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Apr 23, 2012 at 6:49 comment added Fedor Petrov Another antidote is the following: if $\det(B-A)=0$, it does not imply that $p(B)=0$. So why should it imply for $B=A$?
Apr 23, 2012 at 2:25 comment added Mariano Suárez-Álvarez @domenico: that's as close to a funny mathematical joke as we are going to get :D
Apr 22, 2012 at 4:44 comment added Angelo To Domenico: That's very good, I had never thought of that.
Apr 21, 2012 at 17:15 comment added domenico fiorenza This false proof is so good I've got used to proposing to my students $q(t)=tr(tI-A)$ as an antidote.
Apr 21, 2012 at 15:48 comment added Angelo It can be made into a correct proof in several ways; unfortunately, they all spoil the pristine elegance of the false proof.
Apr 21, 2012 at 15:46 comment added Martin Brandenburg Somehow this can be made into a correct proof with the Zariski topology.
Apr 21, 2012 at 14:38 history answered Angelo CC BY-SA 3.0