矩阵运算
,1. 化下列矩阵成
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
形
2,,,,,,132,,,,,,,2,1),,,,, 2) ,,,2,,,,5,3,,,222,,,,1,,,,,
2,,000,2,,,,,,,002,,00,,,0,,3), 4) 00,,,,20,(,1)00,,2,,,00(,1),,2,,,,,000,,
22,,,,,,,3,2,32,1,2,3,,5)22,,,,, 4,3,53,2,3,4,,
,,2,,4,2,1,,,,,,,2301,,,,,43,60,22,,,,,,6),, 0620,,,,,
,,10,,100,,
,,3,,31,,2,,200,,
1)对矩阵作初等变换,有 ,,
3222,,,,,,,,,,,,23,,5,,3,,5A,,,,,,,,(,), 22332,,,,,,,,0,-10,-3,,5,3,2,,,,,,,,,
,0,, , = B(,) ,,,32,,,,0,10,3,,,
B(,)即为所求。
2)对矩阵作初等变换,有 ,,
22,,,,,,,,,,10011,,,,,,,,A,,,,,,,,,(,), ,,0,0,,,,,,,,,,22222,,,,,00,(,1)1,,,1,,,,,,,,,,
100,,,, ,0,0(,) = B, ,,
2,,00,,,,,
B(,)即为所求。
2,,,,,00,,,的行列式因子为 00,,
,,2,00(,1),,
3)因为23 ,(,,1)=1, =, = , ,(,,1)DDD 12 3
所以
DD232,(,,1)= 1, = = , = = , ,(,,1)ddd1 2 3 DD12从而
2,,100,,,,,00,,,,,,0,(,,1)0A = B, (,),(,)00,, ,,,,22,,,00(,1)00(1),,,,,,,
B(,)即为所求。
2,,000,,,200,,,0,,4)因为的行列式因子为 ,,20,(,1)00,,2,,,,,000,,
2244,(,,1),(,,1)=1, =,(,,1), = , = , DDDD 12 3 4 所以
DDD22324,(,,1),(,,1),(,,1)= 1,= = ,= = ,= = , dddd1 2 3 4 DDD123
从而
2,,000,,,200,,,0,,A(,), ,,20,(,1)00,,2,,,,,000,,
2,,100,,,
0,(,,1)00,,,(,) = B, ,,00,(,-1)0,,22,,000,(,-1),,
B(,)即为所求。
5)对矩阵作初等变换,有 ,,
22,,,,,,,3,2,32,1,2,3,,22,,,,, (,),4,3,53,2,3,4,,,,2,,4,2,1,,,,,,
A
22,,3,,22,,1,,2,, 22, 4,,33,,2,,2,,
,,2,,2,,21,,
4232,,,,,7,,6,,,2,,4,,50,,2 , ,,1,,10,,
,,001,,
3232,,,,,,,,,1,,,2,,4,,50,, , 0,,10,,
,,001,,
32,,,,,,,,100,, , 0,,10,,
,,001,,
100,,,, ,0,,10(,)= B, ,,
32,,00,,,,,,1,,
B(,)即为所求。
6)对矩阵作初等变换,有 ,,
,2301,,,,,43,60,22,,,,,,,,A(,), 0620,,,,,,,10,,100,,,,3,,31,,2,,200,,
0001,,,,,000,,22,,,,,, 00,2,0,,,,10,,100,,,,01,,000,,
0001,,,,,000,0,,
,,, 00,2,0,,
,,10000,,
,,01,,000,,
0001,,,,,20000,,,,
,,, 00,00,,
,,10000,,
,,01,,000,,
00010,,,,20000,,,,
,,, 00,00,,
,,10000,,
,,01,,000,,
10000,,,,20,,000,,
,,,, 00,00,,
000,,10,,
,,00001,,,,
在最后一个行列式中
32,(,,1)=1, =,(,,1), = , DDD 34 5
所以
DD254,(,,1),(,,1)===1, = =, = =。 ddddd1 2 3 4 5 DD34
故所求标准形为
10000,,,,01000,,
B,,(,) = 。 00100,,
000,(,,1)0,,
,,20000,(,,1),,
2.求下列矩阵的不变因子: ,,
,,100,,,,2,10,,,,,,0,,10,,1)0,,2,1 2) ,,,,00,,1,,,,00,,2,,,,543,,2,,
