编辑
2023-11-04
数学
00
请注意,本文编写于 161 天前,最后修改于 154 天前,其中某些信息可能已经过时。

目录

题目1
题目2
题目3

题目1

给定线性空间R3\mathbb{R}^{3}的两组基

ε1=(1,0,1),ε2=(2,1,0),ε3=(1,1,1)\varepsilon_{1}=(1,0,1)^{\top},\varepsilon_{2}=(2,1,0)^{\top},\varepsilon_{3}=(1,1,1)^{\top}
η1=(1,2,1),η2=(2,2,1),η3=(2,1,1)\eta_{1}=(1,2,-1)^{\top},\eta_{2}=(2,2,-1)^{\top},\eta_{3}=(2,-1,-1)^{\top}

定义R3\mathbb{R}^{3}的线性变换A\mathscr{A}A(εi)=ηi(i=1,2,3)\mathscr{A}(\varepsilon_i)=\eta_i(i=1,2,3) 试求:

(1)从基ε1,ε2,ε3\varepsilon_1,\varepsilon_2,\varepsilon_3到基η1,η2,η3\eta_1,\eta_2,\eta_3的过渡矩阵;

(2)A\mathscr{A}在基ε1,ε2,ε3\varepsilon_1,\varepsilon_2,\varepsilon_3下的矩阵;

(3)A\mathscr{A}在基η1,η2,η3\eta_1,\eta_2,\eta_3下的矩阵

解:

(1) 从基ε1,ε2,ε3\varepsilon_1,\varepsilon_2,\varepsilon_3到基η1,η2,η3\eta_1,\eta_2,\eta_3的过渡矩阵为

P=(ε1,ε2,ε3)1(η1,η2,η3)=(211011101)1(122221111)=12(433233215)P=(\varepsilon_1,\varepsilon_2,\varepsilon_3)^{-1}(\eta_1,\eta_2,\eta_3)= \begin{pmatrix} -2&1&1\\0&1&1\\1&0&1\end{pmatrix}^{-1}\begin{pmatrix}1&2&2\\2&2&-1\\-1&-1&-1\end{pmatrix}=\frac{1}{2}\begin{pmatrix}-4&-3&3\\2&3&3\\2&1&-5\end{pmatrix}

(2) A(ε1,ε2,ε3)=(η1,η2,η3)=(ε1,ε2,ε3)P\mathscr{A}(\varepsilon_1,\varepsilon_2,\varepsilon_3)=(\eta_1,\eta_2,\eta_3)=(\varepsilon_1,\varepsilon_2,\varepsilon_3)P,故A\mathscr{A}在基ε1,ε2,ε3\varepsilon_1,\varepsilon_2,\varepsilon_3下的矩阵为PP

(3) A(η1,η2,η3)=A((ε1,ε2,ε3)P)=(η1,η2,η3)P\mathscr{A}(\eta_1,\eta_2,\eta_3)=\mathscr{A}((\varepsilon_1,\varepsilon_2,\varepsilon_3)P)=(\eta_1,\eta_2,\eta_3)P,故A\mathscr{A}在基η1,η2,η3\eta_1,\eta_2,\eta_3下的矩阵为PP

题目2

ε1,ε2,ε3,ε4\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon_4是4维线性空间VV的一组基,线性变换A\mathscr{A}在这组基下的矩阵为

(1021121312552212)\left(\begin{matrix}1&0&2&1\\-1&2&1&3\\1&2&5&5\\2&-2&1&-2\end{matrix}\right)

(1)求A\mathscr{A}在基η1=ε12ε2+ε4,η2=3ε2ε3ε4,η3=ε3+ε4,η4=2ε4\eta_1=\varepsilon_1-2\varepsilon_2+\varepsilon_4,\eta_2=3\varepsilon_2-\varepsilon_3-\varepsilon_4,\eta_3=\varepsilon_3+\varepsilon_4,\eta_4=2\varepsilon_4下的矩阵;

(2)求A\mathscr{A}的值域与核

解:

(1) 从ε1,ε2,ε3,ε4\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon_4η1,η2,η3,η4\eta_1,\eta_2,\eta_3,\eta_4的过渡矩阵PP

P=(1000230001101112)P=\begin{pmatrix}1&0&0&0\\-2&3&0&0\\0&-1&1&0\\1&-1&1&2\end{pmatrix}

因此,A\mathscr{A}η1,η2,η3,η4\eta_1,\eta_2,\eta_3,\eta_4的矩阵为

A=P1AP=(1000230001101112)1(1021121312552212)(1000230001101112)=(23322343103103831634034030178)A'=P^{-1}AP=\begin{pmatrix}1&0&0&0\\-2&3&0&0\\0&-1&1&0\\1&-1&1&2\end{pmatrix}^{-1}\begin{pmatrix}1&0&2&1\\-1&2&1&3\\1&2&5&5\\2&-2&1&-2\end{pmatrix}\begin{pmatrix}1&0&0&0\\-2&3&0&0\\0&-1&1&0\\1&-1&1&2\end{pmatrix}=\begin{pmatrix}2&-3&3&2\\\frac23&-\frac43&\frac{10}3&\frac{10}3\\\frac83&-\frac{16}3&\frac{40}3&\frac{40}3\\0&1&-7&-8\end{pmatrix}

