## 张祖锦2023年数学专业真题分类70天之第64天
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1450、 8、 $\displaystyle V$ 是复数域上的 $\displaystyle n$ 维线性空间, $\displaystyle \varphi$ 是 $\displaystyle V$ 上的线性变换. 证明:
\begin\{aligned\} \exists\ f(x)\in\mathbb\{C\}[x],\mathrm\{ s.t.\} \varphi f(\varphi)=\sigma, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \sigma$ 可对角化, 并且 $\displaystyle \varphi$ 与 $\displaystyle \sigma$ 有相同的特征多项式. (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 $\displaystyle \varphi$ 的特征多项式为 $\displaystyle |x \mathscr\{E\}-\varphi|=\prod\_\{k=1\}^s (x-\lambda\_k)^\{n\_k\}$, 其中 $\displaystyle \lambda\_k$ 互异, $\displaystyle n\_k\geq 1$, $\displaystyle \sum\_\{k=1\}^s n\_k=n$, 则有直和分解
\begin\{aligned\} V=\oplus\_\{k=1\}^s V\_k, V\_k=\ker(\varphi-\lambda\_k\mathscr\{E\})^\{n\_k\}.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle 0$ 是 $\displaystyle \varphi$ 的特征值, 则考虑同余式
\begin\{aligned\} g(x)\equiv \lambda\_k, \mod (x-\lambda\_k)^\{n\_k\}, 1\leq k\leq s,\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle 0$ 不是 $\displaystyle \varphi$ 的特征值, 则考虑同余式
\begin\{aligned\} &g(x)\equiv \lambda\_k, \mod (x-\lambda\_k)^\{n\_k\}, 1\leq k\leq s,\\\\ &g(x)\equiv 0, \mod x.\qquad(II)' \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由中国剩余定理知 $\displaystyle (II)$ 或 $\displaystyle (II)'$ 有解 $\displaystyle g(x)$, 它满足
\begin\{aligned\} g(\varphi)\alpha=\lambda\_k\alpha, \forall\ \alpha\in V\_k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle g(\varphi)$ 在 $\displaystyle V\_k$ 的任一组基下的矩阵为 $\displaystyle \lambda\_k E\_\{n\_k\}$, $\displaystyle g(\varphi)$ 在由 $\displaystyle V\_k, 1\leq k\leq s$ 的基拼接而成的基下的矩阵为 $\displaystyle \mathrm\{diag\}(\lambda\_1E\_\{n\_1\},\cdots,\lambda\_s E\_\{n\_s\})$. 从而 $\displaystyle g(\varphi)$ 的特征多项式就是 $\displaystyle \prod\_\{k=1\}^s (x-\lambda\_k)^\{n\_k\}$, 与 $\displaystyle \varphi$ 的相同. 再者,
\begin\{aligned\} g(x)=0\Rightarrow \exists\ f(x),\mathrm\{ s.t.\} g(x)=xf(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sigma=g(\varphi)=\varphi f(\varphi)$ 满足题设.跟锦数学微信公众号. [在线资料](
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1451、 1、 每题 10 分, 共 90 分. (1)、 设 $\displaystyle \alpha\_1,\cdots,\alpha\_t$ 是一组线性无关的向量,
\begin\{aligned\} \beta\_i=\sum\_\{j=1\}^n a\_\{ij\}\alpha\_j\left(i=1,\cdots,t\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \beta\_1,\cdots,\beta\_t$ 线性无关的充要条件是 $\displaystyle \left|\begin\{array\}\{cccccccccc\}a\_\{11\}&\cdots&a\_\{1t\}\\\\ \vdots&&\vdots\\\\ a\_\{t1\}&\cdots&a\_\{tt\}\end\{array\}\right|\neq 0$. (山东大学2023年高等代数与常微分方程考研试题) [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) / $\displaystyle \beta\_1,\cdots,\beta\_t \mbox\{线性无关\}$
\begin\{aligned\} &\Leftrightarrow 0=\sum\_\{i=1\}^t x\_i\beta\_i=(\beta\_1,\cdots,\beta\_t)X =(\alpha\_1,\cdots,\alpha\_t)A^\mathrm\{T\} X\mbox\{只有零解\}\\\\ \Leftrightarrow&A^\mathrm\{T\} X\mbox\{只有零解\}\Leftrightarrow t=\mathrm\{rank\} (A^\mathrm\{T\})=\mathrm\{rank\} A\Leftrightarrow \left|\begin\{array\}\{cccccccccc\}a\_\{11\}&\cdots&a\_\{1t\}\\\\ \vdots&&\vdots\\\\ a\_\{t1\}&\cdots&a\_\{tt\}\end\{array\}\right|\neq 0. \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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---
1452、 (2)、 设 $\displaystyle A$ 是数域 $\displaystyle \mathbb\{K\}$ 上 $\displaystyle m\times n$ 矩阵, $\displaystyle V$ 是 $\displaystyle \mathbb\{K\}$ 上满足 $\displaystyle AX=0$ ($X$ 是 $\displaystyle n\times s$ 未知矩阵, $\displaystyle O$ 是 $\displaystyle m\times s$ 零矩阵) 的全体 $\displaystyle n\times s$ 矩阵组成的集合. 问: 对矩阵普通加法以及数与矩阵的乘法, $\displaystyle V$ 是否构成线性空间? 若 $\displaystyle V$ 构成线性空间, 求其维数并给出一组基. (山东大学2023年高等代数与常微分方程考研试题) [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 $\displaystyle V\subset \mathbb\{K\}^\{n\times s\}$ 及
\begin\{aligned\} X\_1,X\_2\in V, k\_1,k\_2\in\mathbb\{K\}\Rightarrow& A(k\_1X\_1+k\_2X\_2)=k\_1AX\_1+k\_2AX\_2=0\\\\ \Rightarrow& k\_1X\_1+k\_2X\_2\in V \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V$ 是 $\displaystyle \mathbb\{K\}^\{n\times s\}$ 的子空间, 而是线性空间. 设 $\displaystyle \mathrm\{rank\} A=r$, 则可设 $\displaystyle \eta\_1,\cdots,\eta\_\{n-r\}$ 是 $\displaystyle Ax=0$ 的基础解系. 于是
\begin\{aligned\} &X=(X\_1,\cdots,X\_s)\in V\Leftrightarrow \forall\ 1\leq i\leq s, AX\_i=0 \Leftrightarrow X\_i=\sum\_\{j=1\}^\{n-r\}x\_\{ij\}\eta\_j\\\\ \Leftrightarrow& X=\sum\_\{i=1\}^s (0,\cdots,X\_i,\cdots,0) =\sum\_\{i=1\}^s (0,\cdots, \sum\_\{j=1\}^\{n-r\}x\_\{ij\}\eta\_j,\cdots,0)\\\\ &=\sum\_\{i=1\}^s \sum\_\{j=1\}^\{n-r\} x\_\{ij\}(0,\cdots,\eta\_j,\cdots,0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle V$ 有一组基 $\displaystyle (0,\cdots,\eta\_j,\cdots,0)$, 其中 $\displaystyle \eta\_j$ 位于第 $\displaystyle i$ 列. 而 $\displaystyle \dim V=s(n-r)$.跟锦数学微信公众号. [在线资料](
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1453、 (3)、 设 $\displaystyle \alpha\_1=(1,1,1)^\mathrm\{T\}, \alpha\_2=(1,1,2)^\mathrm\{T\}, \alpha\_3=(1,2,3)^\mathrm\{T\}$. 试证: $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 为线性空间 $\displaystyle \mathbb\{R\}^3$ 的一组基, 并用两种方法求向量 $\displaystyle \alpha=(6,9,14)^\mathrm\{T\}$ 在该组基下的坐标. (山东大学2023年高等代数与常微分方程考研试题) [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 $\displaystyle |\alpha\_1,\alpha\_2,\alpha\_3|=\left|\begin\{array\}\{cccccccccc\}1&1&1\\\\ 1&1&2\\\\ 1&2&3\end\{array\}\right|=-1\neq 0$ 知 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性无关, 而是 $\displaystyle \mathbb\{R\}^3$ 的一组基. 设 $\displaystyle \alpha=x\_1\alpha\_1+x\_2\alpha\_2+x\_3\alpha\_3$, 则由
\begin\{aligned\} (\alpha\_1,\alpha\_2,\alpha\_3,\alpha)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1&6\\\\ 1&1&2&9\\\\ 1&2&3&14\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&1\\\\ 0&1&0&2\\\\ 0&0&1&3\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle x\_1=1,x\_2=2,x\_3=3$. 也可另解如下:
\begin\{aligned\} X=(\alpha\_1,\alpha\_2,\alpha\_3)^\{-1\}\alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\3\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