,,,,001010,,,,,,,,,,,,,01,,2001,,,, 4) ,,,,,,,,1,,20000,,,,,,,,,,,,200000,,,,,,,3)
,,1000,,,,0,,200,,5) ,,00,,10,,,,000,,2,,
1)所给矩阵的右上角的二阶子式为1,所以其行列式因子为
3 (,,2)=1, =1, = , DDD 12 3
故该矩阵的不变因子为 ,,
3 (,,2)==1, =。 ddd1 2 3
2)因为所给矩阵的右上角的三阶子式为-1,所以其行列式因子为
432 ===1, =, DDDD3 2 14 ,,2,,3,,4,,5故矩阵的不变因子为 ,,
432 ===1, =。 1 2 3 4 dddd,,2,,3,,4,,5
,,03)当时,有
2 = = ,22D4 ,,,,,,,,,,(,,,),,
,,,,,,,,,,
且在矩阵中有一个三阶子式 ,,
,10
,,,2,(,,,) = , ,01
0,,,,
于是由
, = 1, D3 2,(,,,),,
可得
= 1, D3
故该矩阵的不变因子为 ,,
222 ===1, = ,,。 (,,,),,dddd1 2 3 4
,,0当时,由
24 (,,,)(,,,)=1, =1, = , = , DDDD 12 3 4
从而
D224(,,,)(,,,) ==1, = , = = 。 dddd1 2 3 4 D3
4)因为所给矩阵的左上角三阶子式为1,所以其行列式因子为
4(,,2) =1, =1, =1, = , DDDD 12 3 4 从而所求不变因子为
4(,,2) ===1, = 。 dddd1 2 3 4
5)因为所给矩阵的四个三阶行列式无公共非零因式,所以其行列式因子为
22 (,,1)(,,4)=1, = , DD3 4
故所求不变因子为
22(,,1)(,,4) ===1, = 。 dddd1 2 3 4
3.
证明
住所证明下载场所使用证明下载诊断证明下载住所证明下载爱问住所证明下载爱问
:
,000?0a,,n,,,,100?0an,1,,,,,,n,2010?0a,,,n,3,,,001?0a
,,???????,,2,,,0000?a,,1,,0000?1,a,,的不变因子是
1,1,,1f(,), ?,,,,,,n,1个
nn,1,,,,,,,aaaf(,)其中= 。 11nn,
因为
,000?0an,,100?0an-1,n0,10?0a-2,(,) = , Dnn00,1?0a-3
???????
,0000?a2
0000?,1,,a1
按最后一列展开此行列式,得
,n1,n2(,,)a,,a,?,,a,,a(,)f(,) == , Dn12,n1n
nn,1,,a,,a,,?,a,,a= , f(,)12n,2n,1n
n,1M(,1)D因为矩阵左下角的阶子式= ,所以= 1,从而 ,,n,1n,1n,1
D== „ = = 1, DD 12 n,2故所给矩阵的不变因子为
d== „ = = 1, dd1 2 n,1
n,n1,,a,,?,a,,ad = = , f(,)1,n1nn
即证。
4. 设A是数域P上一个n,n阶矩阵, 证明A与相似 A'
设
aa?a,,11121n,,aa?a,,21222n A = , ,,????,,,,aa?an1n2nn,,
aa?a,,1121n1,,aa?a,,1222n2则 = , A',,????,,,,aa?a1n2nnn,,因为A与相似的充分必要条件是它们有相同的不变因子,所以只需证明与A',E,A
有相同的不变因子即可 ,E,A'
注意到与对应的级子式互为转置, 因而对应的级子式相等, 故 ,E,A,E,A'kk
与 ,E,A,E,A'有相同的各级行列式因子, 从而有相同的不变因子, 即证A与相似 A'
5. 设
,00,,,, A ,= 10,,
,,01,,,
kA求。
因为
k(k,1),,kk,1k,2k,,k,,,,00,,2,,,,kk,1 ,k = 0,,, 10,,,,
k,,,,01,00,,,,,,,所以
k'kk',,,,,,10,10,,,,00,,,,,,,,,,,,k,,, = = = A0,10,110,,,,,,,,,,,,,,,,,,01,00,00,,,,,,,,,,,,,
'kk(1),,,,1,2kkk,,kk,,,,,,00,,2,,k,1k,1kk,, = = 。 ,k,0k0,,,,,,kk,(1)kk,2k,1k,,,,00,,k,,,,2,,,,
6. 求下列复系数矩阵的若尔当标准形:
120,,131616,,,,,,1)020 2) ,5,7,6,,,,,,,,,6,8,7,2,2,1,,,,
