Proving S^n = 0 using Shift Operator Properties

[TranscendArcu]

Homework Statement

http://desmond.imageshack.us/Himg810/scaled.php?server=810&filename=screenshot20120131at923.png&res=medium

The Attempt at a Solution

So in particular I want to look at the last part of this problem. That is, "Show that S^n = 0"

I know that dim(KerS^k) = k and therefore, dim(ImS^k)= n-k. If k=n, dim(ImS^k)= n- n = 0 which implies that ImS^k = \left\{ 0 \right\}.

I'm having trouble drawing from this the conclusion that S^n = 0. Is S an isomorphism? If so, does that mean that the only thing that can map to 0 in the codomain is 0 in the domain?

 

[Dick]

You mean ImS^n = \left\{ 0 \right\}. That means for all vectors x, S^n(x)=0. Doesn't that mean S^n=0? Of course S isn't an isomorphism! It has a nonzero kernel. Not sure what's bothering you here.