习题 1.2
1. 解:因为对R
的任一向量(
),按对应规则
都有R
中
惟一确定的向量与之对应,所以
是R
的一个变换.
(1) 关于
轴的对称变换;
(2) 关于
轴的对称变换;
(3) 关于原点的对称变换;
(4) 到
轴的投影变换;
(5) 到
轴的投影变换.
2. 解: (1) 不是.因为
(
)=
+
≠k
(
)+k
=
+
(2) 不是.因为
(
)=
≠k
(
)+k
(3) 不是.因为取 x=(1 , 0 , 0 ) ,
时,
(k x)=(k
,0, 0)≠k
( x)= k(1, 0, 0)=(k, 0, 0)
(4) 是.因为 设x=(
) , y=(
)
(k
x+k
y)=
EMBED Equation.3
=k
(x)+k
( y)
(5) 是.因为
(
)=
=k
(f
(x))+k
EMBED PBrush
(6) 是.因为
(
)=
= k
(f
(x))+k
EMBED PBrush
(7) 不是.因为 设x=(
) , y=(
)
(k
x+k
y)= (
≠k
(x)+k
EMBED PBrush ( y)
=
=(
.
3. 解:
EMBED Equation.3 (
+β)=
EMBED Equation.3 [
EMBED Equation.3 (
)+
EMBED Equation.3 (β)
EMBED Equation.3 (k
)=
EMBED Equation.3 (k(x
, x
))
EMBED Equation.3 (
)
所以
EMBED Equation.3 是线性变换.同理可证
EMBED Equation.3 也是线性变换.
(
EMBED Equation.3 +
EMBED Equation.3 )(
)= (
EMBED Equation.3 +
EMBED Equation.3 )[(x
, x
)]
=
EMBED Equation.3 [(x
, x
)]+
EMBED Equation.3 [(x
, x
)]
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 (
)=
EMBED Equation.3 [
EMBED Equation.3 (
)]=
EMBED Equation.3 [( x
, -x
)]=(- x
, -x
)
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 (
)=
EMBED Equation.3 [
EMBED Equation.3 (
)]=
EMBED Equation.3 [( x
, -x
)]=( x
, x
) .
4. 证:(1)因
(A)+
(B)
k
(A)
故
是线性变换.
(2)
(A)B+A
(B)
(AB)
5. 解:令
即可.
6. 证:设
,则
(
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 -
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 )(f(x))
=
EMBED Equation.3 [
EMBED Equation.3 (f(x))]-
EMBED Equation.3 [
EMBED Equation.3 (f(x))]
=
EMBED Equation.3 [xf(x)]-
EMBED Equation.3 [f(x)]
故
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 -
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 是恒等变换.
7. 证:设
,则
,由于
EMBED Equation.3 (e
)+
EMBED Equation.3 (e
)=
EMBED Equation.3 (e
+e
)=e
+e
EMBED Equation.3 (e
)-
EMBED Equation.3 (e
)=
EMBED Equation.3 (e
-e
)=e
-e
所以,
EMBED Equation.3 (e
)=e
,
EMBED Equation.3 (e
)= e
于是
EMBED Equation.3 (α)=k
EMBED Equation.3 (e
)+k
EMBED Equation.3 (e
)
= k
EMBED Equation.3 (e
)+k
EMBED Equation.3 (e
)=
EMBED Equation.3 (α)
故
EMBED Equation.3 =
EMBED Equation.3 .
8. 解:(1) 因为
在
平面上,其投影不变,故有
(i)=i ,
(j)=j ,
又
垂直
平面,则
, 得
(
(i),
(j),
(k))=(
,
,
)
所求矩阵为A=
.
(2) 因为
,
所以, 所求矩阵为 A=
.
(3) 由
的定义知,
(i)=
((1 ,0 ,0 ))= ( 2 ,0 ,1)
(j)=
((0 ,1, 0 ))= ( -1, 1 , 0)
(k)=
((0 ,0 ,1))= ( 0 ,1 , 0)
有 (
(i),
(j),
(k))=(
所求矩阵为 A=
.
(4) 据题设:
则
=(
)
=
=
=(
)
=
=(
)
=
=(
)
=
=(
)
=
= (
)
=
于是
(
,
,
,
,
,
)
,
所求矩阵为
D=
9. 解:(1) (
)=(
)
=(
)C
所求矩阵为 B=C
AC=
(2) (
)=(
)
=(
)C
所求矩阵为 B=C
AC =
(3) (
)=(
)
=(
)C
所求矩阵为 B=C
AC
=
10. 解:由定义知
所以,所求矩阵为
.