308,,45,2,,,,,,3,163) 4) ,2,21,,,,,,,,,1,11,20,5,,,,
37,31,12,,,,,,,,5) 6) ,2,523,36,,,,,,,,,4,1032,24,,,,11,1,,,4210,,,,,,7) ,3,33 8) ,437,,,,,,,,,317,2,22,,,,
830,14,,033,,,,,,9) ,6,199 10) ,186,,,,
,,,,2,14,10,6,2311,,,,
31001234,,,,,,,,,4,1000123,,,,11) 12) ,,,,71210012,,,,,,,,,7,6,100001,,,,
010?00,,,,001?00,,1,303,,,,,,000?00,2601313),,,, 14) ,,??????,,0,313,,,,,,,1208000?01,,,,,,100?00,,
,,1,,11,,,,1,20,, 1)设原矩阵为A ,则 2,,,,,, = 0,,20,E,A0,20,,,,,,,,22,1,,,1,20,,,, ,,
,,100,,
,, , 0,,20,,,,(,1)(1,),,0,,,12,,
,,100,,,(,1)(1,,) , ,0,3,2,,,,0,,20,,
100,,,, ,010 , ,,
,,,,,00(,1)(,1)(,2),,于是A的初等因子是
, , , ,,1,,1,,2故A的若尔当标准形为
100,,,, = 。 J0,10,,
,,002,,
2)设原矩阵为A ,则
,1,,1,10,,,13,16,16,,,,,, ,,,,,2,,16 = ,E,A5,76,,,,
,,,,68,7,,21,,,,7,,,,
100,,100,,,,,, ,0,(,1)(,,3),10,,14,, 0,16,3,,,,22,,,,0,(,1)00,3(,1),,,,,,,,
100,,,, ,010 , ,,
2,,00(,,1)(,,3),,于是A的初等因子是
2 (,,1), , ,,3
故A的若尔当标准形为
,300,,,, = 。 J010,,
,,011,,
3) 设原矩阵为A ,则
,1,,12,,,30,8,,,,,,,,,,2,,1,6 = ,E,A,3,1,6,,,,
,,,,20,5,,20,5,,,,100100,,,,,,,,2 ,0,(,1)0,0,,1,2,(,1) ,,,,
2,,,,,,00(,1)0,2(,1),1,,,,,100,,,, ,0,,10 , ,,
2,,00(,1),,,
2于是A的初等因子是(,,1), ,故A的若尔当标准形为 ,,1
,100,,,, = 。 J0,10,,
,,01,1,,
4) 设原矩阵为A ,则
,,4,52,,,,1,52,,,,,,2,,2,1,,, = ,E,A,,2,1,,,,,,,,01,1,11,1,,,,,
1,,31100,,,,,,,,2 ,,,,,2,1,0,,,2,2,,1 ,,,,,,,,,,01,101,1,,,,100,,100,,,,,, ,01,,1, , 010,,,,23,,,,,,0(,1)000(,1),,,,
3于是A的初等因子是(,,1),故A的若尔当标准形为
100,,,, = 。 J110,,
,,011,,
5) 设原矩阵为A ,则
,,3,73,,,,72,,,,,,2,,5,2,,, = ,E,A0,5,,,,
,,,,,110,3,,,410,3,,,,
,1170,,117,6,,,,,, ,,,0,,5,2 0,5,2,,,,2,,,,,110,3,,,,,0,17,7,,6,3,,,,
100,,,, , , 010,,2,,00(,1)(,1),,,,
于是A的初等因子是, , , 从而A的若尔当标准形为 ,,1,,i,,i
100,,,, = 。 J0,i0,,
,,00i,,
6) 设原矩阵为A ,则
,,,11,21,1,2,,,,,,,,,,, = ,E,A,3,3,6,3,3,6,,,,
,,,,,22,42,2,4,,,,,,
100,,100,,,,,,2 ,,,,,0,,,2,2, ,,0(2)2,,,,
,,,,,02,,00,,,,,
100,,,, ,, , 00,,
,,00(,2),,,,
于是A的初等因子是, , , 从而A的若尔当标准形为 ,,,,2
000,,,, = 。 J000,,
,,002,,
7) 设原矩阵为A ,则
,,,1,111,1,1,,,,,,,,,,, = ,E,A3,3,3,3,33,,,,
,,,,22,2,222,,,,,,
100,,100,,,,,, ,0,,3,, , 00,,,,32,,,,0,,3,00,,,,,,,
2, ,故A的若尔当标准形为 ,,
