,\cdots,A^\{m-1\}$ 线性无关, 是 $\displaystyle V$ 的一组基, $\displaystyle \dim V=m\leq n$.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)/
---
1442、 9、 (20 分) 在数域 $\displaystyle \mathbb\{P\}$ 上的线性空间
\begin\{aligned\} \mathbb\{P\}[x]\_3=\left\\{a\_0+a\_1x+a\_2x^2; a\_0,a\_1,a\_2\in\mathbb\{P\}\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上定义
\begin\{aligned\} f\_t: \mathbb\{P\}[x]\_3\to \mathbb\{P\}, p(x)\mapsto p(t), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle t=0,-1,1$. (1)、 证明: $\displaystyle f\_0,f\_\{-1\}, f\_1$ 是对偶空间 $\displaystyle L(\mathbb\{P\}[x]\_3,\mathbb\{P\})$ 的一组基; (2)、 求 $\displaystyle \mathbb\{P\}[x]\_3$ 的一组基 $\displaystyle p\_1(x),p\_2(x),p\_3(x)$, 使得 $\displaystyle f\_0,f\_\{-1\},f\_1$ 是它的对偶基. (南京理工大学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)、 显然 $\displaystyle f\_t\in L(\mathbb\{P\}[x]\_3,\mathbb\{P\})$. 由
\begin\{aligned\} &f\_0(1)=1, f\_0(x)=0, f\_0(x^2)=0; f\_\{-1\}(1)=1, f\_\{-1\}(x)=-1, f\_\{-1\}(x^2)=1;\\\\ &f\_1(1)=1, f\_1(x)=1, f\_1(x^2)=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f\_0,f\_\{-1\},f\_1$ 在 $\displaystyle \mathbb\{P\}[x]\_3$ 的基 $\displaystyle 1,x,x^2$ 及 $\displaystyle \mathbb\{P\}$ 的基 $\displaystyle 1$ 下的矩阵为
\begin\{aligned\} f\_0(1,x,x^2)=1\cdot A\_0, A\_0=(1,0,0);\\\\ f\_\{-1\}(1,x,x^2)=1\cdot A\_\{-1\}, A\_\{-1\}=(1,-1,1);\\\\ f\_1(1,x,x^2)=1\cdot A\_1, A\_1=(1,1,1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle k\_0f\_0+k\_\{-1\}f\_\{-1\}+k\_1f\_1=0$, 则
\begin\{aligned\} (0,0,0)=(k\_0f\_0+k\_\{-1\}f\_\{-1\}+k\_1f\_1)(1,x,x^2) =k\_0A\_0+k\_\{-1\}A\_\{-1\}+k\_1A\_1. \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\} \left|\begin\{array\}\{cccccccccc\}A\_0\\\\A\_\{-1\}\\\\A\_1\end\{array\}\right|=\left|\begin\{array\}\{cccccccccc\}1&0&0\\\\ 1&-1&1\\\\ 1&1&1\end\{array\}\right|=-2\neq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A\_0,A\_\{-1\},A\_1$ 线性无关, 而 $\displaystyle k\_0=k\_\{-1\}=k\_1=0$. 故 $\displaystyle f\_0,f\_\{-1\},f\_1$ 线性无关. 又设 $\displaystyle f\in L(\mathbb\{P\}[x]\_3,\mathbb\{P\})$, 则
\begin\{aligned\} \exists\ \alpha,\beta,\gamma\in\mathbb\{P\},\mathrm\{ s.t.\} f(1,x,x^2)=1\cdot (\alpha,\beta,\gamma). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle A\_0,A\_\{-1\},A\_1$ 线性无关知
\begin\{aligned\} \exists\ |\ x\_i\in\mathbb\{P\},\mathrm\{ s.t.\} &f(1,x,x^2)=(\alpha,\beta,\gamma)=x\_0A\_0+x\_\{-1\}A\_\{-1\}+x\_1A\_1\\\\ =&x\_0f\_0(1,x,x^2)+x\_\{-1\}f\_\{-1\}(1,x,x^2)+x\_1f\_1(1,x,x^2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f=x\_0f\_0+x\_\{-1\}f\_\{-1\}+x\_1f\_1$. 这就证明了 $\displaystyle f\_0,f\_\{-1\}, f\_1$ 是对偶空间 $\displaystyle L(\mathbb\{P\}[x]\_3,\mathbb\{P\})$ 的一组基. (2)、 设
\begin\{aligned\} p\_i(x)=a\_\{i1\}+a\_\{i2\}x+a\_\{i3\}x^2, i=1,2,3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的对偶基为 $\displaystyle f\_0,f\_\{-1\},f\_1$, 则
