南京师范大学2021年硕士研究生入学考试高等代数试卷及参考答案

1.(20分)计算行列式.

(1) 计算4阶行列式∣2−4−35−314−272534−3−26∣.\begin{vmatrix} 2 & -4 & -3 & 5\\ -3 & 1 & 4 & -2 \\ 7 & 2 & 5 & 3 \\ 4 & -3 & -2 & 6\end{vmatrix}.∣∣∣∣∣∣∣∣2−374−412−3−345−25−236∣∣∣∣∣∣∣∣.

参考答案：

原式
=∣2−4−35−314−272534−3−26∣=−∣−314−22−4−3572534−3−26∣=∣1−34−2−42−352753−34−26∣=∣1−34−20−1013−3013−370−5100∣=∣−1013−313−37−5100∣=∣−10−7−313237−500∣=(−5)(−1)1+3∣−7−3237∣=−5(−49+69)=−100.\begin{aligned} &=\begin{vmatrix}2 & -4 & -3 & 5\\ -3 & 1 & 4 & -2 \\ 7 & 2 & 5 & 3 \\ 4 & -3 & -2 & 6\end{vmatrix} \\&=-\begin{vmatrix}-3 & 1 & 4 & -2\\2 & -4 & -3 & 5\\7 & 2 & 5 & 3 \\ 4 & -3 & -2 & 6\end{vmatrix} \\&=\begin{vmatrix}1 & -3 & 4 & -2 \\-4 & 2 & -3 & 5\\2 & 7 & 5 & 3 \\ -3 & 4 & -2 & 6\end{vmatrix} \\&=\begin{vmatrix} 1 & -3 & 4 & -2 \\0 & -10 & 13 & -3\\0 & 13 & -3 & 7 \\ 0 & -5 & 10 & 0\end{vmatrix} \\&=\begin{vmatrix}-10 & 13 & -3\\13 & -3 & 7 \\ -5 & 10 & 0\end{vmatrix} \\&=\begin{vmatrix} -10 & -7 & -3\\13 & 23 & 7\\ -5 & 0 & 0\end{vmatrix}\\&=(-5)(-1)^{1+3}\begin{vmatrix}-7 & -3\\ 23 & 7 \end{vmatrix} \\&=-5(-49+69) \\&=-100.\end{aligned}=∣∣∣∣∣∣∣∣2−374−412−3−345−25−236∣∣∣∣∣∣∣∣=−∣∣∣∣∣∣∣∣−32741−42−34−35−2−2536∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣1−42−3−32744−35−2−2536∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣1000−3−1013−5413−310−2−370∣∣∣∣∣∣∣∣=∣∣∣∣∣∣−1013−513−310−370∣∣∣∣∣∣=∣∣∣∣∣∣−1013−5−7230−370∣∣∣∣∣∣=(−5)(−1)1+3∣∣∣∣−723−37∣∣∣∣=−5(−49+69)=−100.

(2)计算n阶行列式∣2aa20⋯0012aa2⋯00012a⋯00⋮⋮⋮⋮⋮000⋯2aa2000⋯12a∣.\begin{vmatrix} 2a & a^2 & 0&\cdots & 0& 0 \\ 1 & 2a & a^2& \cdots & 0& 0 \\ 0& 1 & 2a & \cdots & 0& 0 \\ \vdots & \vdots & \vdots & & \vdots &\vdots \\0& 0& 0 & \cdots & 2a & a^2 \\ 0& 0 & 0 & \cdots & 1& 2a \end{vmatrix} .∣∣∣∣∣∣∣∣∣∣∣∣∣2a10⋮00a22a1⋮000a22a⋮00⋯⋯⋯⋯⋯000⋮2a1000⋮a22a∣∣∣∣∣∣∣∣∣∣∣∣∣.