1454、 (4)、 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{K\}$ 上的线性空间, $\displaystyle \sigma$ 是 $\displaystyle V$ 的一个线性变换, $\displaystyle f(x),g(x)\in\mathbb\{P\}[x], h(x)=f(x)g(x)$. 证明: (1)、 $\displaystyle \ker f(\sigma)+\ker g(\sigma)\subset \ker h(\sigma)$; (2)、 若 $\displaystyle \left(f(x),g(x)\right)=1$, 则 $\displaystyle \ker h(\sigma)=\ker f(\sigma)\oplus \ker g(\sigma)$. (山东大学2023年高等代数与常微分方程考研试题) [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 f(\sigma)\Rightarrow f(\sigma)\alpha=0\Rightarrow h(\sigma)\alpha=f(\sigma)g(\sigma)\alpha=0\Rightarrow \alpha\in \ker h(\sigma) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \ker f(\sigma)\subset \ker h(\sigma)$. 同理, $\displaystyle \ker g(\sigma)\subset \ker h(\sigma)$. 故
\begin\{aligned\} \ker f(\sigma)+\ker g(\sigma)\subset \ker h(\sigma). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由 $\displaystyle (f,g)=1$ 知 $\displaystyle \exists\ u,v,\mathrm\{ s.t.\} uf+vg=1$. 于是
\begin\{aligned\} \alpha\in \ker h(\sigma)&\Rightarrow \alpha=u(\sigma)f(\sigma)\alpha+v(\sigma)g(\alpha)\alpha \in \ker g(\sigma)+\ker f(\sigma). \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 $\displaystyle \ker h(\sigma)=\ker f(\sigma)+\ker g(\sigma)$. 又由
\begin\{aligned\} \alpha\in \ker f(\sigma)\cap \ker g(\sigma) &\Rightarrow f(\sigma)\alpha=0=g(\sigma)\alpha\\\\ &\Rightarrow \alpha=u(\sigma)f(\sigma)\alpha+v(\sigma)g(\alpha)\alpha=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \ker h(\sigma)=\ker f(\sigma)\oplus\ker g(\sigma)$.跟锦数学微信公众号. [在线资料](
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---
1455、 3、 (15 分) $\displaystyle \alpha\_1,\cdots,\alpha\_s,\beta$ 的秩与 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 的秩相同. 证明: $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 线性表出, 且表法唯一的充分必要条件为 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 线性无关. (山西大学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 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_s)$, 则由题设, $\displaystyle \mathrm\{rank\} (A,\beta)=\mathrm\{rank\} A$, 而 $\displaystyle Ax=\beta$ 有解, 此即 $\displaystyle \beta$ 可由 $\displaystyle \alpha\_1,\cdots,\alpha\_s$ 线性表出, 且表法唯一 $\displaystyle \Leftrightarrow Ax=\beta$ 有唯一解 $\displaystyle \Leftrightarrow \mathrm\{rank\} A=s\Leftrightarrow \alpha\_1,\cdots,\alpha\_s$ 线性无关.跟锦数学微信公众号. [在线资料](
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1456、 5、 (20 分) $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&2\\\\ 2&2&2\\\\ 2&2&2\end\{array\}\right)$. $\displaystyle C(A)$ 为可与 $\displaystyle A$ 交换的实矩阵所构成的集合. (1)、 证明: $\displaystyle C(A)$ 为线性空间 $\displaystyle \mathbb\{R\}^\{3\times 3\}$ 的子空间. (2)、 求 $\displaystyle C(A)$ 的基和维数. (山西大学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\} &B,C\in C(A)\Rightarrow AB=BA, AC=CA\\\\ \Rightarrow& A(kB+lC)=kAB+lAC=kBA+lCA=(kB+lC)A\\\\ \Rightarrow& kB+lC\in C(A) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle C(A)$ 是 $\displaystyle \mathbb\{R\}^\{3\times 3\}$ 的线性子空间. (2)、 设 $\displaystyle \alpha\_1=(1,1,1)^\mathrm\{T\}$, 则 $\displaystyle A\alpha\_1=6\alpha\_1$. 又由 $\displaystyle A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)$ 知 $\displaystyle Ax=0$ 的基础解系为 $\displaystyle \alpha\_2=(-1,1,0)^\mathrm\{T\}, \alpha\_3=(-1,0,1)^\mathrm\{T\}$. 设
\begin\{aligned\} P=(\alpha\_1,\alpha\_2,\alpha\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1\\\\ 1&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\}
则 $\displaystyle P$ 可逆, 且 $\displaystyle P^\{-1\}AP=\mathrm\{diag\}(6,0,0)$. 对 $\displaystyle B\in C(A)$, 设 $\displaystyle \tilde\{B\}=P^\{-1\}BP=(\tilde\{b\}\_\{ij\})$, 则
\begin\{aligned\} &AB=BA\Leftrightarrow P^\{-1\}AP\cdot P^\{-1\}BP=P^\{-1\}BP\cdot P^\{-1\}AP\\\\ \Leftrightarrow& \tilde\{B\}\mathrm\{diag\}(6,0,0)=\mathrm\{diag\}(6,0,0)\tilde\{B\} \Leftrightarrow \tilde\{B\}=\left(\begin\{array\}\{cccccccccccccccccccc\}a&&\\\\ &b&c\\\\ &d&e\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 $\displaystyle \dim C(A)=5$, 且 $\displaystyle C(A)$ 有一组基
\begin\{aligned\} PE\_\{11\}P^\{-1\}, PE\_\{22\}P^\{-1\}, PE\_\{23\}P^\{-1\}, PE\_\{32\}P^\{-1\}, PE\_\{33\}P^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1457、 8、 (15 分) $\displaystyle \sigma,\tau$ 为 $\displaystyle n$ 维线性空间 $\displaystyle V$ 的线性变换, $\displaystyle \sigma^2=\sigma$. 证明: $\displaystyle \sigma\tau=\tau\sigma$ 的充分必要条件是 $\displaystyle \sigma$ 的值域 $\displaystyle \mathrm\{im\} \sigma$ 与核 $\displaystyle \ker \sigma$ 是 $\displaystyle \tau$ 的不变子空间. (山西大学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) / $\displaystyle V\_0=\left\\{\alpha\in V; \sigma\alpha=0\right\\}=\ker \sigma$, $\displaystyle V\_1=\left\\{\alpha\in V; \sigma\alpha=\alpha\right\\}=\mathrm\{im\}\sigma$. (1)、 由
\begin\{aligned\} \alpha\in V&\Rightarrow \alpha=\left\[\alpha-\sigma(\alpha)\right\]+\sigma(\alpha)\in V\_0+V\_1,\\\\ \alpha\in V\_0\cap V\_1&\Rightarrow \sigma(\alpha)=0, \sigma(\alpha)=\alpha\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 $\displaystyle V=V\_0\oplus V\_1$. (2)、 $\displaystyle \Rightarrow$: 若 $\displaystyle \sigma\tau=\tau\sigma$, 则由
\begin\{aligned\} \alpha\in V\_0&\Rightarrow \sigma(\alpha)=0 \Rightarrow \sigma\left(\tau(\alpha)\right) =\tau\left(\sigma(\alpha)\right) =\tau(0)=0\\\\ &\Rightarrow \tau(\alpha)\in V\_0,\\\\ \alpha\in V\_1&\Rightarrow \sigma(\alpha)=\alpha \Rightarrow \sigma\left(\tau(\alpha)\right) =\tau\left(\sigma(\alpha)\right)=\tau(\alpha)\\\\ &\Rightarrow \tau(\alpha)\in V\_1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_0,V\_1$ 是 $\displaystyle \tau$ 不变的. (3)、 $\displaystyle \Leftarrow$: 设 $\displaystyle \alpha\_1,\cdots,\alpha\_r$ 是 $\displaystyle V\_0$ 的一组基, $\displaystyle \alpha\_\{r+1\},\cdots,\alpha\_n$ 是 $\displaystyle V\_1$ 的一组基, 则由第 1 步知 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 是 $\displaystyle V$ 的一组基. 为证 $\displaystyle \sigma\tau=\tau\sigma$, 只需验证它们在一组基下的作用相同. 事实上,
\begin\{aligned\} 1\leq i\leq r&\Rightarrow \tau(\alpha\_i)\in V\_0\Rightarrow \sigma\left(\tau(\alpha\_i)\right) =0=\tau\left(\sigma(\alpha\_i)\right),\\\\ r+1\leq i\leq n&\Rightarrow \tau(\alpha\_i)\in V\_1\Rightarrow \sigma\left(\tau(\alpha\_i)\right)=\tau(\alpha\_i)=\tau\left(\sigma(\alpha\_i)\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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---
