/sci/ - Science & Math

>>11591199
Bingo.
You can also use the same trick to clean up the earlier parts of the proof.
So, for [math]i \neq j[/math], we have [math]f(E_{ij}) = f(E_{ii}E_{ij}) = f(E_{ij}E_{ii})=f(0)=0=\lambda ~ trace(E_{ij})[/math] for any complex [math]\lambda[/math]. With the case [math]i=j[/math] you've just proved, this completes the proof across the whole base.