参考答案：

D1=∣2a∣=2a;D_1=|2a|=2a;D1=∣2a∣=2a;

D2=∣2aa212a∣=4a2−a2=3a2;D_2=\begin{vmatrix} 2a & a^2 \\ 1& 2a \end{vmatrix} =4a^2-a^2=3a^2;D2=∣∣∣∣2a1a22a∣∣∣∣=4a2−a2=3a2;

Dn=2aDn−1+1⋅(−1)2+1⋅∣a200⋯00012aa2⋯000⋮⋮⋮⋮⋮⋮000⋯12aa2000⋯012a∣=2aDn−1−a2Dn−2,\begin{aligned} D_n&=2aD_{n-1}+\\& 1\cdot(-1)^{2+1}\cdot \begin{vmatrix} a^2 & 0 & 0&\cdots & 0& 0& 0 \\ 1 & 2a & a^2& \cdots& 0 & 0& 0 \\ \vdots & \vdots & \vdots & & \vdots &\vdots&\vdots \\0& 0& 0 & \cdots & 1 & 2a & a^2 \\ 0& 0 & 0 & \cdots & 0& 1& 2a \end{vmatrix} \\&=2aD_{n-1}-a^2D_{n-2},\end{aligned}Dn=2aDn−1+1⋅(−1)2+1⋅∣∣∣∣∣∣∣∣∣∣∣a21⋮0002a⋮000a2⋮00⋯⋯⋯⋯00⋮1000⋮2a100⋮a22a∣∣∣∣∣∣∣∣∣∣∣=2aDn−1−a2Dn−2,

于是Dn−aDn−1=a(Dn−1−aDn−2)=⋯=an−2(D2−aD1)=an,\begin{aligned}D_n-aD_{n-1}&=a(D_{n-1}-aD_{n-2})\\&=\cdots=a^{n-2}(D_2-aD_1)=a^n,\end{aligned}Dn−aDn−1=a(Dn−1−aDn−2)=⋯=an−2(D2−aD1)=an,

两边同时除以an,a^n,an,得Dnan−Dn−1an−1=1\frac{D_n}{a^n}-\frac{D_{n-1}}{a^{n-1}}=1anDn−an−1Dn−1=1

从而{Dnan}\{\frac{D_n}{a^n}\}{anDn}是以2为首项，1为公差的等差数列，

即Dnan=2+(n−1)=n+1,Dn=(n+1)an,\frac{D_n}{a^n}=2+(n-1)=n+1,D_n=(n+1)a^n,anDn=2+(n−1)=n+1,Dn=(n+1)an,n∈N∗.n\in N^*.n∈N∗.

2.(20分)记g(x)=∑i=0n−1xi(n>1),g(x)=\sum\limits_{i=0}^{n-1} x^i(n>1),g(x)=i=0∑n−1xi(n>1),证明g(x)g(x)g(x)在有理数域QQQ上不可约的充要条件是nnn为素数.

参考答案：

充分性：nnn为素数，(x−1)g(x)=xn−1.(x-1)g(x)=x^n-1.(x−1)g(x)=xn−1.

令x=y+1,x=y+1,x=y+1,得yg(y+1)=(y+1)n−1=yn+nyn−1+⋯+Cnkyn−k+⋯+ny.\begin{aligned}yg(y+1)&=(y+1)^n-1\\&=y^n+ny^{n-1}+\cdots+C_n^ky^{n-k}\\&+\cdots+ny.\end{aligned}yg(y+1)=(y+1)n−1=yn+nyn−1+⋯+Cnkyn−k+⋯+ny.

g(x)=g(y+1)=yn−1+nyn−2+⋯+Cnkyn−k−1+⋯+n.\begin{aligned}g(x)&=g(y+1)\\&=y^{n-1}+ny^{n-2}+\cdots+C_n^ky^{n-k-1}\\&+\cdots+n.\end{aligned}g(x)=g(y+1)=yn−1+nyn−2+⋯+Cnkyn−k−1+⋯+n.

Cnk=n(n−1)⋯(n−k+1)k!,1≤k<n.C_n^k=\frac{n(n-1)\cdots (n-k+1)}{k!} ,1 \le k<n.Cnk=k!n(n−1)⋯(n−k+1),1≤k<n.

由于(n,k!)=1,(n,k!)= 1, (n,k!)=1,

因此k!∣(n−1)⋯(n−k+1).k!|(n-1)\cdots(n-k+1).k!∣(n−1)⋯(n−k+1).

从而n∣Cnk,1≤k<n.n|C_n^k,1\le k< n . n∣Cnk,1≤k<n.

