We need to prove T(nullS)⊂nullS
Take arbitrary x∈T(nullS)
Since x∈T(nullS), there exists y∈nullS such that x=Ty
Then Sx=STy. Since ST=TS, we have Sx=TSy
Sy=0, because y∈nullS, then Sx=TSy=T0=0, that is x∈nullS.
Since we take arbitrary x∈T(nullS), we have T(nullS)⊂nullS
Comments