Matrix Algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:17:28Z http://mathoverflow.net/feeds/question/103298 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103298/matrix-algebras Matrix Algebras zacarias 2012-07-27T11:58:55Z 2012-07-27T13:04:23Z <p>I'm reading the papaer "On the Reduction of a Matrix to Diagonal Form" of Epstein and Flanders (Amer. Math. Monthly 62, (1955). 168–171.</p> <p>Let $S$ denote the trace function.</p> <p>The authors stated that a well-known result in the theory of algebras of matrices is: </p> <p>A matrix algebra $\mathbb U$ over a field $\mathbb F$ of characteristice zero is semisimples if and only if $S(XY)=0$, for fixed $X\in\mathbb U$ and all $Y\in\mathbb U$,implies $X=0$.</p> <p>The authors do not give references. Does anyone know where I can find a proof of this result?</p> http://mathoverflow.net/questions/103298/matrix-algebras/103304#103304 Answer by Bugs Bunny for Matrix Algebras Bugs Bunny 2012-07-27T13:04:23Z 2012-07-27T13:04:23Z <p>Any book on Ring Theory covering Artin-Wedderburn's Theorem :-))</p> <p>To be fair, you need some dexterity in using it:</p> <p>If $U$ is not semisimple, then it has nonzero nilpotent ideal $I$. All elements $x\in I$ will have zero trace since $x^n=0$. Hence $S(Iy)=0$ for any $y$.</p> <p>In the opposite direction, let $I$ be the kernel of the form $=S(xy)$. Clearly, $I$ is an ideal and $S(x^n)=0$ for all $x\in I$ and natural $n$. In characteristic zero, this means that such $x$ is nilpotent, i.e. $I$ is a nil ideal. It must be zero, since $U$ is semisimple.</p>