1458、 3、 (20 分) 已知向量组 $\displaystyle (I)$ 为
\begin\{aligned\} \beta\_1=(0,1,-1)^\mathrm\{T\}, \beta\_2=(s,2,1)^\mathrm\{T\}, \beta\_3=(t,1,0)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
向量组 $\displaystyle (II)$ 为
\begin\{aligned\} \alpha\_1=(1,2,-3)^\mathrm\{T\}, \alpha\_2=(3,0,1)^\mathrm\{T\}, \alpha\_3=(9,6,-7)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
向量组 $\displaystyle (I)$ 与 $\displaystyle (II)$ 有相同的秩, 且 $\displaystyle \beta\_3$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出, 求 $\displaystyle s,t$ 的值. (陕西师范大学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\} &(\alpha\_1,\alpha\_2,\alpha\_3,\beta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&3&9&t\\\\ 2&0&6&1\\\\ -3&1&-7&0\end\{array\}\right)\\\\ \to& \left(\begin\{array\}\{cccccccccccccccccccc\}1&3&9&t\\\\ 0&-6&-12&1-2t\\\\ 0&10&20&3t\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}1&3&9&t\\\\ 0&1&2&\frac\{2t-1\}\{6\}\\\\ 0&0&0&\frac\{5-t\}\{3\}\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及题设’$\beta\_3$ 可由 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性表出‘知 $\displaystyle t=5$. 再者, 上式及题设蕴含
\begin\{aligned\} \mathrm\{rank\}(\beta\_1,\beta\_2,\beta\_3)=\mathrm\{rank\}(\alpha\_1,\alpha\_2,\alpha\_3)=2. \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\} (\beta\_1,\beta\_2,\beta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}0&s&5\\\\ 1&2&1\\\\ -1&1&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&s&5\\\\ 1&2&1\\\\ 0&3&1\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&1\\\\ 0&3&1\\\\ 0&s-15&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 $\displaystyle s=15$.跟锦数学微信公众号. [在线资料](
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---
1459、 6、 (25 分) 已知
\begin\{aligned\} &\alpha\_1=(1,2,1,-2), \alpha\_2=(2,3,1,0), \alpha\_3=(1,2,2,-3);\\\\ &\beta\_1=(1,1,1,1), \beta\_2=(1,0,1,-1), \beta\_3=(1,3,0,-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 $\displaystyle W\_1=L(\alpha\_1,\alpha\_2,\alpha\_3)$ 的基与维数; (2)、 求 $\displaystyle W\_2=L(\beta\_1,\beta\_2,\beta\_3)$ 的基与维数; (3)、 求 $\displaystyle W\_1+W\_2$ 及 $\displaystyle W\_1\cap W\_2$ 的基与维数. (陕西师范大学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\} (\alpha\_1^\mathrm\{T\}, \alpha\_2^\mathrm\{T\}, \alpha\_3^\mathrm\{T\})=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&1\\\\ 2&3&2\\\\ 1&1&2\\\\ -2&0&-3\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&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\}
知 $\displaystyle W\_1$ 有一组基 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$, $\displaystyle \dim W\_1=3$. 又由
\begin\{aligned\} (\beta\_1^\mathrm\{T\},\beta\_2^\mathrm\{T\},\beta\_3^\mathrm\{T\})=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 1&0&3\\\\ 1&1&0\\\\ 1&-1&-4\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&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\}
知 $\displaystyle W\_2$ 有一组基 $\displaystyle \beta\_1,\beta\_2,\beta\_3$, $\displaystyle \dim W\_2=3$. 再由
\begin\{aligned\} C\equiv (\alpha\_1^\mathrm\{T\}, \alpha\_2^\mathrm\{T\}, \alpha\_3^\mathrm\{T\},\beta\_1^\mathrm\{T\},\beta\_2^\mathrm\{T\},\beta\_3^\mathrm\{T\})=&\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&1&1&1&1\\\\ 2&3&2&1&0&3\\\\ 1&1&2&1&1&0\\\\ -2&0&-3&1&-1&-4\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-2&0&5\\\\ 0&1&0&1&0&-1\\\\ 0&0&1&1&0&-2\\\\ 0&0&0&0&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 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\beta\_2$ 线性无关, 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\} \beta\_1=-2\alpha\_1+\alpha\_2+\alpha\_3, \beta\_3=5\alpha\_1-\alpha\_2-2\alpha\_3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\beta\_2$ 是 $\displaystyle W\_1+W\_2$ 的一组基, $\displaystyle \dim(W\_1+W\_2)=4$. 最后, $\displaystyle Cx=0$ 的基础解系为
\begin\{aligned\} (2,-1,-1,1,0,0)^\mathrm\{T\}, (-5,1,2,0,0,1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\gamma\in W\_1\cap W\_2\Leftrightarrow \gamma=\sum\_i x\_i\alpha\_i=-\sum\_j x\_\{3+j\}\beta\_j\\\\ \Leftrightarrow& \gamma=\sum\_j x\_\{3+j\}\beta\_j, Cx=0 \Leftrightarrow \gamma\in L(\beta\_1,\beta\_3). \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 $\displaystyle \beta\_1,\beta\_3$ 是 $\displaystyle W\_1\cap W\_2$ 的一组基, $\displaystyle \dim(W\_1\cap W\_2)=2$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1460、 7、 (20 分) 设 $\displaystyle A,B$ 都是 $\displaystyle n$ 级实矩阵, 并设 $\displaystyle \lambda$ 为 $\displaystyle BA$ 的非零特征值, 以 $\displaystyle V\_\lambda^\{BA\}$ 表示 $\displaystyle BA$ 关于 $\displaystyle \lambda$ 的特征子空间. 证明: (1)、 $\displaystyle \lambda$ 也是 $\displaystyle AB$ 的特征值; (2)、 $\displaystyle \dim(V\_\lambda^\{AB\})=\dim(V\_\lambda^\{BA\})$. (陕西师范大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}AB&A\\\\ 0&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -B&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0&A\\\\ 0&BA\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 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}AB&A\\\\ 0&0\end\{array\}\right)\sim \left(\begin\{array\}\{cccccccccccccccccccc\}0&A\\\\ 0&BA\end\{array\}\right)$. 从而
\begin\{aligned\} &\left|\begin\{array\}\{cccccccccc\}\lambda E-AB&-A\\\\ 0&\lambda E\end\{array\}\right|=\left|\begin\{array\}\{cccccccccc\}\lambda E&-A\\\\ 0&\lambda E-BA\end\{array\}\right|\\\\ \Rightarrow&|\lambda E-AB|\cdot \lambda^n=\lambda^n |\lambda E-BA| \Rightarrow |\lambda E-AB|=|\lambda E-BA|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle |\lambda E-BA|=0\Rightarrow |\lambda E-AB|=0\Rightarrow \lambda$ 是 $\displaystyle AB$ 的特征值. (2)、 由第 1 步知 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E-AB&A\\\\ 0&\lambda E\end\{array\}\right)\sim \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E&-A\\\\ 0&\lambda E-BA\end\{array\}\right)$. 再由 $\displaystyle \lambda\neq 0$ 知
