Let be a matrix with characteristic polynomial

A) Prove that the rank of is or .

B) If the rank of is 5 is diagonalizable?

A) I understand why rank must be less than because is clearly not signular since it has as an eigenvalue, however im having trouble actually proving that the rank must be or .

B) I think i may have this part. my thought is since Rank = Dim of the kernal must be . Which means that the geometric multiplicity of the eigenvalue is where as its algebreic multiplicity is and since they are not equal it is not diagonalizable. If i understand right, it would be diagonalizable if was of rank , correct?

Thanks for the help!