Spaces with operators notes

   Let E be a normed vector space.  Elements of L(E,E), here L(E,E) refers to a linear operator, are also called operators on E.  Let S  be a set of operators on E.  By an S-invariant subspace F we mean a subspace such that for every A in S we have AF is contained in F (i.e, if x is an element in F and A is an element in F, then Ax is an element of F). 

   An operator B is said to commute with S if AB = BA for all A an element of S.  If B commutes with S, then both the kernel of B and its image are S-invariant subspaces.  

Proof:  If x is an element of E and Bx = 0.  then ABx = BAx = 0 for all A an element of S, so the kernel of B is S-invariant.  Note: need to show if x is in ker B then Ax is also in ker B, but A(Bx) = B(Ax) = 0 means that  on the right side B(Ax) = 0 implies that Ax is in ker (B), this relation arises from the commutative relation AB = BA where the left side of our given equation provisions the A(Bx) = A(0) = 0 which ensures through associative properties that we have B(Ax) = 0 ).  We can see similarly the case for the im(B) being S-invariant.  Also recall the definition of kernel in this context meaning ker(B) = {x: Bx =0}.  It is also worth noting that the zero vector is linearly mapped to the zero vector (obvious statement I believe for normed vector spaces and linear maps...or here coupled with Banach spaces and their algebras).