\begin\{aligned\} \delta\_\{ij\}=&f\_i\left(p\_j(x)\right) =a\_\{j1\}f\_i(1)+a\_\{j2\}f\_i(x)+a\_\{j3\}f\_i(x^2)\\\\ =&\left(f\_i(1),f\_i(x),f\_i(x^2)\right)\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{j1\}\\\\a\_\{j2\}\\\\a\_\{j3\}\end\{array\}\right) =A\_i\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{j1\}\\\\a\_\{j2\}\\\\a\_\{j3\}\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\}
此即
\begin\{aligned\} E=\left(\begin\{array\}\{cccccccccccccccccccc\}A\_0\\\\A\_\{-1\}\\\\A\_1\end\{array\}\right)A^\mathrm\{T\}, A=(a\_\{ij\})\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}A\_0\\\\A\_\{-1\}\\\\A\_1\end\{array\}\right)^\{-1\} =\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&-\frac\{1\}\{2\}&\frac\{1\}\{2\}\\\\ 0&\frac\{1\}\{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\}
故
\begin\{aligned\} p\_1(x)=1-x^2, p\_2(x)=\frac\{x^2-x\}\{2\}, p\_3(x)=\frac\{x^2+x\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1443、 6、 设 $\displaystyle V\_1,V\_2,V\_3$ 是 $\displaystyle n$ 维线性空间 $\displaystyle V$ 的子空间, 且 $\displaystyle V\_1\subset V\_3$. 证明:
\begin\{aligned\} V\_1+(V\_2\cap V\_3)=(V\_1+V\_2)\cap V\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南京师范大学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\} &V\_1\subset V\_1\cap V\_2, V\_1\subset V\_3\Rightarrow V\_1\subset (V\_1+V\_2)\cap V\_3,\\\\ &V\_2\cap V\_3\subset V\_2\subset V\_1+V\_2, V\_2\cap V\_3\subset V\_3\Rightarrow V\_2\cap V\_3\subset (V\_1+V\_2)\cap V\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_1+(V\_2\cap V\_3)\subset(V\_1+V\_2)\cap V\_3$. 又由
\begin\{aligned\} &\dim\left\[V\_1+(V\_2\cap V\_3)\right\]=\dim V\_1+\dim(V\_2\cap V\_3) -\dim\left\[V\_1\cap (V\_2\cap V\_3)\right\]\\\\ =&\dim V\_1+\dim (V\_2\cap V\_3)-\dim (V\_1\cap V\_2)\\\\ =&\dim (V\_1+V\_2)-\dim V\_2+\dim (V\_2\cap V\_3)\\\\ &\left(\dim V\_1+\dim V\_2=\dim (V\_1+V\_2)+\dim (V\_1\cap V\_2)\right)\\\\ =&\dim (V\_1+V\_2)-\left\[\dim V\_2-\dim (V\_2\cap V\_3)\right\]\\\\ =&\dim (V\_1+V\_2)-\left\[\dim (V\_2+V\_3)-\dim V\_3\right\]\\\\ =&\dim (V\_1+V\_2)+\dim V\_3-\dim (V\_2+V\_3)\\\\ =&\dim (V\_1+V\_2+V\_3)+\dim \left\[(V\_1+V\_2)\cap V\_3\right\]-\dim (V\_2+V\_3)\\\\ \stackrel\{V\_1\subset V\_3\}\{=\}&\dim \left\[(V\_1+V\_2)\cap V\_3\right\] \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_1+(V\_2\cap V\_3)=(V\_1+V\_2)\cap V\_3$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1444、 7、 设 $\displaystyle f(x)$ 是实系数多项式. 证明:
\begin\{aligned\} W=\left\\{f(x); f(1)=0, \deg f(x)\leq n\mbox\{或\} f(x)=0\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle \mathbb\{R\}[x]$ 的一个线性子空间, 并求它的一组基. (南京师范大学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 $\displaystyle V$ 关于加法和数量乘法封闭知 $\displaystyle V$ 是 $\displaystyle \mathbb\{R\}[x]\_n$ 的一个线性子空间, 而是 $\displaystyle \mathbb\{R\}$ 上的线性空间. 由