(2) 对AA行化简得到

A=(1021121312552212)(1021023402340234)(10210132200000000)A=\begin{pmatrix}1&0&2&1\\-1&2&1&3\\1&2&5&5\\2&-2&1&-2\end{pmatrix}\sim\begin{pmatrix}1&0&2&1\\0&2&3&4\\0&2&3&4\\0&-2&-3&-4\end{pmatrix}\sim\begin{pmatrix}1&0&2&1\\0&1&\frac{3}{2}&2\\0&0&0&0\\0&0&0&0\end{pmatrix}

AA的主元列为1、2列,因此,R(A)=span(ε1,ε2)R(\mathscr{A})=\operatorname{span}(\varepsilon_1,\varepsilon_2)

任取k1ε1+k2ε2+k3ε3+k4ε4Ker(A)k_1\varepsilon_1+k_2\varepsilon_2+k_3\varepsilon_3+k_4\varepsilon_4\in\operatorname{Ker}(\mathscr{A})

也即

(10210132200000000)(k1k2k3k4)=0\begin{pmatrix}1&0&2&1\\0&1&\frac{3}{2}&2\\0&0&0&0\\0&0&0&0\end{pmatrix}\begin{pmatrix}k_1\\k_2\\k_3\\k_4\end{pmatrix}=0

解得

{k1=2k3k4k2=32k32k4k3是自由变量k4是自由变量\begin{cases} k_1 = -2k_3 -k_4\\ k_2 = -\frac{3}{2}k_3-2k_4\\ k_3 \text{是自由变量}\\ k_4 \text{是自由变量}\\ \end{cases}

也即

Ker(A)=(2k3k4)ε1+(32k32k4)ε2+k3ε3+k4ε4=span(2ε132ε2+ε3,ε1+2ε2+ε4)\operatorname{Ker}(\mathscr{A})=(-2k_3 -k_4)\varepsilon_1+(-\frac{3}{2}k_3-2k_4)\varepsilon_2+k_3\varepsilon_3+k_4\varepsilon_4=\operatorname{span}(-2\varepsilon_1-\frac{3}{2}\varepsilon_2+\varepsilon_3,-\varepsilon_1+2\varepsilon_2+\varepsilon_4)

题目3

nn维线性空间R[x]n\mathbb{R}[x]_n中,定义线性变换D(f(x))=f(x)\mathscr{D}(f(x))=f^{\prime}(x),其中f(x)R[x]nf(x)\in \mathbb{R}[x]_n. 求D\mathscr{D}的值域与核。

解:

任取 f(x)=a0+a1x+a2x2++an1xn1R[x]nf(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n-1}x^{n-1}\in \mathbb{R}[x]_n, 则

D(f(x))=f(x)=a1+2a2x++(n1)an1xn2\mathscr{D}(f(x))=f^{\prime}(x)=a_{1}+2a_{2}x+\cdots+(n-1)a_{n-1}x^{n-2}

a0,a1,a2,,an1a_0,a_1,a_2,\cdots,a_{n-1}的任意性,R(D)=R[x]n1R(\mathscr{D})=\mathbb{R}[x]_{n-1}.

D(f(x))=0\mathscr{D}(f(x))=0 当且仅当a1=a2==an1=0a_{1}=a_{2}=\cdots=a_{n-1}=0,故Ker(D)=R\operatorname{Ker}(\mathscr{D})=\mathbb{R}

题目4

A,BA,Bnn阶矩阵,并且AB=BAAB=BA。证明:

(1)如果AAnn个互异的特征值,则BB相似于对角矩阵。

(2)如果AABB均为可对角化矩阵,则存在可逆矩阵PP使得P1APP^{-1}APP1BPP^{-1}BP同时为对角矩阵

证明:

(1)如果AAnn个互异的特征值λ1,,λn\lambda_{1},\cdots,\lambda_{n},则存在可逆矩阵PP使得

A=P(λ1λn)P1A=P\left(\begin{matrix}\lambda_{1}&&\\&\ddots&\\&&\lambda_{n}\end{matrix}\right)P^{-1}

也即

AB=P(λ1λn)P1B=BA=BP(λ1λn)P1AB=P\left(\begin{matrix}\lambda_{1}&&\\&\ddots&\\&&\lambda_{n}\end{matrix}\right)P^{-1}B=BA=BP\left(\begin{matrix}\lambda_{1}&&\\&\ddots&\\&&\lambda_{n}\end{matrix}\right)P^{-1}

等式两边同时左乘P1P^{-1}右乘PP,则

(λ1λn)P1BP=P1BP(λ1λn)\left(\begin{matrix}\lambda_{1}&&\\&\ddots&\\&&\lambda_{n}\end{matrix}\right)P^{-1}BP=P^{-1}BP\left(\begin{matrix}\lambda_{1}&&\\&\ddots&\\&&\lambda_{n}\end{matrix}\right)

因为对角线元素互异的对角阵的可交换矩阵一定是对角阵,P1BPP^{-1}BP为对角阵,也即BB相似于对角阵

(2) 该命题为高等代数中的“同时相似对角化问题”, 详见https://mp.weixin.qq.com/s?__biz=MzU4MTg0NjEzOQ%3D%3D&mid=2247492990&idx=2&sn=da46d2701c4c7d24c83d555a3648781d&chksm=fd43f06aca34797ccad03d57e091bf5f60b3a3997a356c85a7fd7bd091bb12c98232b202d664&scene=27 的例4

CleanShot 2023-11-04 at 20.05.07@2x.png

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