\begin\{aligned\} &\mathrm\{rank\}(\lambda E-AB)+n=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E-AB&\\\\ &\lambda E\end\{array\}\right)\\\\ =&\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E-AB&A\\\\ &\lambda E\end\{array\}\right) =\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E&-A\\\\ 0&\lambda E-BA\end\{array\}\right)\\\\ =&\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda E&\\\\ &\lambda E-BA\end\{array\}\right)=n+\mathrm\{rank\}(\lambda E-BA). \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\} \dim(V\_\lambda^\{AB\})=n-\mathrm\{rank\}(\lambda E-AB)=\mathrm\{rank\}(\lambda E-BA)=\dim(V\_\lambda^\{BA\}). \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)/
---
1461、 4、 已知 $\displaystyle \gamma\_1,\cdots,\gamma\_n$ 中无零向量. 证明: $\displaystyle \gamma\_1,\cdots,\gamma\_n$ 线性相关的充要条件是至少存在 $\displaystyle \gamma\_t$ ($2\leq t\leq n$) 被 $\displaystyle \gamma\_1,\cdots,\gamma\_\{t-1\},\gamma\_\{t+1\},\cdots,\gamma\_n$ 线性表示. (上海财经大学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)、 $\displaystyle \Leftarrow$: 若存在 $\displaystyle \gamma\_t$ ($2\leq t\leq n$) 被 $\displaystyle \gamma\_1,\cdots,\gamma\_\{t-1\},\gamma\_\{t+1\},\cdots,\gamma\_n$ 线性表示, 则
\begin\{aligned\} &\gamma\_t=x\_1\gamma\_1+\cdots+x\_\{t-1\}\gamma\_\{t-1\}+x\_\{t+1\}\gamma\_\{t+1\}+\cdots+x\_n\gamma\_n\\\\ \Rightarrow&x\_1\gamma\_1+\cdots+x\_\{t-1\}\gamma\_\{t-1\}+1\gamma\_t+x\_\{t+1\}\gamma\_\{t+1\}+\cdots+x\_n\gamma\_n=0. \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 $\displaystyle \gamma\_1,\cdots,\gamma\_n$ 线性相关. (2)、 $\displaystyle \Rightarrow$: 由 $\displaystyle \gamma\_1,\cdots,\gamma\_n$ 线性相关知存在不全为 $\displaystyle 0$ 的数 $\displaystyle x\_i$, 使得
\begin\{aligned\} x\_1\gamma\_1+\cdots+x\_n\gamma\_n=0.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} x\_2=\cdots=x\_n=0&\left\\{\begin\{array\}\{llllllllllll\}\stackrel\{(I)\}\{\Rightarrow\} x\_1\gamma\_1=0\\\\ \stackrel\{x\_i\mbox\{不全为 $\displaystyle 0$\}\}\{\Rightarrow\}x\_1\neq 0\end\{array\}\right.\\\\ &\ \Rightarrow\gamma\_1=0,\mbox\{与题设矛盾\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle x\_2,\cdots,x\_n$ 不全为 $\displaystyle 0$. 不妨设 $\displaystyle x\_t\neq 0, 2\leq t\leq n$, 则
\begin\{aligned\} \gamma\_t=-\frac\{x\_1\}\{x\_t\}\gamma\_1-\cdots-\frac\{x\_\{t-1\}\}\{x\_t\}\gamma\_\{t-1\} -\frac\{x\_\{t+1\}\}\{x\_t\}\gamma\_\{t+1\}-\cdots-\frac\{x\_n\}\{x\_t\}\gamma\_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](
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---
1462、 9、 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n$ 维线性空间, $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle V$ 的一组基. 再设
\begin\{aligned\} V\_1=&\left\\{k(\varepsilon\_1+\cdots+\varepsilon\_n); k\in\mathbb\{P\}\right\\},\\\\ V\_2=&\left\\{\sum\_\{i=1\}^n k\_i\varepsilon\_i; \sum\_\{i=1\}^n k\_i=0\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)、 $\displaystyle V\_1,V\_2$ 都是 $\displaystyle V$ 的子空间; (2)、 $\displaystyle V=V\_1\oplus V\_2$; (3)、 令 $\displaystyle V$ 上的线性变换 $\displaystyle \sigma$ 在基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 下的矩阵为置换矩阵 $\displaystyle P$, 即 $\displaystyle P$ 的每一行每一列只有一个元素为 $\displaystyle 1$, 其余元素为 $\displaystyle 0$. 证明: $\displaystyle V\_1,V\_2$ 都是 $\displaystyle \sigma$ 的不变子空间. (上海财经大学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)、 $\displaystyle V\_1=L(\varepsilon\_1+\cdots+\varepsilon\_n)$, 而是 $\displaystyle V$ 的子空间. 由
\begin\{aligned\} &\sum\_\{i=1\}^n k\_i\alpha\_i\in V\_2, \sum\_\{i=1\}^n l\_i\alpha\_i\in V\_2, \sum\_\{i=1\}^n k\_i=\sum\_\{i=1\}^n l\_i=0\\\\ \Rightarrow&\sum\_\{i=1\}^n (kk\_i+ll\_i) =k\sum\_\{i=1\}^n k\_i+l\sum\_\{i=1\}^n l\_i=k0+l0=0\\\\ \Rightarrow&k\left(\sum\_\{i=1\}^n k\_i\alpha\_i\right) +l\left(\sum\_\{i=1\}^n l\_i\alpha\_i\right) =\sum\_\{i=1\}^n (kk\_i+ll\_i)\alpha\_i \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知 $\displaystyle V\_2$ 是 $\displaystyle V$ 的子空间. (2)、 由
\begin\{aligned\} &V\ni \alpha=\sum\_\{i=1\}^n k\_i\alpha\_i =\frac\{\sum\_\{j=1\}^n k\_j\}\{n\}\sum\_\{i=1\}^n \alpha\_i+\sum\_\{i=1\}^n \left(k\_i-\frac\{\sum\_\{j=1\}^n k\_j\}\{n\}\right)\alpha\_i \in V\_1+V\_2,\\\\ &\alpha\in V\_1\cap V\_2 \Rightarrow \alpha=k\sum\_\{i=1\}^n \alpha\_i =\sum\_\{i=1\}^n k\_i\alpha\_i, \sum\_\{i=1\}^n k\_i=0\\\\ &\Rightarrow \sum\_\{i=1\}^n (k-k\_i)\alpha\_i=0 \Rightarrow k-k\_i=0\Rightarrow 0=\sum\_\{i=1\}^n k\_i=nk\\\\ &\Rightarrow k=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\}
即知 $\displaystyle V=V\_1\oplus V\_2$. (3)、 由题设, 存在 $\displaystyle n$ 阶排列 $\displaystyle i\_1\cdots i\_n$ 使得
\begin\{aligned\} \sigma(\alpha\_j)=\alpha\_\{i\_j\}, 1\leq j\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} &\sigma(\alpha\_1+\cdots+\alpha\_n)=\sigma(\alpha\_1)+\cdots+\sigma(\alpha\_n)\\\\ =&\alpha\_\{i\_1\}+\cdots+\alpha\_\{i\_n\}=\alpha\_1+\cdots+\alpha\_n\in V\_1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle V\_1$ 是 $\displaystyle \sigma$ 的不变子空间. 再者, 对 $\displaystyle \sum\_\{j=1\}^n k\_j=0$,
\begin\{aligned\} \sigma\left(\sum\_\{j=1\}^n k\_j\alpha\_j\right) =\sum\_\{j=1\}^n k\_j\sigma(\alpha\_j) =\sum\_\{j=1\}^n k\_j\alpha\_\{i\_j\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
还是 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 的线性组合, 且组合系数的和还是 $\displaystyle 0$. 故 $\displaystyle \sigma\left(\sum\_\{j=1\}^n k\_j\alpha\_j\right)\in V\_2$. 这就证明了 $\displaystyle V\_2$ 也是 $\displaystyle \sigma$ 的不变子空间.跟锦数学微信公众号. [在线资料](
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---
1463、 (6)、 (20 分) 设 $\displaystyle \sigma$ 是有限维线性空间 $\displaystyle W$ 上的线性变换, 记 $\displaystyle \det\sigma$ 在 $\displaystyle W$ 的任意一组基下矩阵的行列式. 现设 $\displaystyle V$ 是 $\displaystyle n$ 维复线性空间, $\displaystyle \mathscr\{A\}$ 是 $\displaystyle V$ 上的线性变换, 把数域从复数域缩小到实数域, 则 $\displaystyle V$ 称为实数域上 $\displaystyle 2n$ 维实线性空间, 记为 $\displaystyle V\_0$. 这是 $\displaystyle \mathscr\{A\}$ 可看成 $\displaystyle V\_0$ 上的实线性变换, 记为 $\displaystyle \mathscr\{A\}\_0$. (6-1)、 若 $\displaystyle e\_1,\cdots,e\_n$ 为 $\displaystyle V$ 的一组基, 证明:
\begin\{aligned\} e\_1,\cdots,e\_n, \mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为 $\displaystyle V\_0$ 的一组基, 其中 $\displaystyle \mathrm\{ i\}$ 为虚数单位; (6-2)、 若 $\displaystyle \mathscr\{A\}$ 在基 $\displaystyle e\_1,\cdots,e\_n$ 下的矩阵为 $\displaystyle A$, 求 $\displaystyle \mathscr\{A\}\_0$ 在基