\begin\{aligned\} &\sum\_\{i=1\}^n k\_i(x^i-1)=0\\\\ \Rightarrow&\left(-\sum\_\{i=1\}^n k\_i\right)+\sum\_\{i=1\}^n k\_ix^i=0\\\\ \Rightarrow&k\_i=0\left(1\leq i\leq n\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \left\\{x^i-1, 1\leq i\leq n\right\\}$ 线性无关, 且 $\displaystyle \subset V$. 又由
\begin\{aligned\} f\in V&\Rightarrow f(1)=0\\\\ &\Rightarrow\mbox\{设 $\displaystyle f(x)=\sum\_\{i=0\}^n a\_ix^i$, 则 $\displaystyle \sum\_\{i=1\}^n a\_i=f(1)=0$\}\\\\ &\Rightarrow f(x)=\sum\_\{i=1\}^n a\_ix^i+a\_0 -\sum\_\{i=1\}^n a\_i =\sum\_\{i=1\}^n a\_i(x^i-1) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \left\\{x^i-1, 1\leq i\leq n\right\\}$ 是 $\displaystyle V$ 的一组基, $\displaystyle \dim V=n$.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)/
---
1445、 6、 (20 分) 设 $\displaystyle \mathscr\{A\}$ 是 $\displaystyle n$ 维复线性空间 $\displaystyle V$ 上的线性变换, $\displaystyle 0$ 是 $\displaystyle \mathscr\{A\}$ 的特征多项式的 $\displaystyle r$ 重根. 证明: $\displaystyle \mathscr\{A\}^r$ 的秩等于 $\displaystyle n-r$. (南开大学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 $\displaystyle \mathscr\{A\}$ 的 Jordan 标准形为
\begin\{aligned\} J=\mathrm\{diag\}\left(J\_\{n\_1\}(0),\cdots,J\_\{n\_s\}(0), \cdots\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle J\_\{n\_i\}(0)$ 为以 $\displaystyle 0$ 为对角元的 $\displaystyle n\_i$ 阶 Jordan 块, $\displaystyle \cdots$ 为对角元非零的 Jordan 块, 则由 $\displaystyle J\_\{n\_i\}(0)^t$ 的形式知当且仅当 $\displaystyle t\geq \max\_\{1\leq i\leq s\}n\_i$ 时, 所有的 $\displaystyle J\_\{n\_i\}(0)^t$ 均为 $\displaystyle 0$. 此时, 其它 Jordan 块的对角元非零, 而 $\displaystyle \mathrm\{rank\}(\mathscr\{A\}^t)=n-r$. 于是
\begin\{aligned\} r=\sum\_\{i=1\}^s n\_i\geq \max\_\{1\leq i\leq s\}n\_i\Rightarrow \mathrm\{rank\}(\mathscr\{A\}^r)=n-r. \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)/
---
1446、 8、 (10 分) 设 $\displaystyle V$ 是 $\displaystyle n$ 维欧氏空间 $\displaystyle V$, $\displaystyle \mathscr\{A\},\mathscr\{B\},\mathscr\{C\}$ 是 $\displaystyle V$ 上的线性变换, 且满足下述条件: (1)、 $\displaystyle \mathscr\{A\}^2\alpha=0$, 对任意 $\displaystyle \alpha\in V$; (2)、 $\displaystyle (\mathscr\{A\}\alpha,\beta)=(\alpha,\mathscr\{B\}\beta)$, 对任意 $\displaystyle \alpha,\beta\in V$; (3)、 $\displaystyle \mathscr\{C\}=\mathscr\{A\}\mathscr\{B\}+\mathscr\{B\}\mathscr\{A\}$. 证明: $\displaystyle V$ 可以写成子空间 $\displaystyle \mathscr\{C\}^\{-1\}(0)$, $\displaystyle \mathscr\{A\} V$ 以及 $\displaystyle \mathscr\{B\} V$ 的直和. (南开大学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 $\displaystyle V$ 的一组标准正交基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 后, 设
\begin\{aligned\} \mathscr\{A\}(\varepsilon\_1,\cdots,\varepsilon\_n)=&(\varepsilon\_1,\cdots,\varepsilon\_n)A,\\\\ \mathscr\{B\}(\varepsilon\_1,\cdots,\varepsilon\_n)=&(\varepsilon\_1,\cdots,\varepsilon\_n)B . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由 (2) 及