11. 解 : 因为
所以,所求矩阵为
.
12. 解: (
,
,
)=(
)
(
)=(
,
,
)
= (
,
,
) C
B=C
AC=
=
.
13. 解:(1) (
,
,
) = (
) C ,
过渡矩阵为
C=(
)
(
,
,
)
=
=
(2) (
,
,
)=(
,
,
) = (
) C
故
在基
下的矩阵就是 C.
(3) (
(
,
(
),
(
) ) = (
,
,
) = (
) C
=(
,
,
) C= (
,
,
) C
故
在基
下的矩阵仍为C.
14. 解:
(1) 由于
故
EMBED Equation.3 在该基下的矩阵为
类似地,可得
EMBED Equation.3 在该基下的矩阵为
.
由于
EMBED Equation.3 =
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 ,所以
EMBED Equation.3 在该基下的矩阵为
同理,可得
EMBED Equation.3 在该基下的矩阵为
(2)由于由简单基E11,E12,E21,E22改变为给定基E1,E2,E3,E4的过渡矩阵为
于是,
EMBED Equation.3 在给定基下的矩阵为
15. 解: (1)将题给关系式写成矩阵形式为
(
(
,
(
),
(
) )
即
由于
,所以有
(
EMBED PBrush
故
在基(II)下的矩阵
(2)因为
(
EMBED PBrush
所以
在基(I)下的坐标为(3,5,9).
16. 解:(1)取
的简单基1,x,x2,则有
从简单基改变到基f1,f2,f3和g1,g2,g3的过渡阵分别为
,
故有
(g
, g
, g
)=
(1, x, x
)C=
即
在基(II)下的矩阵
(2)因为
所以
(f(x))=
.
17. 证:设
在给定基下的矩阵为
,并设C为从旧基到新基的过渡矩阵,由于
在任一组基下的矩阵相同,则有
,即AC=CA,根据“A与一切满秩矩阵可变换”性质,即可定出A必为数量矩阵
.
18. 解:由基
到基
的过渡矩阵为
故
下的矩阵为
.那么,
+ ,
,
, (
+ )在基
下的矩阵分别为
,
,
,
.
19. 证:设有可逆方阵P与Q,使 B=P
AP , D=Q
CQ 则
=
=
=
即
与
相似.
20. 证:设
,
,则A,B的行向量的极大无关组中分别含有
个行向量,设分别为
和
,则A的每个行向量均可由
线性
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
示,B的每个行向量均可由
线性表示.又可A+B的每个行向量是A与B的相应行向量的和,故A+B的每个行向量均可由
,
线性表示.因此A+B的行向量组的极大无关组中所含向量的个数不超过
,即
.
21. 证:设
,则
,
所以
,
,…,
.这就说明B的列向量
都是以A为系数矩阵的齐次方程组的解.由于
,所以解空间的维数为
,从而知
的极大无关组所含向量的个数
,即
,因此有
.
22. 证:设A,B为同一数域上的
与
阶矩阵,显然,方程组BX=
的解向量X也满足方程组
,记
,
则
,于是
即
.
又由于
因此
.
23. 证:由上题知,
,现在只需证明
即可.
考虑线性方程组
,设
是方程组的一组解,将
两边左乘XT,得
,即
,所以
,即
.于是
即有
,故有
,并且有
即有
.
注:对复矩阵A,上式不一定成立.例如
,
.由于
故
.此时,相应的关系式应为
.
24. 证:必要性.由上题已证得,充分性只要在AX=
两边左乘AT即可.
25. 证:(1)因为
,故
,不妨设A的前n行线性无关,且构成的n阶满秩方阵为A1,后
行构成的矩阵为A2,则
所以
,但
,故
.
(2) 同理可证.
26. 解:(1)
,
;
(2)
,
;
(3)
,
.
27. 证:因为
,但
,故m阶方阵C的秩
,所以C是降秩的.
28. 解:先求矩阵A的特征值和特征向量为
,
,
故
的特征值和特征向量为
,
,
,
,
.
29. 解:(1)
,
,
,
,
.
(2)
,
,
,
(3)
,
,
,
;
(4)
,
,
,
,
,
.
以上分别求出了
在不同基下所对应矩阵A的特征值和特征向量,则类似于上题的方法,可求出
不同基下所对应的特征值和特征向量.