又n∤1,n2∤n,n\nmid 1,n^2\nmid n,n∤1,n2∤n,

因此g(y+1)g(y+1)g(y+1)在有理数域QQQ上不可约，从而g(x)g(x)g(x)在有理数域QQQ上不可约。

必要性：(反证法)若nnn不是素数，因此n=n1n2,n=n_1n_2,n=n1n2,其中 $ 0<n_i<n,i=1,2.$

于是
g(x)=(1+x+⋯+2n1−1)+(xn1+xn1+1+⋯+x2n1−1)+(x2n1+x2n1+1+⋯+x3n1−1)+⋯+(x(n2−1)n1+x(n2−1)n1+1+⋯+xn2n1−1)=(1+x+⋯+2n1−1)⋅(1+xn1+x2n1+⋯+x(n2−1)n1).\begin{aligned} g(x)=&(1+x+\cdots+2^{n_1-1})\\&+(x^{n_1} +x^{n_1+1}+\cdots+x^{2n_1-1})\\&+(x^{2n_1}+x^{2n_1+1}+\cdots+x^{3n_1-1})\\&+\cdots\\&+(x^{(n_2-1)n_1}+x^{(n_2-1)n_1+1}\\&+\cdots+x^{n_2n_1-1}) \\&=(1+x+\cdots+2^{n_1-1})\cdot\\&(1+x^{n_1}+x^{2n_1} +\cdots+x^{(n_2-1)n_1}). \end{aligned}g(x)=(1+x+⋯+2n1−1)+(xn1+xn1+1+⋯+x2n1−1)+(x2n1+x2n1+1+⋯+x3n1−1)+⋯+(x(n2−1)n1+x(n2−1)n1+1+⋯+xn2n1−1)=(1+x+⋯+2n1−1)⋅(1+xn1+x2n1+⋯+x(n2−1)n1).
因此g(x)g(x)g(x)在QQQ上可约.

3.(15分)证明：如果AAA为n(n≥2)n(n\ge2)n(n≥2)阶矩阵，则
r(A∗)={n,r(A)=n;1,r(A)=n−1;0,r(A)<n−1.r(A^*)= \left\{ \begin{aligned} n&,r(A)=n;\\ 1&,r(A)=n-1;\\ 0&,r(A)< n-1.\\ \end{aligned} \right.r(A∗)=⎩⎪⎨⎪⎧n10,r(A)=n;,r(A)=n−1;,r(A)<n−1.

参考答案：

若rank(A)=n,rank(A)=n,rank(A)=n,则∣A∣≠0.|A|\ne0.∣A∣=0.由于AA∗=∣A∣EAA^*=|A|EAA∗=∣A∣E
因此∣A∣∣A∗∣=∣A∣n≠0.|A||A^*|=|A|^n\ne0.∣A∣∣A∗∣=∣A∣n=0.从而∣A∗∣≠0.|A^*|\ne0.∣A∗∣=0.于是rank(A∗)=n.rank(A^*)=n.rank(A∗)=n.

若rank(A)=n−1,rank(A)=n-1,rank(A)=n−1,则AAA至少有一个n−1n-1n−1阶子式不等于0，从而AAA有一个元素的代数余子式不等于0.于是A∗≠0.A^*\ne0.A∗=0.由于∣A∣=0,|A|=0,∣A∣=0,因此AA∗=∣A∣E=0.AA^*=|A|E=0.AA∗=∣A∣E=0. 从而rank(A)+rank(A∗)≤n.rank(A) + rank(A^*)\le n.rank(A)+rank(A∗)≤n.于是rank(A∗)≤n−rank(A)=n−(n−1)=1.rank(A^*)\le n- rank(A)=n-(n-1)=1.rank(A∗)≤n−rank(A)=n−(n−1)=1.
由于A∗≠0,A^*\ne0,A∗=0,因此rank(A∗)=1.rank(A^*)=1.rank(A∗)=1.

若rank(A)<n−1,rank(A)< n-1,rank(A)<n−1,则AAA的所有n−1n-1n−1阶子式都等于0,从而A∗=0.A^*=0.A∗=0.于是rank(A∗)=0.rank(A^*)=0.rank(A∗)=0.

4.(20分)线性方程组
{a11x1+a12x2+⋯+a1nxn=0;a21x1+a22x2+⋯+a2nxn=0;⋯⋯an−1,1x1+an−2,2x2+⋯+an−1,nxn=0.\left\{ \begin{aligned} &a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n=0 ;\\ &a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=0 ;\\ &\cdots\cdots\\ &a_{n-1,1}x_1+a_{n-2,2}x_2+\cdots+a_{n-1,n}x_n=0.\\ \end{aligned} \right.⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧a11x1+a12x2+⋯+a1nxn=0;a21x1+a22x2+⋯+a2nxn=0;⋯⋯an−1,1x1+an−2,2x2+⋯+an−1,nxn=0.
的系数矩阵为AAA，设MiM_iMi是矩阵AAA划掉第iii列剩下的n−1n-1n−1阶矩阵的行列式，证明：