\begin\{aligned\} (\mathscr\{A\}\varepsilon\_i,\varepsilon\_j)=&\left(\sum\_k a\_\{ki\}\varepsilon\_k,\varepsilon\_j\right)=a\_\{ji\},\\\\ (\varepsilon\_i,\mathscr\{B\}\varepsilon\_j)=&\left(\varepsilon\_i,\sum\_k b\_\{kj\}\varepsilon\_k\right)=b\_\{ij\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle B=A^\mathrm\{T\}$,
\begin\{aligned\} \dim \mathrm\{im\}\mathscr\{A\}=\mathrm\{rank\} A=\mathrm\{rank\} (A^\mathrm\{T\})=\dim \mathrm\{im\} \mathscr\{B\} . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再者,
\begin\{aligned\} \dim \ker \mathscr\{C\}&=n-\mathrm\{rank\}(AB+BA) =n-\mathrm\{rank\}(AA^\mathrm\{T\}+A^\mathrm\{T\} A)\\\\ &=n-\mathrm\{rank\}\left((A,A^\mathrm\{T\})\left(\begin\{array\}\{cccccccccccccccccccc\}A^\mathrm\{T\}\\\\A\end\{array\}\right)\right) =n-\mathrm\{rank\} (A,A^\mathrm\{T\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由
\begin\{aligned\} &\alpha\in \mathrm\{im\} \mathscr\{A\}\cap \mathrm\{im\} \mathscr\{B\}\Rightarrow \alpha=\mathscr\{A\}\beta=\mathscr\{B\}\gamma\\\\ \Rightarrow&(\alpha,\alpha)=(\mathscr\{A\}\beta,\mathscr\{B\}\gamma) =\left(\mathscr\{A\}^2\beta,\gamma\right)=0\Rightarrow \alpha=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mathrm\{im\} \mathscr\{A\}\cap \mathrm\{im\} \mathscr\{B\}=\left\\{0\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\}
于是
\begin\{aligned\} 2\mathrm\{rank\} A&=\mathrm\{rank\} A+\mathrm\{rank\} B=\dim \mathrm\{im\} \mathscr\{A\}+\dim \mathrm\{im\} \mathscr\{B\}\\\\ &=\dim(\mathrm\{im\} \mathscr\{A\}\oplus \mathrm\{im\} \mathscr\{B\})\\\\ &=\dim L(\mathscr\{A\} \varepsilon\_1,\cdots,\mathscr\{A\} \varepsilon\_n,\mathscr\{B\}\varepsilon\_1,\cdots,\mathscr\{B\}\varepsilon\_n)\\\\ &=\mathrm\{rank\}(A,A^\mathrm\{T\}), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \dim V&=n=n-\mathrm\{rank\}(A,A^\mathrm\{T\})+\mathrm\{rank\}(A,A^\mathrm\{T\})\\\\ &=\dim \ker \mathscr\{C\}+2\mathrm\{rank\} A\\\\ &=\dim \ker \mathscr\{C\}+\dim\mathrm\{im\}\mathscr\{A\}+\dim \mathrm\{im\} \mathscr\{B\} . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 由维数公式知
\begin\{aligned\} &\dim(\ker \mathscr\{C\}+\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\})\\\\ =&\dim \ker \mathscr\{C\}+\dim(\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\})\\\\ &-\dim \left\[\ker \mathscr\{C\}\cap (\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\})\right\]\\\\ =&\dim \ker \mathscr\{C\}+\dim \mathrm\{im\} \mathscr\{A\}+\dim \mathrm\{im\} \mathscr\{B\}\\\\ &-\dim \left\[\ker \mathscr\{C\}\cap (\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\})\right\].\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \alpha=\mathscr\{A\}\beta+\mathscr\{B\}\gamma\in \ker \mathscr\{C\}$, 则
