线性代数同济第四版答案第四章课后答案

??1????b1,a3??b2,a3?1?3? b3?a3?b1?b2????b1,b1??b2,b2?53?????4???1??0?故正交化后得 (b1,b2,b3)????1???1?13?12313?1??5?3?5? 3??5?4??5?

2．下列矩阵是不是正交阵:

??1?1?(1) ??2?1??3?121??3?1?; (2) ?2??1???1??9??8?9?4???9?8914??9?4?． ??97??9??11294?9解 (1) 第一个行向量非单位向量,故不是正交阵．

(2) 该方阵每一个行向量均是单位向量，且两两正交，故为正交阵．

3．设A与B都是n阶正交阵，证明AB也是正交阵． 证明 因为A,B是n阶正交阵，故A?1?AT，B?1?BT

T?1?1(AB)(AB)?BTATAB?BAAB?E 故AB也是正交阵．

4．求下列矩阵的特征值和特征向量:

?1(1)???2?1??1???; (2)?24??3?213?a1?3???a23?; (3)????6??a?n?????a1???a2?an?,(a1?0).

并问它们的特征向量是否两两正交? 解 (1) ① A??E?1??2?14???(??2)(??3)

故A的特征值为?1?2,?2?3．

② 当?1?2时,解方程(A?2E)x?0,由

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??1?1??11???1? 得基础解系???(A?2E)??P1????~????? 2??2?00??1?所以k1P1(k1?0)是对应于?1?2的全部特征值向量．

当?2?3时,解方程(A?3E)x?0,由

??2(A?3E)????2?1???1?~?2???0?1?1???? 得基础解系?P?2??2?0??1?所以k2P2(k2?0)是对应于?3?3的全部特征向量．

?1?3???③ [P1,P2]?PTP?(?1,1)??0 12?2?21??故P1,P2不正交．

1??21??3336?????(??1)(??9)

(2) ① A??E?23故A的特征值为?1?0,?2??1,?3?9．

② 当?1?0时，解方程Ax?0，由

?1?A??2?3?2133??3?6??~?1??0?0?2103???1????1?得基础解系P1???1?

?1?0????故k1P1(k1?0)是对应于?1?0的全部特征值向量.

当?2??1时,解方程(A?E)x?0，由

?2?A?E??2?3?2233??3?7??~?2??0?0?2003???1????1?得基础解系P2??1?

?0?0????故k2P2(k2?0)是对应于?2??1的全部特征值向量

当?3?9时，解方程(A?9E)x?0，由

??8?A?9E??2?3?2?833??3??3???1??0??0110~?1????1??2??1?1???得基础解系P3? ??2?2??0??????1?故k3P3(k3?0)是对应于?3?9的全部特征值向量．

??1???T③ [P1,P2]?P1P2?(?1,?1,1)?1??0，

?0???18

?1????2??1?T[P2,P3]?P2P3?(?1,1,0)?0， ?2????????1?????[P1,P3]?PT1P3?(?1,?1,1)????1??2?1??0， 2??1??所以P1,P2,P3两两正交．

a1??2a1a2a2???ana22????a1ana2an?an??2(3) A??E?a2a1?ana1

22 =?n??n?1(a12?a2???an)

22 ??n?1???(a12?a2???an)?

n??1?a?a???an21222n??i?1ai, ?2??3????n?0

2当?1??i?1ai时，

2?A??E?

222??a2?a3???an?a2a1??????ana1?a1a2?a1?a3???an?ana2222??????a2an?? ??222?a1?a2???an?1??a1an?a1??an?0?a2?初等行变换~?????

?0?an?an?1??0?00?取xn为自由未知量，并令xn?an，设x1?a1,x2?a2,?xn?1?an?1.

??an??0????0??00019

?a1????a2?故基础解系为P1???

????a??n?当?2??3????n?0时，

?a12??a2a1?A?0?E??????aa?n1a1a2a2?ana2?2??a1an??a2an????2?an?

?a1?初等行变换?0~?????0a20?0???an??0? ????0????? ????可得基础解系

??a2??a1P2??0?????0??an???a2??????0??0??,P??a?,?,P??0n??3?1???????????0????a1综上所述可知原矩阵的特征向量为

?a1??a2?P1,P2,?,Pn??????a?n?a2a1?0????an??0? ???a1??

?2?4?0??1?50????5．设方阵A???2x?2?与???0y0?相似，求x,y.

??4?2?00?4?1?????解 方阵A与?相似，则A与?的特征多项式相同，即

1??A??E????E??2x???2?4?2?1??5??000y??000?4???2?4

?x?4． ???y?5

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