30. 解:(1),(2),(4)为非亏损矩阵(单纯矩阵),其变换矩阵P分别为
(1)
; (2)
;
(4)
.
31. 证 : 设
在给定基下的矩阵为A,则
32. 证:设
,则存在满秩矩阵P与Q,使得
,故有
其中
, 这说明AB与diag(
)相似.
另一方面,有
,说明BA与
相似.不难验证有
故AB与BA有相同的特征多项式,因此有相同的特征值和迹.
33. 证:设A的任一特征值为
,
的对应于
的特征子空间记为
.对
中任意向量Z有
故
,因此
为线性变换
的不变子空间,即
为
中的线性变换,此线性变换的特征向量即为B的特征向量,但它又属于
,由
的定义知它又是A的特征向量,即A与B有公共的特征向量.
34. 证:设A的特征值为
,则A2的特征值为
,由
有
,若所有
,则A+I为满秩矩阵,故由(A+I)(A-I)=A2-I2=0,有A=I.
35. 证:不失一般性,设B非奇异,有AB=B-1(BA)B即AB与BA相似,所以它们有相同的特征多项式.
36. 证:设A为n阶方阵,具其秩为
,由于A2=A,知A的列向量都是A的对应于特征值1的特征向量.因
,故特征值1的几何重复度为r,其代数重复度至少为r.又
的基础解系中的向量个数为
,即A的特征值0的几何重复度为
,其代数重复度不小于
.由于一个n阶矩阵的特征值的代数重复度之和恰为n,故特征值1和0的代数重复度分别为r和
.可见A除了1和0外无其它特征值,而1和0的几何重复度之和为n,故A为非亏损矩阵,所以A相似
.
37. 证:用反证法.若A可相似于对角矩阵,对角元素即为A的特征值,且至少有一个不为0.但是,由于
,于是
,因为
,所以
,故
,即A的特征值都等于0,矛盾.
38. 证:由
,有
,
,从而有
,即X也是
的特征向量.显然
的特征值为
,即为
的多项式.
39. 解:取R3中的自然基
,计算得
(
)=(0 , -2 ,-2 ) ,
(
)=(-2 , 3 ,-1 ) ,
(
)=(-2 , -1 ,3 )
则
在基
下的矩阵为
而A的特征值为
,与之对应的特征向量为
,
,
,则有
,
其中
.由
=(
)C求得
的另一组基为
,
,
,显然
在该基下的矩阵为对角阵
.
40. 解:(1)因为
,
,
,所以
在基1,x,x2下的矩阵
.
(2)由于A原特征值为
,
,相应的特征向量为
,
,
,存在可逆阵
,使
,故所求的基
为
.
41. 解:(1)对任意的
及
,有
=k(
(
))+l(
(β))
故
是线性变换.
(2)取V的简单基
由于
EMBED Equation.3 ,
EMBED Equation.3 ,
所以
在基
下的矩阵为
R的特征值为
,对应的线性无关的特征向量为(1,1,0)T,(0,1,1)T,(0,1,-1)T,令
,
则有
,由(B1,B2,B3)=(A1,A2,A3)C求得V的另一组基为
,
,
,
在该基下的矩阵为
.
42. 证:(1)取Vn的一组基
,设
EMBED Equation.3 (
)=(
)A
EMBED Equation.3 (
)=(
)B
则有
(
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 )(
)=(
)(AB)
(
EMBED Equation.3 +
EMBED Equation.3 )(
)=(
)(A+B)
由
EMBED Equation.3
EMBED PBrush
EMBED Equation.3 =
EMBED Equation.3 +
EMBED Equation.3 ,可得AB=A+B,从而有BTAT=AT+BT.
若1是
EMBED Equation.3 的特征值,则 1也是A的特征值,从而1也是AT的特征值,设AT对应于特征值1的特征向量为
,即
,由(BTAT)
=(AT+BT)
,可得BT
=
+BT
,即
=0,这与
是AT的特征向量矛盾,故1不是
EMBED Equation.3 的特征值.
(2)因
EMBED Equation.3 有几个不同的特征值,所以
EMBED Equation.3 有n个线性无关的特征向量.记
EMBED Equation.3 的对应于特征值
的线性无关的特征向量为X1,X2,…,Xn,即
EMBED Equation.3
(i=1,2,…,n),则X1,X2,…,Xn作为Vn的基时,
EMBED Equation.3 的矩阵A=diag(
).再由AB=A+B及
知
即
EMBED Equation.3 与
EMBED Equation.3 在该基X1,X2,…,Xn下的矩阵都为对角阵.
