第1个回答 2009-05-12
好久没有做抽代的题目了,现在拿第二题练一下笔吧!!(注意我严格证明了这个内积定义不合理的,不像有些人想当然认为就是,完全蒙人。但是又很奇怪既然不是怎么会接下来的第二问第三问呢?因为n=2时连反例都找出来了,至于n≥3的情况会不会就有该结论呢这个我就没有细想了)
1. for any pair A,B,let (A,B)=tr(AB),then we have
(1)any α∈K,(αA,B)=tr(αAB)=αtr(AB)=α(A,B)
(2)(A+B,C)=tr((A+B)C)=tr(AC+BC)=tr(AC)+tr(BC)=(A,C)+(B,C)
(3)when K denotation R,we need to check (A,B)=(B,A) that is tr(AB)=tr(BA),let C=AB,D=(BA),then
Cii=∑(AijBji),So tr(AB)=tr(C)=∑(Cii)=∑∑(AijBji) and(j ,i from 1 to n )
Dii=∑(BijAji),So tr(BA)=tr(D)=∑(Dii)=∑∑(BijAji) consider the sum of i first we have tr(BA)=∑∑(AjiBij),change the flow indicators i and j we have tr(BA)=∑∑(AijBji) So (A,B)=(B,A)
(4) for any A ∈R(n×n),we have (A,A)=tr(AA),let B=AA,
then Bii=∑(AijAji),tr(B)=∑(Bii)=∑∑(AijAji) that can not determine (A,A)≥0,for example n=2,A=[2,1;-5,-2]∈R(2×2),AA=-E,tr(AA)=-2<0,we can also have A∈S, so that can't define an inner product on S,Not to mention R(n×n).