?1?r3?(?1)?r4?0???????0??000?1100100110因此, 为使方程有解, 必须有a1+a2-b1-b2=0, 这时有a2=b1+b2-a1. 方程有一个自由变元x4, 令x4=t, t为任意常数, 则x1=a1-b2+x4=a1-b2+t, x2=b2-t, x3=a2-t, 写成向量形式, 就是
??b2?b1?b2?a1??a2?a1?b1?b2?
a1?b2
#. 设三维向量空间里的两个基底分别为α1,α2,α3与β1,β2,β3, 且
?x1??a1?b2?t??a1?b2??1??x??b?t??b???1????2??t??X??2???2?x3??a2?t??a2???1?????????t??0??1? ?x4????1??1??2???2?2?1?3?2?2?3?????3??2?123 ?3
1) 若向量ξ=2β1-β2+3β3, 求ξ对于基底α1,α2,α3的坐标; 2) 若向量η=2α1-α2+3α3, 求η对于基底β1,β2,β3的坐标.
解: 将两个基底拼成按列分块的矩阵, 即令A=(α1,α2,α3), B=(β1,β2,β3), 则A与B均为三阶方阵. 则按题意知A与B的关系为
其中则 1)
?121???ACB?(?1,?2,?3)?(?1,?2,?3)??133????022??
?121??C???133????022??
即ξ对于基底α1,α2,α3的坐标为3,4,4
2)
由B=AC知A=BC-1, 先求C-1如下:
?2??2??2???B??1??AC??1???2?1??2?3?3?(?1,?2,?3)??1??????????3???3???3??
?121??2??3??3????1??A?4??(?,?,?)?4??3??4??4??A??133123??123??????????022????3???4???4??
?121100??121100???133010??r?054110?1?r2??????????022001???022001??
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r2?r3?1??121100?r2?(?2)?r2?10?110r2?(?5)?r3?r2?(1/2)????????011001/2???????011001/2??????054110???00?111?5/2??
r3?(?1)r3?r1?1000?13/2?r3?(?1)?r2????????01011?2????001?1?15/2?? ?0?13/2??C?1??11?2?????1?15/2?? 求出
则有
?2?????2?1??2?3?3?(?1,?2,?3)??1????3????2??2??2?????????1??1?A??1??BC??1??B?C??1????????3?3???3??????
因此η对基底β1,β2,β3的坐标为11/2, -5, 13/2.
# 求如下矩阵的特征值和特征向量:
?0?1?B??11???1?111??1?5?223/2??2??11/2??11/2???1??B??5??(?,?,?)??5??2?123?????????5/2????3???13/2???13/2??13??32
解: (注: 对于三阶以上矩阵, 没有多少可以解出特征值的好办法, 通常是尝试0,1,2,-1,-2这几个值是否特征值, 通过这样的尝试找出一个特征值之后, 通过因式分解将多项式化为二次方程再解余下的两个根). 1) 特征方程为
??34?A????2?1?; 1)
02??3?2?4??1??0?A???26?2A?12????????4?23??; 3) ?3?a?22a?? 2)
det(A??I)??3??24?1???(3??)(1??)?8?3?4???2?8
解出两个特征值为:
??2?4??5?0?1,2即两个特征值λ1=1, λ2=-5,
对λ1=1, 解齐次线性方程组
?1?4?42?4?1?536???2???2?3??22??5
??4x1?4x2?0??2x1?2x2?0, 容易看出方程有一个自由变元x, 令x=t为任意常数, 则x=x=t, 因此
2
2
1
2
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?x1??1??x??t?1???, 则求得λ1=1对应的特征向量为t(1,1)T. 通解为?2?对λ2=5, 解齐次线性方程组
?2x1?4x2?0??2x1?4x2?0, 此方程也有一个自由变元x2, 令x2=t为任意常数, 则x1??2x2??2t
?x1???2t???2??x???t??t?1????, 则求得λ=5对应的特征向量为t(-2,1)T 因此通解为?2??2
2) 特证方程为
?4r?(?1)?r7??0?7??31det(A??I)??26???2??26???2?4?23???4?23??
c?c?101c1?43?1002?(??7)?26???2?(??7)?424?4??4?23??因此特征值为λ1=λ2=7, λ3=-2. 对于特征值λ1=λ2=7, 解齐次方程
?1??
?100?100c3?(?1)?c2?(??7)?428?4??4??4(??7)2111?4?7???1???41?1??
?100c2?(?1)?c3??4(??7)2110??4(??7)2(??2)?41?2??
?4?2?4?8??4x1?2x2?4x3?0???2x1?x2?2x3?0??4x?2x?4x?0123?