\begin\{aligned\} 0&=(\mathscr\{A\}\mathscr\{B\}+\mathscr\{B\}\mathscr\{A\})(\mathscr\{A\}\beta+\mathscr\{B\}\gamma) =\mathscr\{A\}\mathscr\{B\}\mathscr\{A\}\beta+\mathscr\{B\}\mathscr\{A\}\mathscr\{B\}\gamma, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \left(\mathscr\{A\}\mathscr\{B\}\gamma,\mathscr\{A\}\mathscr\{B\}\gamma\right)=&\left(\mathscr\{B\}\mathscr\{A\}\mathscr\{B\}\gamma,\mathscr\{B\}\gamma\right) =-\left(\mathscr\{A\}\mathscr\{B\}\mathscr\{A\}\beta,\mathscr\{B\}\gamma\right)\\\\ =&-\left(\mathscr\{A\}^2\mathscr\{B\}\mathscr\{A\}\beta,\gamma\right)=-(0,\gamma)=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle \mathscr\{A\}\mathscr\{B\}\gamma=0$;
\begin\{aligned\} \left(\mathscr\{B\}\gamma,\mathscr\{B\}\gamma\right)=\left(\mathscr\{A\}\mathscr\{B\}\gamma,\gamma\right)=(0,\gamma)=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle \mathscr\{B\}\gamma=0$. 同理, $\displaystyle \mathscr\{A\}\beta=0\Rightarrow \alpha=0$. 这就证明了
\begin\{aligned\} \ker \mathscr\{C\}\cap (\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\})=\left\\{0\right\\}.\qquad(III) \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\} (II)\Rightarrow& \dim(\ker \mathscr\{C\}+\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\})\\\\ =&\dim \ker \mathscr\{C\}+\dim \mathrm\{im\} \mathscr\{A\}+\dim \mathrm\{im\} \mathscr\{B\} =n=\dim V. \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 $\displaystyle \ker \mathscr\{C\}+\mathrm\{im\} \mathscr\{A\}+\mathrm\{im\} \mathscr\{B\}=V$. 联合 $\displaystyle (I), (II)$ 即知
\begin\{aligned\} V=\ker \mathscr\{C\}\oplus \mathrm\{im\} \mathscr\{A\}\oplus\mathrm\{im\} \mathscr\{B\} . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里, 最后一步我们利用了如下直和的定理: 设 $\displaystyle V=V\_1+\cdots+V\_s$, 则
\begin\{aligned\} V=\oplus\_\{i=1\}^s V\_i\Leftrightarrow V\_i\cap \sum\_\{j < i\}V\_j=\left\\{0\right\\}, i=2,\cdots,s. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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1447、 (4)、 已知
\begin\{aligned\} V=\left\\{\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&a\\\\ a+3b&c&0\\\\ 0&b-c&a\end\{array\}\right); a,b,c\in\mathbb\{F\}\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
按照通常矩阵的加法和数乘构成线性空间, 则 $\displaystyle \dim V=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学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) / $\displaystyle V$ 有一组基 $\displaystyle E\_\{13\}+E\_\{21\}+E\_\{33\}, 3E\_\{21\}+E\_\{32\}, E\_\{22\}-E\_\{32\}$, 而 $\displaystyle \dim V=3$.跟锦数学微信公众号. [在线资料](
