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.

if determinant of a matrix is root of 1, the matrix has a inverse

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.

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"? 

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.