3??对系数矩阵作行初等变换,
方程有两个自由变元x2,x3, 令x2=s, x4=t, s,t为任意实数， 则
??4?2?4?r1?(?1)?r3??4?2?4???2?1?2??r?0?1?(?1/2)?r2??????00??????00???4?2?4???0? 11x1??x2?x3??s?t22
写成向量形式有
?x1???1??1/2???1?2s?t??x???s??s?1??t?0???2?????????x3????0????1??, ?t??因此特征值λ1=λ2=7对应的特征向量为s(-1/2,1,0)T, t(-1,0,1)T. 对于特征值λ3=-2, 解下面的齐次方程
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对系数矩阵作行初等变换
?5x1?2x2?4x3?0???2x1?8x2?2x3?0??4x?2x?5x?0123?
有一个自由变元x3, 令x3=t为任意常数, 则 x1=x3=t, x2=(1/2)x3=(1/2)t, 写成向量形式, 得
?5?2?4?r1?r2?1?41?r1?(?5)?r2?1?41???28?2??r?5?2?4??r?018?9?1?4?r31?(?1/2)?????????????????????4?25????4?25???0?189?? r2?r31??1?4?10?1?r2?(1/18)?r2?4?r1?????????01?1/2??????01?1/2?????0?0??00??00?
因此特征值λ3=-2对应的特征向量为t(1,1/2,1).
3) 特征方程为
?x1??t??1??x???1t??t?1/2??2??2?????x3????t????1??
1??det(A??I)??1031r2?(?1)?r11??1??2?0?a?22a??3020?(1??)1???a?2022a??
因此A的三个特征值为λ1=1, λ2=2, λ3=2a-1. 对于特征值λ1=1, 解齐次方程
?110r1?3?r3?(??1)01??2?(??1)01??23?a?22a??0?a?12a?? 1??2c2?(1/2)?c12??2??(??1)??(??1)?a?12a??1??/22a??
12r1?(?1/2)?r212?(??1)(??2)?(??1)(??2)1/22a??02a?1??
??(??1)(??2)(??2a?1)?0
对其系数矩阵作初等行变换,
2x3?0??2x3?0??3x?(a?2)x?(2a?1)x?023?1
02?r1?r3?0?3?a?22a?1??0??r?0?2?(?1)?r302?????02??????00??3?a?22a?1???0?
有一个自由变量x2, 令x2=t为任意常数, 则x3=0, x1=(1/3)(a+2)x2-(2a-1)x3=(1/3)(a+2)t, 写成向量形式, 得
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即对于应特征值λ1=1的特征向量为t((a+2)/3,1,0)T. 对于特征值λ2=2, 解齐次方程
?1?x1??13(a?2)t?3(a?2)???t?1??x???t????2????0?x3???????0??
对系数矩阵作初等行变换,
?2x3?0??x1??x2?2x3?0??3x?(a?2)x?(2a?2)x?023?1
02?0??1??1?0??r?01?3?r3?12?????1??????3?a?22a?2???0?a?2??102?r1?(?1)?1r2?(?a?2)?r3?r2?(?1)??????????0?12????????0??00??0??0?2??2a?4?? 0?2?1?2??00??
2方程有一个自由变量x3, 令x3=t为任意常数, 则x1=x2=2x3=2t, 写成向量形式, 得
即对应于特征值λ2=2的特征向量为t(2,2,1)T.
对于特征值λ3=2a-1, 解齐次方程
?x1??2t??2??x???2t??t?2??2???????x3????t????1??
对其系数矩阵作行初等变换
?(2?2a)x1?2x3?0??(2?2a)x2?2x3?0?3x?(a?2)x?x?023?1
这是为了方便起见使矩阵变成一个\倒\的阶梯形, 可以看出x1为自由变元, 令x1=t为任意常
数, 则x2=x1=t, x3=(a-1)x1=(a-1)t, 写成向量形式:
02?r1?(1/2)?1?1a01??2?2a?0??r?0?2?(1/2)2?2a2?????1?1a1???????a?21??a?21??3??3?
r2?(1/3)r3?(?1)?r1??2?aa?20?r2?(?a?2)?r1?000?r3?(?1)?r2?r2?(a?2)?r3??110?????????330??????????????a?21??3??1?a01??
因此, λ3=2a-1对应的特征向量为t(1,1,a-1)T.
# 已知A为n阶方阵且A2=A, 求A的特征值.
解: 设A的一个特征值为λ, 对应的特征向量为X, 则有AX=λX, 又将题意中的条件A2=A代
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?x1??t??1??x???t??t?1??2???????x3????(a?1)t????a?1??