(1)(M1,−M2,⋯,(−1)n−1Mn)′(M_1,-M_2,\cdots,(-1)^{n-1}M_n)'(M1,−M2,⋯,(−1)n−1Mn)′是方程组的一个解；

参考答案：

作nnn级行列式 DDD，它是AAA的第iii行元素与AAA的各行依次排列依次排列成的行列式，即
∣ai1ai2⋯aina11a12⋯a1na21a22⋯a2n⋮⋮⋮ai1ai2⋯ain⋮⋮⋮an−1,1an−1,2⋯ain−1,n∣\begin{vmatrix} a_{i1} & a_{i2} &\cdots & a_{in} \\ a_{11} & a_{12} &\cdots & a_{1n} \\ a_{21} & a_{22} &\cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} &\cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n-1,1} & a_{n-1,2} &\cdots & a_{in-1,n} \end{vmatrix} ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ai1a11a21⋮ai1⋮an−1,1ai2a12a22⋮ai2⋮an−1,2⋯⋯⋯⋯⋯aina1na2n⋮ain⋮ain−1,n∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

由于DDD有两行相同，所以D=0D=0D=0.将DDD按第一行展开，得
D=ai1M1−ai2M2+⋯+(−1)n−1ainMn=0D=a_{i1}M_1-a_{i2}M_2+\cdots+(-1)^{n-1}a_{in}M_n=0D=ai1M1−ai2M2+⋯+(−1)n−1ainMn=0(i=1,2,⋯,n)(i=1,2,\cdots,n)(i=1,2,⋯,n)
即η=(M1,−M2,⋯,(−1)n−1Mn)\eta=(M_1,-M_2,\cdots,(-1)^{n-1}M_n)η=(M1,−M2,⋯,(−1)n−1Mn)是方程组的一个解.

(2)如果r(A)=n−1r(A)=n-1r(A)=n−1，那么方程组的解全是(M1,−M2,⋯,(−1)n−1Mn)′(M_1,-M_2,\cdots,(-1)^{n-1}M_n)'(M1,−M2,⋯,(−1)n−1Mn)′的倍数.

参考答案：

由r(A)=n−1r(A)=n-1r(A)=n−1知，η\etaη为非零解，且方程组的基础解系含有n−r=n−(n−1)=1n-r=n-(n-1)=1n−r=n−(n−1)=1个向量，其通解为kη,k\eta,kη,所以方程组的解全是(M1,−M2,⋯,(−1)n−1Mn)′(M_1,-M_2,\cdots,(-1)^{n-1}M_n)'(M1,−M2,⋯,(−1)n−1Mn)′的倍数.

5.(10分)已知M=(ABB′D)M=\begin{pmatrix}A & B\\ B' & D\end{pmatrix}M=(AB′BD)是nnn阶正定矩阵，其中AAA为r(r<n)r(r< n)r(r<n)阶矩阵，证明A,D,D−B′A−1BA,D,D-B'A^{-1}BA,D,D−B′A−1B都是正定矩阵.

参考答案：

由于MMM正定，因此MMM的所有主子式全大于0.而AAA的各阶顺序主子式是MMM的1,2,⋯,r1,2,\cdots,r1,2,⋯,r阶顺序主子式，因此AAA正定.DDD的所有顺序主子式都是MMM的主子式，因此DDD正定.

(ABB′D)(2)+(−B′A−1)⋅(1)→(AB0D−B′A−1B)(2)+(1)⋅(−A−1B)→(A00D−B′A−1B),\begin{aligned}&\begin{pmatrix}A & B\\ B' & D\end{pmatrix} \underrightarrow{(2)+(-B'A^{-1})\cdot(1)}\\& \begin{pmatrix}A & B\\ 0 & D-B'A^{-1}B\end{pmatrix} \overrightarrow{(2)+(1)\cdot(-A^{-1}B)} \\&\begin{pmatrix}A & 0\\ 0 & D-B'A^{-1}B\end{pmatrix},\end{aligned}(AB′BD)(2)+(−B′A−1)⋅(1)(A0BD−B′A−1B)(2)+(1)⋅(−A−1B)(A00D−B′A−1B),