000,,于是A的初等因子是,, = 。 J000,,
,,010,,
8) 设原矩阵为A ,则
,,4,2,10,13,,7,,,,,,,,4,,3,7,,,34,7 = ,E,A,,,,,,,,3,1,,7,2,,4,10,,,,,100,,100,,,,,,2 ,0,3,5,10,,14,,, 01,4,2,,,,,,,,0,2,2,4,,0,2,2,4,,,,,,
100,,,, ,010, ,,
3,,00(,2),,,
3于是A的初等因子是(,,2), 故A的若尔当标准形为
200,,,, = 。 J120,,
,,012,,
9) 设原矩阵为A ,则
,,,3,31,8,6,,,,,,,,,,, = ,E,A1,8,6,3,3,,,,,,,,,214,10,2,14,10,,,,,,100100,,,,,,,,22 ,0,,,,8,3,6,3,014,,6,,10,,6 ,,,,
,,,,,,,,02,2,202(,1),2,,,,
100,,,, ,010, ,,
2,,00,(,,1),,
2于是A的初等因子是(,,1), ,故A的若尔当标准形为 ,
000,,,, = 。 J0,10,,
,,01,1,,
,,,,8,3014123,,11,,,,10) 设原矩阵为A ,则 ,,,,6,,19,19,= 1,,19,9,E,A,,,,,,8,,,,,623,11,3014,,6,,
100100,,,,,,,, ,0,,4,,,2,0,2 ,,,,
222,,,,,,,02,15,4,4,0,23,4,,19,4,,,,,,,
100,,,, ,, 010,,3,,00,30,8,,,,
3设(,,,)(,,,)(,,,) =, 则由“卡当”公式可解得 ,,30,,8123
33 ,,4,1016,4,10161
233 ,,,4,1016,,4,10162
233 ,,,4,1016,,4,10163
13其中,,,,i,,,,,,,,,. 于是A的初等因子是, , ,故A的若尔当标准形31222
为
,00,,1,, , = 。 J00,,2,,00,,,3
11) 设原矩阵为A ,则
,1000,,,,,3100,,,,,,2,0,,2100,,,,,4100,,, = ,E,A,,,,,0,,,4,,2,1,,,,7121,,,,,,,,761,06,,111,,,,,
10001000,,,,,,,,22,,0(,1)000(,1)00,,,,,, ,,,,0,6,,2,10001,,,,22,,,,0,10(,,1)00,10(,,1)0,,,,
1000,,,,0100,,,, ,,0010,,4,,,000(,1),,
4(,,1)于是A的初等因子是,故 A的若尔当标准形为
1000,,,,1100,, = 。 J,,0110,,,,0011,,12) 设原矩阵为A ,则
,,1,2,3,4,,,,,0,1,2,3,, = , ,E,A,,,00,1,2,,,,000,1,,,
因为三阶子式无公共非零因式,所以的行列式因子为 ,E,A
4 =1, = , DD,E,A,(,,1)3 4
于是
4 (,,1) = , ===1, dddd4 3 2 1
4(,,1)因此A的初等因子是,故 A的若尔当标准形为
1000,,,,1100,, = 。 J,,0110,,,,0011,,13) 设原矩阵为A ,则
,,,130,8,,,,,2,60,13,, = ,E,A,,03,,1,3,,,,1,20,,8,,
1000,,,,,,0,20,2,3,,, ,,,03,1,3,,2,,0,2,10,,9,1,,,,,
1000,,,,,03,10,,, ,,,,0,20,,1,,2,,02,10,,11,10,,,,,
1000,,,,0300,,, ,,,000,,1,,32,,00,,15,33,190,,,,,
1000,,,,0100,,,, ,,00,,10,,,,000(,,1)(,,7,30)(,,7,30),,
,,7,30,,7,30所以A的初等因子是, , , ,故 A的若尔当标准形为 ,,1,,1
1000,,,,0100,, = 。 J,,007,300,,,,0007,30,,
14) 设原矩阵为A ,则
,,10?00,,,,,0,1?00,,
,,00,?00,, = , ,E,A??????,,
,,000?,,1,,,,,100?0,,,
于是有一个阶子式 ,E,An,1
,10?00
,,1?00
n,1 (,1)M = , ,?????n,1
00?,10
00?,,1
所以的行列式因子为 ,E,A
D== „ = = 1, DD 12 n,1
,,10?00
,0,1?00
00,?00D = n??????