\begin\{aligned\} e\_1,\cdots,e\_n, \mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n \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; (6-3)、 求 $\displaystyle \det \mathscr\{A\}$ 与 $\displaystyle \det \mathscr\{A\}\_0$ 的关系. (上海大学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\} &\sum\_\{i=1\}^n x\_ie\_i+\sum\_\{i=1\}^n y\_i\mathrm\{ i\} e\_i=0, x\_i,y\_i\in\mathbb\{R\}\\\\ \Rightarrow&\sum\_\{i=1\}^n (x\_i+\mathrm\{ i\} y\_i)e\_i=0\\\\ \Rightarrow& x\_i+\mathrm\{ i\} y\_i=0, 1\leq i\leq n\Rightarrow x\_i=y\_i=0, 1\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\}
知 $\displaystyle e\_1,\cdots,e\_n, \mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n$ 线性无关. 又由
\begin\{aligned\} \alpha\in V&\Rightarrow \alpha =\sum\_\{i=1\}^n z\_ie\_i\left(z\_i=x\_i+\mathrm\{ i\} y\_i\in\mathbb\{C\}, x\_i,y\_i\in\mathbb\{C\}\right)\\\\ &\Rightarrow \alpha=\sum\_\{i=1\}^n x\_ie\_i+\sum\_\{i=1\}^n y\_i\mathrm\{ i\} e\_i \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle e\_1,\cdots,e\_n, \mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n$ 确为 $\displaystyle V\_0$ 的一组基. (2)、 设 $\displaystyle A=(a\_\{ij\}), a\_\{ij\}=b\_\{ij\}+\mathrm\{ i\} c\_\{ij\}, b\_\{ij\},c\_\{ij\}\in\mathbb\{R\}, B=(b\_\{ij\}), C=(c\_\{ij\})$, 则
\begin\{aligned\} &\mathscr\{A\}(e\_1,\cdots,e\_n)=(e\_1,\cdots,e\_n)A=(e\_1,\cdots,e\_n)(B+\mathrm\{ i\} C)\\\\ =&(e\_1,\cdots,e\_n)B+(\mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n)C. \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\}(\mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n)=\mathrm\{ i\}\left\[\mathscr\{A\}(e\_1,\cdots,e\_n)\right\]\\\\ =&(e\_1,\cdots,e\_n)(-C)+(\mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n)B. \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\} &\mathscr\{A\}(e\_1,\cdots,e\_n, \mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n)\\\\ =&(e\_1,\cdots,e\_n, \mathrm\{ i\} e\_1,\cdots,\mathrm\{ i\} e\_n)\left(\begin\{array\}\{cccccccccccccccccccc\}B&-C\\\\ C&B\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
所求矩阵即为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{ Re\} A&-\mathrm\{ Im\} A\\\\ \mathrm\{ Im\} A&\mathrm\{ Re\} A\end\{array\}\right)$. (3)、 我们给出一个常用的行列式计算公式. 设 $\displaystyle A,B,C,D$ 为同级方阵且 $\displaystyle AC=CA$, 则
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, (3-1)、 若 $\displaystyle A$ 可逆, 则由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\} E&0\\\\ -CA^\{-1\}&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} A&B\\\\ C&D\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} E&-A^\{-1\}B\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\} A&0\\\\ 0&D-CA^\{-1\}B\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\} A&B\\\\ C&D\end\{array\}\right|&=|A|\cdot |D-CA^\{-1\}B|\\\\ &=|A|\cdot |D-A^\{-1\}CB|\left(AC=CA\Rightarrow CA^\{-1\}=A^\{-1\}C\right)\\\\ &=|A(D-A^\{-1\}CB|\left(|AB|=|A|\cdot |B|\right)\\\\ &=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3-2)、 若 $\displaystyle A$ 奇异, 则关于 $\displaystyle \lambda$ 的多项式 $\displaystyle |\lambda E+A|=0$ 至多有 $\displaystyle n$ 个复根 $\displaystyle \lambda\_1,\cdots,\lambda\_n$. 而 \begin\{aligned\} &\forall\ \lambda\not\in \left\\{\lambda\_1,\cdots,\lambda\_n\right\\}, |\lambda E+A|\neq 0\\\\ \Rightarrow&\tilde\{A\}=\lambda E+A\mbox\{可逆, 满足\}\tilde\{A\} C=C\tilde\{A\}\\\\ \Rightarrow&\left|\begin\{array\}\{cccccccccc\} \tilde\{A\}&B\\\\ C&D\end\{array\}\right|=|\tilde\{A\} D-CB|\left(\mbox\{由 (1)\}\right)\\\\ \Rightarrow& \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|=|(\lambda E+A)D-CB|. \qquad(210226: eq)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f(\lambda)\equiv \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|-|(\lambda E+A)D-CB|$. 若 $\displaystyle f(\lambda)\not\equiv 0$, 则 $\displaystyle f(\lambda)$ 是次数 $\displaystyle \leq n$ 的多项式, 至多有 $\displaystyle n$ 个复根. 这与 (210226: eq) 矛盾. 故
\begin\{aligned\} &f(\lambda)\equiv 0, \forall\ \lambda\in\mathbb\{C\}\\\\ \Rightarrow&f(0)=0\Rightarrow \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
回到题目. 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{ Re\} A&-\mathrm\{ Im\} A\\\\ \mathrm\{ Im\} A&\mathrm\{ Re\} A\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{ Re\} A+\mathrm\{ i\} \mathrm\{ Im\} A&-\mathrm\{ Im\} A\\\\ \mathrm\{ Im\} A-\mathrm\{ i\} \mathrm\{ Re\} A&\mathrm\{ Re\} A\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A&-\mathrm\{ Im\} A\\\\ -\mathrm\{ i\} A&\mathrm\{ Re\} A\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \det \left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{ Re\} A&-\mathrm\{ Im\} A\\\\ \mathrm\{ Im\} A&\mathrm\{ Re\} A\end\{array\}\right)&=\det\left\[A\cdot \mathrm\{ Re\} A-(-\mathrm\{ i\} A)(-\mathrm\{ Im\} A)\right\]\\\\ &=\det\left\[A\cdot(\mathrm\{ Re\} A-\mathrm\{ i\} \mathrm\{ Im\} A)\right\]\\\\ &=\det(A\cdot \bar\{A\}) =\det A\cdot \overline\{\det A\}=|\det A|^2. \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 $\displaystyle \det \mathscr\{A\}\_0=|\det \mathscr\{A\}|^2$.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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---
1464、 8、 (20 分)设 $\displaystyle U,V,W$ 是域 $\displaystyle \mathbb\{F\}$ 上的有限维向量空间, 设 $\displaystyle \alpha: U\to V, \beta: V\to W$ 是线性映射, 且满足下列条件: (1)、 $\displaystyle \beta\alpha=\mathscr\{O\}$; (2)、 对任意向量空间 $\displaystyle X$ 与线性映射 $\displaystyle f: V\to X$, 若 $\displaystyle f\alpha=\mathscr\{O\}$, 则存在唯一的线性映射 $\displaystyle \mu: W\to X$, 使得 $\displaystyle f=\mu\beta$. 证明: (1)、 $\displaystyle \beta$ 是满映射; (2)、 $\displaystyle W\cong V/\mathrm\{im\} \alpha$. (上海交通大学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 $\displaystyle \beta$ 不是满射, 则 $\displaystyle \dim \mathrm\{im\}\beta=r < n=\dim W$. 取定 $\displaystyle \mathrm\{im\}\beta$ 的一组基 $\displaystyle \eta\_1,\cdots,\eta\_r$, 将其扩充为 $\displaystyle W$ 的一组基 $\displaystyle \eta\_1,\cdots,\eta\_r,\eta\_\{r+1\},\cdots,\eta\_n$. 对题中条件 2 的 $\displaystyle X,f,\mu$, 我们可定义线性映射 $\displaystyle \mu\_1: W\to X$ 使得
\begin\{aligned\} \mu\_1(\eta\_i)=\left\\{\begin\{array\}\{llllllllllll\}\mu(\eta\_i),&1\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\}
