Zareon Posted November 7, 2006 Author Share Posted November 7, 2006 Try to find a counterexample. Try real hard. Hint: There are no such beasts for matrices over the reals. I think you're missing the point I`m making. There's no such thing for matrices (that's what I've just proved), but I can't believe the same holds for a general ring. I will try to find a counterexample. Link to comment Share on other sites More sharing options...
weknowthewor Posted March 2, 2007 Share Posted March 2, 2007 if determinant of a matrix is root of 1, the matrix has a inverse Link to comment Share on other sites More sharing options...
Zareon Posted March 15, 2007 Author Share Posted March 15, 2007 I kinda forgot this topic, but here's the counterexample I promised. Consider the vector space l_2® consisting of the strings (x_1,x_2,...) with x_i real numbers. Then define the linear operators R and L (the Rightshift and the Leftshift) by: R(x_1,x_2,x_3,...)=(0,x_1,x_2,...) L(x_1,x_2,x_3,...)=(x_2,x_3,x_4,...) The set of linear operators on l_2 form a ring ofcourse and LR=1 but RL is not the identity. Link to comment Share on other sites More sharing options...
chri5 Posted February 23, 2018 Share Posted February 23, 2018 Zareon, in "Now suppose AB=I, then the equation Ax=b has a solution for any vector b. Just pick x=Cb, then Ax=A(Cb)=b" should it read "Just pick x=Bb"? Link to comment Share on other sites More sharing options...
John Cuthber Posted February 24, 2018 Share Posted February 24, 2018 2 hours ago, chri5 said: Zareon, in "Now suppose AB=I, then the equation Ax=b has a solution for any vector b. Just pick x=Cb, then Ax=A(Cb)=b" should it read "Just pick x=Bb"? Read the date of the post you are replying to. Link to comment Share on other sites More sharing options...
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