(a) Let $A\in M_{5\times 5}(\mathbb{C})$ s.t: $$dim~N(A) =3$$ $$dim~N(A-I)=2$$ $$p_A(x)=x^2(x-1)^3$$
(b) Let $A\in M_{5\times 5}(\mathbb{C})$ s.t $$dim~N(A) =3$$ $$dim~N(A-I)=0$$ $$p_A(x)=x^4(x-1)$$
I have been asked if A is diagonable, and I think no on both (a) and (b) because the details are not possible, on (a) I can see that the $0$ is eigenvalue with multiplicity of 2 but it appear to have 3 eigenvectors.
in (b) it apear that 1 is eigenvalue with no eigenvectors
Noting that algebraic multiplicity of an eigenvalue is greater than or equal to its geometric multiplicity, we have
(a) : $$p_A(x)=x^2(x-1)^3$$
Algebraic multiplicity of $0 =2 <3$(geometric multiplicity of 0), which is not possible. Hence diaginalization not possible in this case.
(b) : You are right. There is no eigenvector (non zero) corresponding to eigenvalue 1. Hence we have 3+1=4 eigenvectors in total but we require 5 for diagonalization and hence diagonalization is not possible.