于是
(Er0−B′A−1En−r)(ABB′D)(Er−A−1B0En−r)=(A00D−B′A−1B).\begin{aligned}&\begin{pmatrix}E_r & 0\\ -B'A^{-1} & E_{n-r}\end{pmatrix}\begin{pmatrix}A & B\\ B' & D\end{pmatrix}\begin{pmatrix}E_r & -A^{-1}B\\ 0 & E_{n-r}\end{pmatrix}\\&=\begin{pmatrix}A & 0\\ 0 & D-B'A^{-1}B\end{pmatrix}.\end{aligned}(Er−B′A−10En−r)(AB′BD)(Er0−A−1BEn−r)=(A00D−B′A−1B).
由于

(−A−1B)′=−B′(A−1)′=−B′(A′)−1=−B′A−1,\begin{aligned}(-A^{-1}B)'&=-B'(A^{-1})'=-B'(A')^{-1}\\&=-B'A^{-1},\end{aligned}(−A−1B)′=−B′(A−1)′=−B′(A′)−1=−B′A−1,

因此
(ABB′D)≃(A00D−B′A−1B).\begin{pmatrix}A & B\\ B' & D\end{pmatrix}\simeq \begin{pmatrix}A & 0\\ 0 & D-B'A^{-1}B\end{pmatrix}.(AB′BD)≃(A00D−B′A−1B).
从而上式右边的矩阵也是正定矩阵.由(1)得，D−B′A−1BD-B'A^{-1}BD−B′A−1B是正定矩阵.

6.(30分)设FFF为数域，M30(F)M_3^0(F)M30(F)表示FFF上所有迹为0的3阶矩阵组成的集合.

(1)证明：M30(F)M_3^0(F)M30(F)是M3(F)M_3(F)M3(F)的一个子空间，其中M3(F)M_3(F)M3(F)为数域FFF上所有3阶矩阵构成的线性空间；

参考答案：

显然0∈M30(F),0\in M_3^0(F),0∈M30(F),因此M30(F)M_3^0(F)M30(F)非空集.

任取A,B∈M30(F)A,B\in M_3^0(F)A,B∈M30(F)则tr(A)=0,tr(B)=0.tr(A)=0,tr(B)=0.tr(A)=0,tr(B)=0.

从而

tr(A+B)=tr(A)+tr(B)=0,tr(kA)=ktr(A)=0,∀k∈F\begin{aligned}& tr(A+B)=tr(A)+tr(B)=0,\qquad \\& tr(kA)=ktr(A)=0,\qquad \forall k\in F\end{aligned}tr(A+B)=tr(A)+tr(B)=0,tr(kA)=ktr(A)=0,∀k∈F

因此M30(F)M_3^0(F)M30(F)对于矩阵的加法和数乘封闭，于是M30(F)M_3^0(F)M30(F)是M3(F)M_3(F)M3(F)的一个子空间.

(2)求M30(F)M_3^0(F)M30(F)的一组基和维数；

参考答案：

X=(xij)∈M30(F)X=(x_{ij})\in M_3^0(F)X=(xij)∈M30(F)

⇔x11+x22+x33=0\Leftrightarrow x_{11}+x_{22}+x_{33}=0⇔x11+x22+x33=0

⇔X=x11E11+x12E12+x13E13+x21E21+x22E22++x23E23+xn31E31+x32E32−(x11+x22)E33\begin{aligned}\Leftrightarrow X= & x_{11}E_{11}+x_{12}E_{12}+x_{13}E_{13}\\&+x_{21}E_{21}+x_{22}E_{22}++x_{23}E_{23}\\&+x_{n31}E_{31}+x_{32}E_{32}-(x_{11}+x_{22})E_{33} \end{aligned}⇔X=x11E11+x12E12+x13E13+x21E21+x22E22++x23E23+xn31E31+x32E32−(x11+x22)E33

⇔X=x11(E11−E33)+x12E12+x13E13+x21E21+x22(E22−E33)+x23E23+xn31E31+x32E32\begin{aligned}\Leftrightarrow X= & x_{11}(E_{11}-E_{33})+x_{12}E_{12}+x_{13}E_{13} \\ & +x_{21}E_{21}+x_{22}(E_{22}-E_{33})+x_{23}E_{23} \\ & +x_{n31}E_{31}+x_{32}E_{32}\end{aligned}⇔X=x11(E11−E33)+x12E12+x13E13+x21E21+x22(E22−E33)+x23E23+xn31E31+x32E32

由于E11−E33,E12,E13,E21,E22−E33,E23,E31,E_{11}-E_{33},E_{12},E_{13},E_{21},E_{22}-E_{33},E_{23},E_{31},E11−E33,E12,E13,E21,E22−E33,E23,E31,E32E_{32}E32线性无关.