000?,,1
,100?0,
n(,,1)(,,a)(,,a)?(,,a) = =, ,,112n,1
?,ann其中1, a,a,是个次单位根, 所以A的初等因子为 n,112
?,,,a, ,,a, ,,a, , ,,1n,112
故 A的若尔当标准形为
100?00,,,,0a0?00,,1,, 00a?00 =。 J,,2
??????,,
,,000?a0n-2,,,,000?0a,,n-1
上述矩阵的若尔当标准形也可用波尔曼公式求得,留给读者作为练习。 7. 把习
题
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6中各矩阵看成有理数域上矩阵,试写出它们的有理标准形。
120,,,,1)已知A = ,且 020,,
,,,2,2,1,,
,100,,,1,20,,,,,, ,,010 = , ,E,A0,20,,,,32,,,,22,1,,,,00,2,,2,,,,所以A的有理标准形为
00,2,,,, = 。 B101,,
,,012,,
131616,,,,2)已知 A = ,且 ,5,7,6,,
,,,6,8,7,,
,100,,,13,16,16,,,,,,,,010 = , ,E,A5,76,,,,32,,,,68,7,,,,00,,5,3,,,,
所以A的不变因子为
32 d(,),,,,,5,,3d(,),d(,),1, , 312
故A的有理标准形为
00,3,,,, = 。 B105,,
,,01,1,,
308,,,,3,163)已知 A = ,且 ,,
,,,20,5,,
,100,,,30,8,,,,,, ,,0,,10 = , ,E,A,3,1,6,,,,2,,,,20,5,00,,2,,1,,,,
所以A的不变因子为
2 d(,),,,2,,1d(,),1d(,),,,1, , , 312
故A的有理标准形为
,100,,,, = 。 B00,2,,
,,01,1,,
45,2,,,,4)已知 A = ,且 ,2,21,,
,,,1,11,,
,,4,52,,100,,,,,,2,,2,1, = , ,E,A010,,,,3,,,,,,11,100(,1),,,,所以A的不变因子为
332 d(,),(,,1),,,3,,3,,1d(,),d(,),1, , 312
故A的有理标准形为
001,,,, = 。 B10,3,,
,,013,,
37,3,,,,5)已知 A = ,且 ,2,52,,
,,,4,103,,
,,3,73,,100,,,,,,2,,5,2, = , ,E,A010,,,,2,,,,00(,1)(,1),410,3,,,,,,所以A的不变因子为
232 d,(),,(,1,)(,1),,,,,,,1d(,),d(,),1, , 312
故A的有理标准形为
001,,,, = 。 B10,1,,
,,011,,
1,12,,,,6)已知 A = ,且 3,36,,
,,2,24,,
,,11,2100,,,,,,,,,,, = , ,E,A,3,3,600,,,,
,,,,,22,400(,2),,,,,,,
所以A的不变因子为
2 d(,),,,2,d(,),1d(,),,, , , 123故A的有理标准形为
000,,,, = 。 B000,,
,,012,,
11,1,,,,,3,337)已知 A = ,且 ,,
,,,2,22,,
,,1,11100,,,,,,,,,,, = , ,E,A3,3,300,,,,2,,,,22,2,00,,,,,
所以A的不变因子为
2 d(,),,d(,),1d(,),,, , , 312
故A的有理标准形为
000,,,, = 。 B000,,
,,010,,
,4210,,,,8)已知 A = ,且 ,437,,
,,,317,,
,,4,2,10100,,,,,,,,4,,3,7,010 = , ,E,A,,,,
3,,,,00(,2)3,1,,7,,,,,所以A的不变因子为
332 d(,),(,,2),,,6,,12,,8d(,),d(,),1, , 312
故A的有理标准形为
008,,,, = 。 B10,12,,
