I am thinking about the following problem:
Let $V$ be an $n$-dimensional vector space over $K$ and let $T:V\rightarrow V$ be an endomorphism which has $n$ eigenvalues in $K$. Show that if there is a natural number $k\in \mathbb{N}$ with $T^{k+1}=T$ then $T$ is diagonalizable.
In this type of statement I don't know if I should consider the $n$ eigenvalues pairwise distinct or not necessarily. I understand that if it says that there are $n$ eigenvalues, then they are only considering the pairwise different eigenvalues in the counting. So, under this reasoning, if I consider the eigenvalues are pairwise distinct then the $n$ eigenvectors associated to each of those eigenvalues are linearly independent and then V has a basis of eigenvectors, therefore $T$ is diagonalizable. (Without using the fact that $T^{k+1}=T$ for some $k\in \mathbb{N}$.)
Please tell me if my reasoning is correct or not. Is it correct to consider the eigenvalues are pairwise different in this type of statement? Any suggestion or solution are welcome.
The problem is from Karlheinz Spindler's Abstract algebra with applications vol. 1, page 227
No, you don't know the eigenvalues are distinct, just there are $n$ eigenvalues in $K$ (counted with algebraic multiplicity). I think you also need to assume $\operatorname{char}K=0$. Use $T^{k+1}=T$ to get the min poly is a product of distinct linears, hence diagonalizable.