它也满足
\begin\{aligned\} \forall\ \alpha\in W, \mu\_1\beta(\alpha)=\mu\_1\left(\sum\_\{i=1\}^r x\_i\eta\_i\right) =\mu\left(\sum\_\{i=1\}^r x\_i\eta\_i\right)=\mu\beta(\alpha). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
与题中的’唯一性‘矛盾. 故有结论. (2)、 由 $\displaystyle \beta\alpha=\mathscr\{O\}$ 知 $\displaystyle \mathrm\{im\}\alpha\subset \ker \beta$. 往用反证法证明 $\displaystyle \mathrm\{im\}\alpha=\ker \beta$, 而
\begin\{aligned\} &\dim W\overset\{\tiny\mbox\{第1步\}\}\{=\} \dim \mathrm\{im\}\beta=\dim V-\dim \ker \beta\\\\ &=\dim V-\dim \mathrm\{im\} \alpha=\dim(V/\mathrm\{im\} \alpha)\\\\ \Rightarrow& W\cong V/\mathrm\{im\} \alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若不然, $\displaystyle \dim \mathrm\{im\} \alpha=s < \dim \ker \beta=t$. 设 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_s$ 是 $\displaystyle \mathrm\{im\}\alpha$ 的一组基, 将其扩充为 $\displaystyle \ker \beta$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_s, \varepsilon\_\{s+1\},\cdots,\varepsilon\_t$, 再将其扩充为 $\displaystyle V$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_s, \varepsilon\_\{s+1\},\cdots,\varepsilon\_t, \eta\_1,\cdots,\eta\_r$. 取 $\displaystyle X=L(\varepsilon\_t), f: V\to W$ 满足
\begin\{aligned\} f(\varepsilon\_i)=0, 1\leq i\leq s; f(\varepsilon\_i)=\varepsilon\_t, s+1\leq i\leq t; f(\eta\_i)=0, 1\leq i\leq r, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle s+1\leq i\leq t\Rightarrow f(\varepsilon\_i)=\varepsilon\_t, \beta(\varepsilon\_i)=0$, 从而对 $\displaystyle \forall\ \mu: W\to X$, $\displaystyle f(\varepsilon\_i)=\varepsilon\_t\neq 0=\mu\beta(\varepsilon\_i)$. 这表明满足题中条件 2 的 $\displaystyle \mu$ 根本找不到, 矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1465、 3、 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 $\displaystyle n\_1,\cdots,n\_s$. 证明: $\displaystyle p\_1(x),\cdots, p\_s(x)$ 在实数域上线性无关. (首都师范大学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) / 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.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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---
1466、 7、 设 $\displaystyle V$ 是一个 $\displaystyle n$ 维实线性空间, $\displaystyle \mathscr\{A\}: V\to V$ 为一个线性变换, $\displaystyle \mathscr\{A\}$ 的特征多项式为 $\displaystyle (x-p)^k(x-q)^\{n-k\}$, 其中 $\displaystyle p,q$ 是互异实数, $\displaystyle 0 < k < n$. 令
\begin\{aligned\} V\_1=&\left\\{\alpha\in V; (\mathscr\{A\}-p\mathscr\{I\})^k\alpha=0\right\\},\\\\ V\_2=&\left\\{\alpha\in V; (\mathscr\{A\}-q\mathscr\{I\})^\{n-k\}\alpha=0\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \mathscr\{I\}$ 是恒等变换. 证明: $\displaystyle V=V\_1\oplus V\_2$. (首都师范大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 先给出一般结果. 设 $\displaystyle \mathscr\{A\}$ 是数域 $\displaystyle \mathbb\{K\}$ 上的 $\displaystyle n$ 维线性空间 $\displaystyle V$ 的线性变换, $\displaystyle f(\lambda)$ 是 $\displaystyle \mathscr\{A\}$ 的特征多项式, 并且 $\displaystyle f(x)=f\_1(x)f\_2(x)$, $\displaystyle \left(f\_1(x),f\_2(x)\right)=1$, 则
\begin\{aligned\} V\xlongequal[\tiny\mbox\{Cayley\}]\{\tiny\mbox\{Hamilton-\}\} \ker f(\mathscr\{A\})=\ker f\_1(\mathscr\{A\})\oplus \ker f\_2(\mathscr\{A\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (f\_1,f\_2)=1$ 知
\begin\{aligned\} \exists\ u\_1,u\_2\in\mathbb\{K\}[x],\mathrm\{ s.t.\} u\_1f\_1+u\_2f\_2=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\} \alpha\in \ker f(\mathscr\{A\} )&\Rightarrow f(\mathscr\{A\} )\alpha=0\\\\ &\Rightarrow \alpha=f\_2(\mathscr\{A\} )u\_2(\mathscr\{A\} )\alpha +f\_1(\mathscr\{A\} )u\_1(\mathscr\{A\} )\alpha\\\\ &\quad \ \in \ker f\_1(\mathscr\{A\} )+\ker f\_2(\mathscr\{A\} ),\\\\ \alpha\in\ker f\_1(\mathscr\{A\} )\cap \ker f\_2(\mathscr\{A\} )&\Rightarrow f\_1(\mathscr\{A\} )\alpha=f\_2(\mathscr\{A\} )\alpha=0\\\\ &\Rightarrow \alpha=u\_1(\mathscr\{A\} )f\_1(\mathscr\{A\} )\alpha +u\_2(\mathscr\{A\} )f\_2(\mathscr\{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\}
知
\begin\{aligned\} V\xlongequal[\tiny\mbox\{Cayley\}]\{\tiny\mbox\{Hamilton-\}\}\ker f(\mathscr\{A\} )=\ker f\_1(\mathscr\{A\} )\oplus \ker f\_2(\mathscr\{A\} ). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 取 $\displaystyle f\_1(x)=(x-p)^k, f\_2(x)=(x-q)^\{n-k\}$, 则由第 1 步即知结论成立.跟锦数学微信公众号. [在线资料](
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---
1467、 4、 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 维线性空间. (1)、 叙述并证明两个子空间的和的维数公式; (2)、 任意取定 $\displaystyle 0\neq\alpha\in V$, 设
\begin\{aligned\} W=\left\\{\beta\in V; \mbox\{存在 $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{T\}$ 使得 $\displaystyle \mathscr\{T\}\alpha=\beta$\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle W=V$. (四川大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle V\_1,V\_2$ 为某个线性空间 $\displaystyle V$ 的两个有限维线性空间, 则
\begin\{aligned\} \dim V\_1+\dim V\_2=\dim(V\_1+V\_2)+\dim(V\_1\cap V\_2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 设 $\displaystyle \alpha\_1,\cdots,\alpha\_r$ 是 $\displaystyle V\_1\cap V\_2$ 的一组基, 将其扩充为 $\displaystyle V\_1$ 的一组基
\begin\{aligned\} \alpha\_1,\cdots,\alpha\_r,\beta\_1,\cdots,\beta\_\{m-r\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
也将其扩充为 $\displaystyle V\_2$ 的一组基
\begin\{aligned\} \alpha\_1,\cdots,\alpha\_r,\gamma\_1,\cdots,\gamma\_\{n-r\}. \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\} &\sum\_i x\_i\alpha\_i+\sum\_j y\_j\beta\_j+\sum\_k z\_k\gamma\_k=0\\\\ \Rightarrow&V\_1\ni \sum\_i x\_i\alpha\_i+\sum\_j y\_j\beta\_j=-\sum\_k z\_k\gamma\_k\in V\_2\\\\ \Rightarrow&-\sum\_k z\_k\gamma\_k\in V\_1\cap V\_2 \Rightarrow \exists\ k\_i,\mathrm\{ s.t.\} -\sum\_k z\_k\gamma\_k=\sum\_i k\_i\alpha\_i\\\\ \Rightarrow&\sum\_i k\_i\alpha\_i+\sum\_k z\_k\gamma\_k=0\Rightarrow k\_i=0, z\_k=0\\\\ \Rightarrow&\sum\_i x\_i\alpha\_i+\sum\_j y\_j\beta\_j=0\Rightarrow x\_i=0, y\_j=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \alpha\_1,\cdots,\alpha\_r, \beta\_1,\cdots,\beta\_\{m-r\}, \gamma\_1,\cdots,\gamma\_\{n-r\}$ 线性无关, 是 $\displaystyle V\_1+V\_2$ 的一组基. 于是