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1448、 (5)、 $\displaystyle \varphi$ 是线性空间 $\displaystyle V$ 到 $\displaystyle W$ 的线性映射, $\displaystyle e\_1,e\_2,e\_3$ 是 $\displaystyle V$ 的一个基, $\displaystyle \eta\_1,\eta\_2$ 是 $\displaystyle W$ 的一个基, $\displaystyle \varphi$ 在上述两个基下的矩阵是 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&2\\\\ -1&1&0\end\{array\}\right)$, 则 $\displaystyle \ker\varphi=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学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) / 设题中矩阵为 $\displaystyle A$, 则 $\displaystyle A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&1\end\{array\}\right)$, 而 $\displaystyle Ax=0$ 的基础解系为 $\displaystyle (-1,-1,1)^\mathrm\{T\}$, $\displaystyle \ker \varphi=L(-e\_1-e\_2+e\_3)$.跟锦数学微信公众号. [在线资料](
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1449、 7、 设 $\displaystyle V$ 为复数域 $\displaystyle \mathbb\{C\}$ 上的 $\displaystyle n$ 维线性空间, $\displaystyle \varphi\_j: V\to\mathbb\{C\}$ 是非零的线性函数, $\displaystyle j=1,2$. 若不存在 $\displaystyle 0\neq c\in\mathbb\{C\}$, 使得 $\displaystyle \varphi\_1=c\varphi\_2$. 证明: 任意的 $\displaystyle \alpha\in V$ 都可表示为 $\displaystyle \alpha=\alpha\_1+\alpha\_2$ 使得
\begin\{aligned\} \varphi\_1(\alpha)=\varphi\_1(\alpha\_2), \varphi\_2(\alpha)=\varphi\_2(\alpha\_1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(厦门大学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 $\displaystyle \varphi\_j$ 不是非零的值 $\displaystyle \exists\ \beta\in V,\mathrm\{ s.t.\} l\equiv f(\beta)\neq 0$. 于是
\begin\{aligned\} \forall\ \alpha\in V, f(\alpha)=\frac\{f(\alpha)\}\{l\}\cdot l\in L(l). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \dim \mathrm\{im\} \varphi\_j=1$,
\begin\{aligned\} \dim \ker \varphi\_j=n-\dim \mathrm\{im\} \varphi\_j=n-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 我们用反证法证明 $\displaystyle \ker \varphi\_1\neq \ker \varphi\_2$. 若不然, $\displaystyle \ker \varphi\_1=\ker \varphi\_2$ 有一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_\{n-1\}$, 将其扩充为 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 则 $\displaystyle \varphi\_i(\varepsilon\_n)\neq 0$. 对 $\displaystyle \forall\ \alpha=\sum\_i x\_i\varepsilon\_i\in V$,
\begin\{aligned\} \varphi\_1(\alpha)=x\_n\varphi\_1(\varepsilon\_n)=x\_n\frac\{\varphi\_1(\varepsilon\_n)\}\{\varphi\_2(\varepsilon\_n)\}\varphi\_2(\varepsilon\_n) =\frac\{\varphi\_1(\varepsilon\_n)\}\{\varphi\_2(\varepsilon\_n)\}\varphi\_2(\alpha). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与题设矛盾. 故有结论. (3)、 由第 2 步知 $\displaystyle \ker \varphi\_1+\ker \varphi\_2\supsetneq \ker \varphi\_1$,
\begin\{aligned\} & \dim (\ker \varphi\_1+\ker \varphi\_2) > \dim \ker \varphi\_1=n-1\\\\ \Rightarrow& \ker \varphi\_1+\ker \varphi\_2=V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle \forall\ \alpha\in V$, 由上式知
\begin\{aligned\} \alpha=\alpha\_1+\alpha\_2, \alpha\_i\in \ker \varphi\_i, \varphi\_1(\alpha)=\varphi\_1(\alpha\_2), \varphi\_2(\alpha)=\varphi\_2(\alpha\_1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 13:23:16
