document.write( "Question 25058: Hello:\r
\n" ); document.write( "\n" ); document.write( "I am not sure how to answer this question.\r
\n" ); document.write( "\n" ); document.write( "If A and B are nxn matrices, then show that there are no square matrices A and B such that AB-BA=I
\n" ); document.write( "(where I is the Identity).\r
\n" ); document.write( "\n" ); document.write( "I know that if AB = I and BA = I then AB-BA cannot equal I. Sorry I am confused.\r
\n" ); document.write( "\n" ); document.write( "Thanks
\n" ); document.write( "

Algebra.Com's Answer #13443 by kev82(151)\"\" \"About 
You can put this solution on YOUR website!
Hi,\r
\n" ); document.write( "\n" ); document.write( "This is quite easy to prove by considering the trace of both sides. The Trace of a square matrix is simply the sum of it's diagonal elements. Hopefully you can see that if then , and that . If you can't see where these identities come from then please email back.\r
\n" ); document.write( "\n" ); document.write( "Ok, The matrix is given by the formula\r
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "\n" ); document.write( "and the trace of a matrix is \r
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "\n" ); document.write( "So the trace of is\r
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "\n" ); document.write( "Doing a similar calculation for we get\r
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "\n" ); document.write( "Now you can either acccept that these are the same expression ie or you can read my discussion in green of why we're assuming it is.\r
\n" ); document.write( "\n" ); document.write( "Now you've not told me what the elements of and are, if they were matrices(making A a matrix of matrices) then we would be in trouble, because we wouldn't have commutivity, but I'm gonna assume that we are working with some nice subfield of hence . I'm also assuming our matrices are finite which allows us to change the order of summation(else we need to worry about absolute convergence on compact subsets - what fun!) So now you are happy that \r
\n" ); document.write( "\n" ); document.write( "Well this is it, we're done. The trace of the LHS must be zero, and the trace of the RHS (the trace of the identity) definitly isn't zero, it's . So unless (doesn't make sense) there are no matrices that can satisfy this.\r
\n" ); document.write( "\n" ); document.write( "Hope that helps.\r
\n" ); document.write( "\n" ); document.write( "Kev
\n" ); document.write( "
\n" );