43. 证:对任意
,有
EMBED Equation.3 (
.由于
EMBED Equation.3 (
EMBED Equation.3 (
))=
EMBED Equation.3 (
EMBED Equation.3 (
))=
EMBED Equation.3 (λ
EMBED Equation.3 )
所以
EMBED Equation.3
, 故
是
EMBED Equation.3 的不变子空间.
44. 解:(1)
(
)=(
)C
=(
)
B=C
AC =
(2) 先求核
) . 设η=
在基
下的坐标为
(
),
(
在此基下的坐标为(0,0,0,0),于是
A
=
此时A的秩为2,解之,得基础解系
,
作
. 显然,
为核
)的
一组基,故核由
所张成,即
)=Span(
) .
再求值域
(V
) . 由于
(
(e
),
(e
),
(e
),
(e
)) = (
) A
而A的秩为2,所以
(e
),
(e
),
(e
),
(e
)的秩也为2,且
(e
),
(e
)线性无关,故组成
( V
)的基,从而
( V
)=Span(
(e
),
(e
)) .
(3) 由(2)知
是核
)的一组基,易知
为V
的一组基,由于有
(
)=(
)
= (
) D
所以
在此基下的矩阵为
B=D
AD=
(4) (2)知
(e
),
(e
)是值域
(V
)的一组基,又知
(e
),
(e
),
为V
的一组基,有
(
(e
),
(e
),
)=(
)
=(
) T
所以
在此基下的矩阵为
B=T
A T =
.
45. 证:取R3中的自然基
,因为
(
+ )(
)=
(
)+ (
)=(1,0,0)+(0,0,1)
=(1,0,1)
同理有
(
+ )(
)=(2,0,0),
(
+ )(
) =(1,1,0)
这表明
+ 将基
变换成R3中的另一组基
=(1,0,1),
=(2,0,0),
=(1,1,0)(易证它们线性无关).
又因(
+ )(R3)是R3的子空间,而
是(
+ )(R3)的最大无关组,故这个子空间的维数为3,再由习题1.1中第22题的结果知(
+ )(R3)=R3(此时取V2=R3).
46. 解:因为
EMBED Equation.3 [(
)]=
(
[(
)])
=
=(0,0,
)
所以
EMBED Equation.3 的像子空间为
R(
EMBED Equation.3 )
核子空间为
N(
EMBED Equation.3 )
因此,dimR(
EMBED Equation.3 )=1,其一组基为(0,0,1);dim N(
EMBED Equation.3 )=2,其一组基为(0,1,0),(0,0,1).
47. 证 :(1)由
的定义容易验证满足可加性和齐次性,所以它为线性变换.又因
EMBED Equation.3 [(
)]=
[
,
…
推知
EMBED Equation.3 [
,即
EMBED Equation.3
(零变换).
(2)若
[
,
则
=
=…=
=0即
为由一切形如(0,0,…,
)的向量构成的子空间,它是一维子空间,则(0,…,0,1)是它的基.
又由维数关系
dim
(V)+dim
EMBED Equation.3 (θ)=n
便得
(V) 的维数等于 n-1 .
48. 证 :(1)必要性.若
(V)= (V),对任
,则
(V)=
(V) ,故存在
,使
EMBED Equation.3 ,
EMBED Equation.3
EMBED Equation.3 =
EMBED Equation.3 =
,由
的任意性有
= .
同理可证
=
.
充分性.若
= ,
=
, 对任
EMBED Equation.3
(V)
,
EMBED Equation.3
EMBED Equation.3 = (
EMBED Equation.3 )
(V) , 故
(V)
(V) ;
同理可证 (V)
(V).
(2)必要性.若
,对任
,作
EMBED Equation.3 ,因
(
EMBED Equation.3 )=
EMBED Equation.3 -
EMBED Equation.3
EMBED Equation.3 =
EMBED Equation.3 -
EMBED Equation.3 =
,所以,
EMBED Equation.3
EMBED Equation.3
EMBED PBrush
=
,则 (
EMBED Equation.3 )=
,故
EMBED Equation.3 ,由
的任意性有 =
.
同理,通过作
-
, 可得
=
.
充分性.若
= ,
=
, 对任
,由
EMBED Equation.3 (
EMBED Equation.3 )= (
)=
,故
;同理,由任
,可得
.
PAGE
24
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