Use property of eigenvalue space to prove that the differentiation operator D on the space P_n(n1) is not reducible (that is, it is not reduced by any non-trivial pair of complementary subspaces M and N)
Please help

Question

Use property of eigenvalue space to prove that the differentiation operator D on the space P_n(n>1) is not reducible (that is, it is not reduced by any non-trivial pair of complementary subspaces M and N)
Please help

loser66 · Answer

I have no idea to proceed it
@KingGeorge

kinggeorge · Answer

To be totally honest, I have no idea what it means for an operator to be reducible.

loser66 · Answer

it's if D in M , D is not in N. I can prove it if I use M is kernel (D) and N is Range (D)
pick any x(t) = a + bt, apply D x(t) to prove that D in both ker and Rang --> D is not reducible by complement of M, or N
but this problem relate to eigenvalue. don't know how to link

kinggeorge · Answer

I think this one has me stumped as well. I'm just a bit too unfamiliar with the definitions, and I too have no idea how it relates to eignvalues.

loser66 · Answer

it's ok, friend