因此他们是M30(F)M_3^0(F)M30(F)的一组基，从而dim⁡M30(F)=8.\dim M_3^0(F)=8.dimM30(F)=8.

(3)证明：M3(F)=⟨E3⟩⊕M30(F)M_3(F)=\langle E_3 \rangle\oplus M_3^0(F)M3(F)=⟨E3⟩⊕M30(F)其中⟨E3⟩\langle E_3 \rangle⟨E3⟩表示3阶单位矩阵E3E_3E3生成的子空间.

参考答案：

(a)证明M3(F)=⟨E3⟩+M30(F).M_3(F)=\langle E_3 \rangle+ M_3^0(F).M3(F)=⟨E3⟩+M30(F).

任取A=(aij)∈M3(F),A=(a_{ij})\in M_3(F),A=(aij)∈M3(F),

设A1=kE3,A_1=kE_3,A1=kE3,令A2=A−A1,A_2=A-A_1,A2=A−A1,满足tr(A2)=0tr(A_2)=0tr(A2)=0

即(a11+a22+a33)−3k=0.(a_{11}+a_{22}+a_{33})-3k=0.(a11+a22+a33)−3k=0.则k=3−1(a11+a22+a33)k=3^{-1}(a_{11}+a_{22}+a_{33})k=3−1(a11+a22+a33)

令A=A1+A2,A=A_1+A_2,A=A1+A2,其中A1∈⟨E3⟩,A2∈M30(F).A_1\in \langle E_3 \rangle,A_2\in M_3^0(F).A1∈⟨E3⟩,A2∈M30(F).

则A∈⟨E3⟩+M30(F).A\in\langle E_3 \rangle+ M_3^0(F).A∈⟨E3⟩+M30(F).
从而得出M3(F)=⟨E3⟩+M30(F).M_3(F)=\langle E_3 \rangle+ M_3^0(F).M3(F)=⟨E3⟩+M30(F).

(b)证明⟨E3⟩+M30(F)\langle E_3 \rangle+ M_3^0(F)⟨E3⟩+M30(F)是直和，

dim⁡⟨E3⟩+dim⁡M30(F)=1+8=9=dim⁡M3(F)=dim⁡(⟨E3⟩+M30(F)).\begin{aligned}\dim \langle E_3 \rangle&+\dim M_3^0(F)\\&=1+8\\&=9=\dim M_3(F)\\&=\dim (\langle E_3 \rangle+ M_3^0(F)).\end{aligned}dim⟨E3⟩+dimM30(F)=1+8=9=dimM3(F)=dim(⟨E3⟩+M30(F)).
从而⟨E3⟩+M30(F)\langle E_3 \rangle+ M_3^0(F)⟨E3⟩+M30(F)是直和，

综上，M3(F)=⟨E3⟩⊕M30(F).M_3(F)=\langle E_3 \rangle\oplus M_3^0(F).M3(F)=⟨E3⟩⊕M30(F).

7.(15分)设FFF为数域，定义F3F^3F3上的线性变换A\mathscr{A}A，满足A(α)=Aα,α∈F3,\mathscr{A}(\alpha)=A\alpha,\alpha\in F^3,A(α)=Aα,α∈F3,其中A=(210021002)A=\begin{pmatrix}2 & 1&0\\ 0 & 2&1\\0&0&2\end{pmatrix}A=⎝⎛200120012⎠⎞求A\mathscr{A}A的所有不变子空间.

参考答案：

(1)平凡子空间：0,F3F^3F3显然.

(2)AAA的特征多项式为f(λ)=∣λE−A∣=(λ−2)3f(\lambda)=|\lambda E-A|=(\lambda-2)^3f(λ)=∣λE−A∣=(λ−2)3f(λ)=0,f(\lambda)=0,f(λ)=0,则AAA的特征值为λ=2\lambda=2λ=2(三重).

解(λE−A)X=0(\lambda E-A)X=0(λE−A)X=0求出基础解系为ε1=(1,0,0)′.\varepsilon_1=(1,0,0)'.ε1=(1,0,0)′.对应的特征子空间为L(ε1),L(\varepsilon_1),L(ε1),则L(ε1)L(\varepsilon_1)L(ε1)是A\mathscr{A}A的不变子空间.