\begin\{aligned\} \dim(V\_1+V\_2)+\dim(V\_1\cap V\_2)=&\left\[r+(m-r)+(n-r)\right\]+r\\\\ &=m+n=\dim V\_1+\dim V\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 将 $\displaystyle \alpha\_1=\alpha$ 扩充为 $\displaystyle V$ 的一组基 $\displaystyle \alpha\_1,\cdots,\alpha\_n$, 对 $\displaystyle \forall\ \beta\in V$, 取 $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{T\}$ 使得
\begin\{aligned\} \mathscr\{T\}\alpha\_1=\beta, \mathscr\{T\}\alpha\_i=0\left(2\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\}
则 $\displaystyle \beta=\mathscr\{T\}\alpha\_1=\mathscr\{T\}\alpha\in W$. 这就证明了 $\displaystyle V\subset W$. 按定义, $\displaystyle W\subset V$. 故 $\displaystyle W=V$.跟锦数学微信公众号. [在线资料](
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---
1468、 5、 设 $\displaystyle \mathbb\{F\}$ 为数域, $\displaystyle C(M)$ 表示数域 $\displaystyle \mathbb\{F\}$ 上所有与数域 $\displaystyle \mathbb\{F\}$ 上的方阵 $\displaystyle M$ 可交换的矩阵组成的集合. 设
\begin\{aligned\} f(x)=(x-1)^2(x+2)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 证明: $\displaystyle C(M)$ 关于矩阵加法和数乘是 $\displaystyle \mathbb\{F\}$ 上的一个线性空间; (2)、 以 $\displaystyle f(x)$ 为特征多项式的 $\displaystyle \mathbb\{F\}$ 上的所有矩阵在相似意义下可以分几类? 并针对每一类写出一个具体的方阵. (3)、 设 $\displaystyle A$ 是 $\displaystyle \mathbb\{F\}$ 上的方阵, 且特征多项式为 $\displaystyle f(x)$, 求 $\displaystyle \dim C(A)$; (4)、 设 $\displaystyle B$ 是 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle 3$ 阶方阵且满足 $\displaystyle f(B)=0$, 求 $\displaystyle \dim C(B)$. (四川大学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\} &B,C\in C(M)\Rightarrow MB=BM, MC=CM\\\\ \Rightarrow& M(kB+lC)=kMB+lMC=kBM+lCM=(kB+lC)M\\\\ \Rightarrow& kB+lC\in C(M) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle C(M)$ 是复线性空间 $\displaystyle M\_n(\mathbb\{C\})$ 的线性子空间. (2)、 $\displaystyle 4$ 类. 它们的 Jordan 标准形 $\displaystyle J$ 只有以下四种:
\begin\{aligned\} &J\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&&\\\\ &&-2&\\\\ &&&-2\end\{array\}\right), J\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &1&&\\\\ &&-2&\\\\ &&&-2\end\{array\}\right),\\\\ &J\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&&\\\\ &&-2&1\\\\ &&&-2\end\{array\}\right), J\_4=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &1&&\\\\ &&-2&1\\\\ &&&-2\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 由于与 $\displaystyle A$ 可交换的矩阵全体与 $\displaystyle J$ 可交换的矩阵全体线性同构, 我们知
\begin\{aligned\} \dim C(A)=\dim C(M)=\left\\{\begin\{array\}\{llllllllllll\}4+4=8,&J=J\_1,\\\\ 2+4=6,&J=J\_2,\\\\ 4+2=6,&J=J\_3,\\\\ 2+2=4,&J=J\_4.\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\}
(4)、 设 $\displaystyle \left\\{\lambda,\mu\right\\}=\left\\{1,-2\right\\}$, 则 $\displaystyle B$ 的 Jordan 标准形 $\displaystyle J$ 可能为
\begin\{aligned\} &J\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda&&\\\\ &\lambda&\\\\ &&\mu\end\{array\}\right)\Rightarrow \dim C(B)=4+1=5,\\\\ &J\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda&1&\\\\ &\lambda&\\\\ &&\mu\end\{array\}\right)\Rightarrow \dim C(B)=2+1=3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1469、 3、 设
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&1&-1&1\\\\ 2&1&-1&2&-3\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}3&-2&-1&1&-2\\\\ 2&-5&1&-2&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\}
(1)、 再设
\begin\{aligned\} V=\left\\{x\in\mathbb\{P\}^5; Ax=0\right\\}, W=\left\\{x\in\mathbb\{P\}^5; Bx=0\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle V\cap W$ 的一组基; (2)、 证明
\begin\{aligned\} V=\left\\{y\in\mathbb\{P\}^4; y=\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)x, x\in\mathbb\{P\}^5\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是 $\displaystyle \mathbb\{P\}^4$ 的一组基, 求 $\displaystyle V$ 的一组基与维数. (太原理工大学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\} \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&\frac\{1\}\{2\}&-\frac\{7\}\{8\}\\\\ 0&1&0&\frac\{1\}\{2\}&-\frac\{5\}\{8\}\\\\ 0&0&1&-\frac\{1\}\{2\}&\frac\{5\}\{8\}\\\\ 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\}
知取 $\displaystyle x\_4,x\_5$ 为自由变量后, $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)x=0$ 的基础解系为
\begin\{aligned\} \eta\_1=(-1,-1,1,2,0)^\mathrm\{T\}, \eta\_2=(7,5,-5,0,8)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由
\begin\{aligned\} x\in V\cap W&\Leftrightarrow Ax=0, Bx=0\Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)x=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \eta\_1,\eta\_2$ 就是 $\displaystyle V\cap W$ 的一组基. (2)、 $\displaystyle V=\mathrm\{im\} \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 是 $\displaystyle \mathbb\{P\}^4$ 的线性子空间. 由 $\displaystyle (I$) 知 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 的前三列
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\3\\\\2\end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\-2\\\\-5\end\{array\}\right), \xi\_3= \left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\1\\\\-2\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
就是 $\displaystyle V=\mathrm\{im\} \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 的一组基, $\displaystyle \dim V=3$, 且 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 的后两列可由 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 线性表示为 [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\} \frac\{1\}\{2\}(\xi\_1+\xi\_2-\xi\_3), \frac\{1\}\{8\}(-7\xi\_1-5\xi\_2+5\xi\_3). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1470、 4、 设 $\displaystyle A\in\mathbb\{P\}^\{n\times n\}$, $\displaystyle E$ 为 $\displaystyle n$ 阶单位矩阵,
\begin\{aligned\} V\_1=\left\\{x\in\mathbb\{P\}^n; (A-E)x=0\right\\}, V\_2=\left\\{x\in \mathbb\{P\}^n; (A+E)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\}