,,016,,
033,,,,9)已知 A = ,且 ,186,,
,,2,14,10,,
,100,,,3,3,,,,,,,,010 = , ,E,A1,8,6,,,,2,,,,,214,10,00,(,,1),,,,
所以A的不变因子为
232 d,(),,,(,1),,,2,,,d(,),d(,),1, , 312
故A的有理标准形为
000,,,, = 。 B10,1,,
,,01,2,,
830,14,,,,,6,19910)已知 A = ,且 ,,
,,,6,2311,,
,,8,3014,,100,,,,,,6,,19,19, = , ,E,A010,,,,3,,,,00,30,8,,,623,11,,,,
3d,(),,,30,,8, , d(,),d(,),1312所以A的不变因子为
故A的有理标准形为
008,,,, = 。 B10,30,,
,,010,,
3100,,,,,4,100,,11)已知 A = ,且 ,,7121,,,,,7,6,10,,
,,,31001000,,,,,,,,,,41000100,,,,, = , ,E,A,,,,,,,,,71210010,,,,4,,,,,761,000(,1),,,,
所以A的不变因子为
4432d(,),d(,),d(,),1d(,),(,,1),,,4,,6,,4,,1, , 1234
故A的有理标准形为
000,1,,,,1004,, =。 B,,010,6,,,,0014,,
1234,,,,0123,,12)已知 A = ,且 ,,0012,,,,0001,,
,,1,2,3,4,,,,,0,1,2,3,, = , ,E,A,,,00,1,2,,,,000,1,,,
4因为= ,=(三阶子式的公因式是零次多项式),所以A的不变DD1,E,A,(,,1)4 3
因子为
4432d(,),d(,),d(,),1d(,),(,,1),,,4,,6,,4,,1 , , 1234故A的有理标准形为
000,1,,,,1004,, =。 B,,010,6,,,,0014,,
1,303,,,,,26013,,13)已知 A = ,且 ,,0,313,,,,,1208,,
,,,130,8,,,,,2,60,13,, = ,E,A,,03,,1,3,,,,1,20,,8,,
1000,,,,0100,,,, ,,00,,10,,32,,000,,15,,33,,19,,
所以A的不变因子为
32d(,),1d(,),d(,),1d(,),,,15,,33,,19, ,, 3124
故A的有理标准形为
1000,,,,00019,, =。 B,,010,33,,,,00115,,
010?00,,,,001?00,,
,,000?00,,14)已知 A = ,且 ??????,,
,,000?01,,,,100?00,,
,,1
,,1
,?n2n,1nD,,(,1),,,1 = = =, ,E,An?,1
,,1
0,
又因为
,1
,,1n,1n(,1),,,1=, ??
,,1
所以D,1。这意味着A的不变因子为 n-1
n d,d,?,d,1d,, , ,,112n,1n
故A的有理标准形为
00?01,,,,10?00,,
,, = 。 B01?00,,
?????,,
,,00?10,,
.
1. A是n维线性空间上V的线性变换。
1) 若A在V的某基下的矩阵A是某多项式d(,)d(,)的伴侣阵, 则A的最小多项式是;
2)设A的最高次的不变因子是d(,)d(,), 则A的最小多项式是。
1) 设A在V的基,,,,?,,下的矩阵为A, 多项式 12n
n,1 d(,),,n,a,,?,a,,a, 1n,1n则d(,)的伴随矩阵为
00?0,a,,n,1,,10?0,a,,n,2
,, A ,01?0,an,3,,
?????,,
,,,00?1a1,,故A的不变因子为
d,d,?,d,1d,d(,), , 12n,1n这意味着d(,)d(,)是A的最高次的不变因子, 因此, 也是A的初等因子的最小公倍式, 从
而它是A的最小多项式,即证.。
2)由1)的证明即知A的最高次的不变因子是d(,), 也是A的最小多项式.。