victoranderson said:
The reason for the eigenvalue is zero implies dim (Ker D) = n+1
is because n+1-fold composition is the zero map?
I think I stuck in here so I cannot give a correct explanation
If a linear map is
nilpotent (ie, a finite iterate is the zero map), then one can show that the only eigenvalue is zero. That's part (d).
For part (e): there are a number of arguments one can run, but they all involve a proof by contradiction, in which one assumes that D is diagonalizable and shows that this is inconsistent with, for example, D(x) = 1.
There are various consequences of a matrix A being diagonalizable. Firstly, by definition, there exists an invertible Q such that QAQ^{-1} is diagonal.
But since the only eigenvalue of D is zero, we have that if it is diagonalizable then there is an invertible Q such that
<br />
QDQ^{-1} = 0<br />
and it should be obvious that this means that D = 0.
Secondly, if A is diagonalizable then there exists a basis of eigenvectors.
Again, the only eigenvalue of D is zero, so if D is diagonalizable then D(v) = 0 for all v \in \mathcal{P}_n (since then v is a linear combination of eigenvectors whose corresponding eigenvalue is zero). Thus again D = 0.
Either of these arguments is sufficient to establish that if D is diagonalizable then it is the zero map, which is equivalent to saying that \ker D = \mathcal{P}_n or that \dim \ker D = n+1.