当前位置:首页 > 线性代数答案第三,四章 课后答案
T?a1?T?a2?A?B????aT?nTT??1???1T??b1??????TTT?b2???2???2?????A???B??
????????T?T?T????bn???s???s?因此 R?A?B??R?A??R?B? 12.证明 ?若B组线性无关 令B?(b1,?,br)A?(a1,?,as)则有B?AK
由定理知R(B)?R(AK)?min{R(A),R(K)}?R(K) 由B组:b1,b2,?,br线性无关知R(B)?r,故R(K)?r. 又知K为r?s阶矩阵则R(K)?min{r,s}
由于向量组B:b1,b2,?,br能由向量组A:a1,a2,?,as线性表示,则r?s ?min{r,s}?r
综上所述知r?R(K)?r即R(K)?r.
?若R(k)?r
令x1b1?x2b2???xrbr?0,其中xi为实数i?1,2,?,r
?x1???则有(b1,b2,?,br)????0
?x??r??x1???又(b1,?,br)?(a1,?,as)K,则(a1,?,as)K????0
?x??r????由于a1,a2,?,as线性无关,所以K?????x1??x2??0 ???xr??
?k11x1?k21x2???kr1xr?0?kx?k22x2???kr2xr?0?121?????????????即 ? (1)
?k1rx1?k2rx2???krrxr?0???????????????k1sx1?k2sx2???krsxr?0由于R(K)?r则(1)式等价于下列方程组:
?k11x1?k21x2???kr1xr?0??k12x1?k22x2???kr2xr?0 ?
???????????????k1rx1?k2rx2???krrxr?0k11由于
k21k22?k2r??kr1kr2?krr?0
k12?k1r?所以方程组只有零解x1?x2???xr?0.所以b1,b2,?,br线性无关, 证毕.
13.证明 集合V成为向量空间只需满足条件: 若??V,??V,则????V
若??V,??R,则???V
V1是向量空间,因为:
??(?1,?2,?,?n)T?1??2????n?0 ?1??2????n?0
T??(?1,?2,?,?n)T????(?1??1,?2??2,?,?n??n)
且(?1??1)?(?2??2)???(?n??n)
?(?1??2????n)?(?1??2????n)?0 故????V1
??R,???(?1,?2,?,?n)
??1???2?????n??(?1??2????n)???0?0故???V1
V2不是向量空间,因为:
(?1??1)?(?2??2)???(?n??n)
?(?1??2????n)?(?1??2????n)?1?1?2故????V2
??R,???(??1,??2,?,??n)
??1???2?????n??(?1??2????n)???1??
故当??1时,???V2 14.证明 设A?(a1,a2,a3)
0A?a1,a2,a31110111?(?1)0?111010101??2?0 1于是R(A)?3故线性无关.由于a1,a2,a3均为三维,且秩为3, 所以a1,a2,a3为此三维空间的一组基,故由a1,a2,a3所生成的向量空间 就是R.
15.证明 设V1?x?k1a1?k2a2k1,k1?R
3??V2??x??1?1??2?2?1,?1?R?
任取V1中一向量,可写成k1a1?k2a2, 要证k1a1?k2a2?V2,从而得V1?V2 由k1a1?k2a2??1?1??2?2得
?k1??k1??k2?k?2?k2?2?1??2??1?2?1?k1?k2??
?3?1??2???1??2?k1?3?1??2上式中,把k1,k2看成已知数,把?1,?2看成未知数
20D1??11?2?0 ??1,?2有唯一解
?V1?V2
同理可证: V112?V1 (?D2?10?0)
故V1?V2
12316.解 由于a1,a2,a3??111??6?0 032即矩阵(a1,a2,a3)的秩为3
故a31,a2,a3线性无关,则为R的一个基. 设v1?k1a1?k2a2?k3a3,则
?k?1?2k2?3k3?5?k1?2??k?1?k2?k3?0??k2?3 ??3k?2?2k3?7?k3??1故v1?2a1?3a2?a3 设v2??1a1??2a2??3a3,则
??1?2?2?3?3??9?k?1?3???1??2??3??8???k2??3 ??3?2?2?3??13??k3??2故线性表示为
v2?3a1?3a2?2a3
??1?8102???117..解 (1)A??245?1?初等行变换?~?0??386?2????0
041?34000?1??4?0??
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