证明: (1)、 $\displaystyle A^2=E\Leftrightarrow \mathrm\{rank\}(A+E)+\mathrm\{rank\}(A-E)=n$; (2)、 $\displaystyle \mathbb\{P\}^n=V\_1\oplus V\_2\Leftrightarrow A^2=E$. (太原理工大学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\} &\left(\begin\{array\}\{cccccccccccccccccccc\}A+E&0\\\\ 0&A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A+E&A-E\\\\ 0&A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}2E&A-E\\\\ E-A&A-E\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}2E&0\\\\ E-A&(A-E)+\frac\{1\}\{2\}(E-A)(E-A)\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&A^2-E\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\{rank\}(A+E)+\mathrm\{rank\}(A-E)=&\mathrm\{rank\}(E\_n)+\mathrm\{rank\}(A^2-E)\\\\ =&n+\mathrm\{rank\}(A^2-E). \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\} A^2=E\Leftrightarrow \mathrm\{rank\}(A^2-E)=0\Leftrightarrow \mathrm\{rank\}(A+E)+\mathrm\{rank\}(A-E)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 (2-1)、 $\displaystyle \Leftarrow$: 由
\begin\{aligned\} x\in V\_1\cap V\_2\Rightarrow x=Ax=-x\Rightarrow x=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_1\cap V\_2=\left\\{0\right\\}$. 又由
\begin\{aligned\} &\dim(V\_1+V\_2) =\dim V\_1+\dim V\_2-\dim(V\_1\cap V\_2)\\\\ =&\left\[n-\mathrm\{rank\}(A-E)\right\]+\left\[n-\mathrm\{rank\}(A+E)\right\]\overset\{\tiny\mbox\{第1步\}\}\{=\} 2n-n=n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle V\_1+V\_2=\mathbb\{P\}^n$. 联合 $\displaystyle V\_1\cap V\_2=\left\\{0\right\\}$ 知 $\displaystyle \mathbb\{P\}^n=V\_1\oplus V\_2$. (2-2)、 $\displaystyle \Rightarrow$: 由 $\displaystyle \mathbb\{P\}^n=V\_1\oplus V\_2$ 知
\begin\{aligned\} n=\dim V\_1+\dim V\_2=\left\[n-\mathrm\{rank\}(A-E)\right\]+\left\[n-\mathrm\{rank\}(A+E)\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\}(A+E)+\mathrm\{rank\}(A-E)=n$. 由第 1 步即知 $\displaystyle A^2=E$.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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---
1471、 5、 设 $\displaystyle V$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle 4$ 维线性空间, $\displaystyle \varepsilon\_1,\varepsilon\_2,\varepsilon\_3,\varepsilon\_4$ 是 $\displaystyle V$ 的一组基, $\displaystyle V$ 上的线性变换 $\displaystyle \mathscr\{A\}$ 满足
\begin\{aligned\} \mathscr\{A\}(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3,\varepsilon\_4)=(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3,\varepsilon\_4)A, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1&2\\\\ 0&1&0&0\\\\ 2&-3&1&-1\\\\ 1&4&-2&-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)、 证明: $\displaystyle L(\varepsilon\_1,\varepsilon\_3,\varepsilon\_4)$ 为 $\displaystyle \mathscr\{A\}$ 子空间; (2)、 设 $\displaystyle W$ 是包含 $\displaystyle \varepsilon\_1$ 的 $\displaystyle \mathscr\{A\}$ 子空间, 证明: $\displaystyle L(\varepsilon\_1,\varepsilon\_3,\varepsilon\_4)\subset W$. (太原理工大学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,
\begin\{aligned\} \mathscr\{A\}\varepsilon\_1=\varepsilon\_1+2\varepsilon\_3+\varepsilon\_4, \mathscr\{A\}\varepsilon\_3=-\varepsilon\_1+\varepsilon\_3-2\varepsilon\_4, \mathscr\{A\}\varepsilon\_4=2\varepsilon\_1-\varepsilon\_3-\varepsilon\_4 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle L(\varepsilon\_1,\varepsilon\_3,\varepsilon\_4)$ 为 $\displaystyle \mathscr\{A\}$ 子空间. (2)、 由
\begin\{aligned\} A^2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&9&-6&1\\\\ 0&1&0&0\\\\ 3&-12&1&4\\\\ -4&5&-1&5\end\{array\}\right), A^3=\left(\begin\{array\}\{cccccccccccccccccccc\}-10&30&-9&7\\\\ 0&1&0&0\\\\ 9&-2&-10&1\\\\ -1&32&-7&-12\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\}
及题设知
\begin\{aligned\} \mathscr\{A\}\varepsilon\_1=\varepsilon\_1+2\varepsilon\_3+\varepsilon\_4, \mathscr\{A\}^2\varepsilon\_1=\varepsilon\_1+3\varepsilon\_3-4\varepsilon\_4, \mathscr\{A\}^3\varepsilon\_1=-10\varepsilon\_1+9\varepsilon\_3-\varepsilon\_4. \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\}\varepsilon\_1,\mathscr\{A\}^2\varepsilon\_1,\mathscr\{A\}^3\varepsilon\_1)=(\varepsilon\_1,\varepsilon\_3,\varepsilon\_4)B, B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-10\\\\ 2&3&9\\\\ 1&-4&-1\end\{array\}\right), |B|=154,\\\\ \Rightarrow&(\varepsilon\_1,\varepsilon\_3,\varepsilon\_4)=(\mathscr\{A\}\varepsilon\_1,\mathscr\{A\}^2\varepsilon\_1,\mathscr\{A\}^3\varepsilon\_1)B^\{-1\}. \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 $\displaystyle \varepsilon\_1,\varepsilon\_3,\varepsilon\_4$ 可由 $\displaystyle \mathscr\{A\}\varepsilon\_1,\mathscr\{A\}^2\varepsilon\_1,\mathscr\{A\}^3\varepsilon\_1\in W$ 线性表出. 故 $\displaystyle L(\varepsilon\_1,\varepsilon\_3,\varepsilon\_4)\subset W$.跟锦数学微信公众号. [在线资料](
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---
1472、 5、 已知向量组满足
\begin\{aligned\} \left\\{\begin\{array\}\{rrrrrrrrrrrrrrrr\} \beta\_1= \alpha\_2+\alpha\_3+\cdots+\alpha\_m,\\\\ \beta\_2=\alpha\_1+\alpha\_3+\cdots+\alpha\_m,\\\\ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots,\\\\ \beta\_m=\alpha\_1+\alpha\_2+\cdots+\alpha\_\{m-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\}
求证: $\displaystyle \beta\_1,\cdots,\beta\_m$ 与 $\displaystyle \alpha\_1,\cdots,\alpha\_m$ 的秩相同. (天津大学2023年高等代数考研试题) [线性空间与线性变换 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&\cdots&1\\\\ 1&0&\cdots&1\\\\ \vdots&\vdots&&\vdots\\\\ 1&1&\cdots&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\}
则只要证明 $\displaystyle |A|\neq 0$, 则 $\displaystyle \beta\_1,\cdots,\beta\_m$ 与 $\displaystyle \alpha\_1,\cdots,\alpha\_m$ 可相互线性表出, 而等价, 具有相同的秩. (2)、 为此, 我们先求一般的
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}x&y&\cdots&y&y\\\\ y&x&\cdots&y&y\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ y&y&\cdots&x&y\\\\ y&y&\cdots&y&x\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\}
事实上, 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&E+BA\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E+AB&0\\\\ 0&E\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\}
即知
\begin\{aligned\} |E+BA|=|E+AB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle x\neq y$ 时,
\begin\{aligned\} |A|=&|(x-y)E+yee^\mathrm\{T\}|\left(e=(1,\cdots,1)^\mathrm\{T\}\right)\\\\ =&|(x-y)E|\cdot \left|E+\frac\{y\}\{x-y\}ee^\mathrm\{T\}\right| = (x-y)^n \left(1+\frac\{y\}\{x-y\}e^\mathrm\{T\} e\right)\\\\ =&(x-y)^n \left(1+\frac\{ny\}\{x-y\}\right) =(x-y)^n \left\[x+(n-1)y\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上式两端都是关于 $\displaystyle x,y$ 的连续函数, 在 $\displaystyle x\neq y$ 时相等而恒等. (3)、 由第 2 步知 $\displaystyle |A|=-(n-1)\neq 0$. 而结论确实成立.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 13:23:57