A\mathscr{A}A是F3F^3F3上的线性变换，在F3F^3F3的标准正交基ε1=(1,0,0)′,ε2=(0,1,0)′,ε3=(0,0,1)′\varepsilon_1=(1,0,0)',\varepsilon_2=(0,1,0)',\varepsilon_3=(0,0,1)'ε1=(1,0,0)′,ε2=(0,1,0)′,ε3=(0,0,1)′下得矩阵为AAA.

设WWW是AAA的非零不变子空间，

dim⁡W=1,\dim W=1,dimW=1,

∀kε2∈L(ε2),A(kε2)=kAε2=kε1+2kε2∉L(ε2),k∈F,∀kε3∈L(ε3),A(kε3)=kAε3=kε2+2kε3∉L(ε3),k∈F.\begin{aligned} &\forall k\varepsilon_2\in L(\varepsilon_2),\mathscr{A}(k\varepsilon_2)=kA\varepsilon_2=k\varepsilon_1+2k\varepsilon_2 \\& \notin L(\varepsilon_2),k\in F, \\&\forall k\varepsilon_3\in L(\varepsilon_3),\mathscr{A}(k\varepsilon_3)=kA\varepsilon_3=k\varepsilon_2+2k\varepsilon_3\\& \notin L(\varepsilon_3),k\in F. \end{aligned}∀kε2∈L(ε2),A(kε2)=kAε2=kε1+2kε2∈/L(ε2),k∈F,∀kε3∈L(ε3),A(kε3)=kAε3=kε2+2kε3∈/L(ε3),k∈F.

所以，L(ε2),L(ε3)L(\varepsilon_2),L(\varepsilon_3)L(ε2),L(ε3)不是A\mathscr{A}A的不变子空间.

dim⁡W=2,\dim W=2,dimW=2,

∀α=aε1+bε2∈L(ε1,ε2),A(α)=Aα=(2a+b)ε1+2bε2∈L(ε1,ε2),a,b∈F,∀α=aε2+bε3∈L(ε2,ε3),A(α)=Aα=aε1+(2a+b)ε2+2bε3∉L(ε2,ε3),a,b∈F,∀α=aε1+bε3∈L(ε1,ε3),A(α)=Aα=2aε1+bε2+2bε3∉L(ε1,ε3),a,b∈F,\begin{aligned} &\forall \alpha=a\varepsilon_1+b\varepsilon_2\in L(\varepsilon_1,\varepsilon_2), \\&\mathscr{A}(\alpha)=A\alpha=(2a+b)\varepsilon_1+2b\varepsilon_2\in L(\varepsilon_1,\varepsilon_2),\\& a,b\in F, \\&\forall \alpha=a\varepsilon_2+b\varepsilon_3\in L(\varepsilon_2,\varepsilon_3), \\&\mathscr{A}(\alpha)=A\alpha=a\varepsilon_1+(2a+b)\varepsilon_2+2b\varepsilon_3\\&\notin L(\varepsilon_2,\varepsilon_3),a,b\in F, \\&\forall \alpha=a\varepsilon_1+b\varepsilon_3\in L(\varepsilon_1,\varepsilon_3), \\&\mathscr{A}(\alpha)=A\alpha=2a\varepsilon_1+b\varepsilon_2+2b\varepsilon_3\notin L(\varepsilon_1,\varepsilon_3),\\& a,b\in F, \end{aligned}∀α=aε1+bε2∈L(ε1,ε2),A(α)=Aα=(2a+b)ε1+2bε2∈L(ε1,ε2),a,b∈F,∀α=aε2+bε3∈L(ε2,ε3),A(α)=Aα=aε1+(2a+b)ε2+2bε3∈/L(ε2,ε3),a,b∈F,∀α=aε1+bε3∈L(ε1,ε3),A(α)=Aα=2aε1+bε2+2bε3∈/L(ε1,ε3),a,b∈F,

所以，L(ε1,ε2)L(\varepsilon_1,\varepsilon_2)L(ε1,ε2)是A\mathscr{A}A的不变子空间,L(ε2,ε3),L(ε1,ε3)L(\varepsilon_2,\varepsilon_3),L(\varepsilon_1,\varepsilon_3)L(ε2,ε3),L(ε1,ε3)不是A\mathscr{A}A的不变子空间.

