?x1x2x3? 解 设A??x2x4x5??
??xxx?356? 因为?1?6对应的特征向量为p1?(1? 1? 1)T? 所以有
x1?x2?x3?6?1??1???A?1??6?1?? 即?x2?x4?x5?6 ???①? ?1??1???????x3?x5?x6?6 ?2??3?3是A的二重特征值, 根据实对称矩阵的性质定理知R(A?3E)?1? 利用①可推出
x3??11?x1?3x21?A?3E??x2x4?3x5?~?x2x4?3x5??
????xxx?3xxx?3?5656?3??3因为R(A?3E)?1? 所以x2?x4?3?x5且x3?x5?x6?3? 解之得
x2?x3?x5?1? x1?x4?x6?4?
因此
?411?A??141??
?114??? 21? 设a?(a1? a2? ???? an)T ? a1?0? A?aaT? (1)证明??0是A的n?1重特征值?
证明 设?是A的任意一个特征值? x是A的对应于?的特征向量? 则有 Ax??x?
?2x?A2x?aaTaaTx?aTaAx??aTax? 于是可得?2??aTa? 从而??0或??aTa?
设?1? ?2? ? ? ?? ?n是A的所有特征值? 因为A?aaT的主对角线性上的元素为a12? a22? ? ? ?? an2? 所以
a12?a22? ? ? ? ?an2?aTa??1??2? ? ? ? ??n?
这说明在?1? ?2? ? ? ?? ?n中有且只有一个等于aTa? 而其余n?1个全为0? 即??0是A的n?1重特征值?
(2)求A的非零特征值及n个线性无关的特征向量? 解 设?1?aTa? ?2? ? ? ? ??n?0?
因为Aa?aaTa?(aTa)a??1a? 所以p1?a是对应于?1?aTa的特征向量?
对于?2? ? ? ? ??n?0? 解方程Ax?0? 即aaTx?0? 因为a?0? 所以aTx?0? 即a1x1?a2x2? ? ? ? ?anxn?0? 其线性无关解为
p2?(?a2? a1? 0? ???? 0)T?
p3?(?a3? 0? a1? ???? 0)T? ? ? ??
pn?(?an? 0? 0? ???? a1)T?
因此n个线性无关特征向量构成的矩阵为
?1 22? 设A??0?0? 解 由
?a1?a(p1, p2, ???,pn)??2?????an42??34?? 求A100? 43???a2a1???0?????????????an?0?? ?????a1?1??42 |A??E|?0?3??4??(??1)(??5)(??5)?
043??得A的特征值为?1?1? ?2?5? ?3??5?
对于?1?1? 解方程(A?E)x?0? 得特征向量p1?(1? 0? 0)T? 对于?1?5? 解方程(A?5E)x?0? 得特征向量p2?(2? 1? 2)T? 对于?1??5? 解方程(A?5E)x?0? 得特征向量p3?(1? ?2? 1)T? 令P?(p1? p2? p3)? 则
P?1AP?diag(1? 5? ?5)??? A?P?P?1? A100?P?100P?1? 因为
?100?diag(1? 5100? 5100)?
所以
?121?1?50?5?P?1??01?2???012??
?021?5?0?21???????50?5??121??11??012? A100??01?2??51005?021??100??0?21?5???????1
?105100?1? ??051000??
?005100??? 23? 在某国? 每年有比例为p的农村居民移居城镇? 有比例为q的城镇居民移居农村? 假设该国总人口数不变? 且上述人口迁移的规律也不变? 把n年后农村人口和城镇人口占总人口的比例依次记为xn和yn(xn?yn?1)?
?x??x? (1)求关系式?n?1??A?n?中的矩阵A?
?yn?1??yn? 解 由题意知
xn?1?xn?qyn?pxn?(1?p)xn?qyn? yn?1?yn?pxn?qyn? pxn?(1?q)yn? 可用矩阵表示为
?xn?1???1?pq??xn??
?y??p1?q??y???n??n?1??1?pq?A???p1?q??
??因此
?x??0.5?? 求?xn??
(2)设目前农村人口与城镇人口相等? 即?0???????y0??0.5??y0??yn??x??x??x?n?x? 解 由?n?1??A?n?可知?n??A?0?? 由
?yn?1??yn??yn?|A??E|?1?p??q?(??1)(??1?p?q)?
p1?q??得A的特征值为?1?1? ?2?r? 其中r?1?p?q?
对于?1?1? 解方程(A?E)x?0? 得特征向量p1?(q? p)T? 对于?1?r? 解方程(A?rE)x?0? 得特征向量p2?(?1? 1)T? 令P?(p1, p2)???q?1?? 则 ??p1? P?1AP?diag(1? r)??? A?P?P?1? An?P?nP?1?
n?q?1??10??q?1?于是 A????0r??p1?
p1??????1?q?1??10??11?
?????n??p?q?p1??0r???pq?n?1q?prnq?qrn?1? ??p?prnp?qrn??
p?q??xn?q?prnq?qrn??0.5?1?? ????p?prnp?qrn??0.5? y?n?p?q????2q?(p?q)rn?1? ??2p?(q?p)rn??
2(p?q)?? 24? (1)设A?? 解 由
?3?2?? 求?(A)?A10?5A9? ???23?|A??E|?3???2?(??1)(??5)?
?23??得A的特征值为?1?1? ?2?5?
1(1, 1)T? 21(?1, 1)T?
对于?1?5? 解方程(A?5E)x?0? 得单位特征向量
21?1?1?? 使得P?1AP?diag(1? 5)???
于是有正交矩阵P??11?2?? 对于?1?1? 解方程(A?E)x?0? 得单位特征向量从而A?P?P?1? Ak?P?kP?1? 因此 ?(A)?P?(?)P?1?P(?10?5?9)P?1 ?P[diag(1? 510)?5diag(1? 59)]P?1 ?Pdiag(?4? 0)P?1
1?1???40?1?11?
?1???????2?11??00?2??11?11?? ??2?2?? ????2??11???2?2???
?212? (2)设A??122?, 求?(A)?A10?6A9?5A8?
?221??? 解 求得正交矩阵为
??1?31P???13?6?20?使得P?1AP?diag(?1? 1? 5)??? A?P?P?1? 于是 ?(A)?P?(?)P?1?P(?10?6?9?5?8)P?1 ?P[?8(??E)(??5E)]P?1
2?? 2??2?? ?Pdiag(1? 1? 58)diag(?2? 0? 4)diag(?6? ?4? 0)P?1 ?Pdiag(12? 0? 0)P?1
??11??1 ?6??2??1 ?2?1??2??3302??12???1?12??2??0???330? ???222?2?0?????1?2?1?2?? ?24?? 25? 用矩阵记号表示下列二次型: (1) f?x2?4xy?4y2?2xz?z2?4yz? 解
?121??x?f?(x, y, z)?242??y??
?121??z??????1?1?2??x?f?(x, y, z)??11?2??y??
??2?2?7??z????? (2) f?x2?y2?7z2?2xy?4xz?4yz? 解
(3) f?x12?x22?x32?x42?2x1x2?4x1x3?2x1x4?6x2x3?4x2x4?
解
?1??1f?(x1, x2, x3, x4)?2??1??113?22310?1??x1??2??x2?? 0??x3???1???x4?