Editing Trace (linear algebra) (section)

===Trace of commutator===
When both {{math|'''A'''}} and {{math|'''B'''}} are {{math|''n'' × ''n''}} matrices, the trace of the (ring-theoretic) [[commutator]] of {{math|'''A'''}} and {{math|'''B'''}} vanishes: {{math|1=tr(['''A''', '''B''']) = 0}}, because {{math|1=tr('''AB''') = tr('''BA''')}} and {{math|tr}} is linear. One can state this as "the trace is a map of [[Lie algebras]] {{math|gl<sub>''n''</sub> → ''k''}} from operators to scalars", as the commutator of scalars is trivial (it is an [[Abelian Lie algebra]]). In particular, using similarity invariance, it follows that the identity matrix is never similar to the commutator of any pair of matrices.

Conversely, any square matrix with zero trace is a linear combination of the commutators of pairs of matrices.<ref group="note">Proof: <math>\mathfrak{sl}_n</math> is a [[semisimple Lie algebra]] and thus every element in it is a linear combination of commutators of some pairs of elements, otherwise the [[derived algebra]] would be a proper ideal.</ref> Moreover, any square matrix with zero trace is [[Unitary representation|unitarily equivalent]] to a square matrix with diagonal consisting of all zeros.