综上，A\mathscr{A}A的所有不变子空间是0,L(ε1),L(ε1,ε2),F3.0,L(\varepsilon_1) , L(\varepsilon_1,\varepsilon_2 ),F^3.0,L(ε1),L(ε1,ε2),F3.

8.(20分)设A=(aij)A=(a_{ij})A=(aij)是nnn阶实对称矩阵，它的nnn个特征值排序成λ1≥λ2≥⋯≥λn,\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_n,λ1≥λ2≥⋯≥λn,证明：

(1)对Rn\mathbb{R}^nRn中任意非零列向量α\alphaα，都有λn≤α′Aαα′α≤λ1;\lambda_n\le\frac{\alpha'A\alpha}{\alpha'\alpha} \le\lambda_1;λn≤α′αα′Aα≤λ1;

参考答案：

因为AAA是nnn级实对称矩阵，所以有nnn级正交矩阵TTT，使得T−1AT=diag{λ1,λ2,⋯,λn}.T^{-1}AT= diag\{\lambda_1,\lambda_2,\cdots,\lambda_n\}.T−1AT=diag{λ1,λ2,⋯,λn}.

任取Rn\mathbb{R}^nRn中一个非零列向量α,\alpha,α,设(Tα)′=(b1,b2,⋯,bn).(T\alpha)'=(b_1,b_2,\cdots,b_n).(Tα)′=(b1,b2,⋯,bn).则

α′Aα=α′Tdiag{λ1,λ2,⋯,λn}T−1α=(T′α)′diag{λ1,λ2,⋯,λn}(T′α)=λ1b12+λ2b22+⋯+λnbn2≤λ1(b12+b22+⋯+bn2)=λ1∣T′α∣2=λ1∣α∣2.\begin{aligned} \alpha'A\alpha&= \alpha'Tdiag\{\lambda_1,\lambda_2,\cdots,\lambda_n\}T^{-1}\alpha \\&= (T'\alpha)'diag\{\lambda_1,\lambda_2,\cdots,\lambda_n\}(T'\alpha) \\&=\lambda_1 b_1^2+\lambda_2b_2^2+\cdots+\lambda_nb_n^2 \\&\le \lambda_1(b_1^2+b_2^2+\cdots+b_n^2) \\&=\lambda_1|T'\alpha|^2 \\&=\lambda_1|\alpha|^2. \end{aligned}α′Aα=α′Tdiag{λ1,λ2,⋯,λn}T−1α=(T′α)′diag{λ1,λ2,⋯,λn}(T′α)=λ1b12+λ2b22+⋯+λnbn2≤λ1(b12+b22+⋯+bn2)=λ1∣T′α∣2=λ1∣α∣2.
同理
α′Aα=λ1b12+λ2b22+⋯+λnbn2≥λn(b12+b22+⋯+bn2)=λn∣α∣2.\begin{aligned} \alpha'A\alpha&=\lambda_1 b_1^2+\lambda_2b_2^2+\cdots+\lambda_nb_n^2 \\&\ge \lambda_n(b_1^2+b_2^2+\cdots+b_n^2) \\&=\lambda_n|\alpha|^2. \end{aligned}α′Aα=λ1b12+λ2b22+⋯+λnbn2≥λn(b12+b22+⋯+bn2)=λn∣α∣2.
因此λn≤α′Aαα′α≤λ1.\lambda_n\le\frac{\alpha'A\alpha}{\alpha'\alpha} \le\lambda_1.λn≤α′αα′Aα≤λ1.

(2)λn≤aii≤λ1(i=1,2,⋯,n).\lambda_n\le a_{ii}\le\lambda_1(i=1,2,\cdots,n).λn≤aii≤λ1(i=1,2,⋯,n).

参考答案：

由于εi′Aεi=aii,\varepsilon_i'A\varepsilon_i=a_{ii},εi′Aεi=aii,且∣εi∣2=1,|\varepsilon_i|^2=1,∣εi∣2=1,因此从(1)的结论立即得到λn≤aii≤λ1(i=1,2,⋯,n).\lambda_n\le a_{ii}\le\lambda_1(i=1,2,\cdots,n).λn≤aii≤λ1(i=1,2,⋯,n).

如果您在看完以后，无论是题目还是参考解答过程有什么问题，希望不吝指出！如果有更好的解题思路或者好的问题分享，欢迎通过后台发送给我们！

–
更多内容请关注微信公众号：小离数学考研