This question is a follow-up to my other question here:
If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.
My problem is as follows:
Given a vector space $V$ over $\mathbb{C}$, and a linear operator $T$ on $V$, show that if $T$ is invertible and $T^k$ is diagonalizable for some $k\geq2$, then $T$ is diagonalizable.
My attempt at a solution:
Because the vector space is over $\mathbb{C}$, $T$ has an eigenvalue (by the fundamental theorem of algebra). Let the eigenvalues of $T$ be denoted by $\lambda_1,\lambda_2,\cdots,\lambda_n$, for some $n\geq1$. Then, observe that for some eigenvector $v$,
$$Tv=\lambda_i v\Rightarrow\lambda_i (Tv)=\lambda_i(\lambda_i v)\Rightarrow T(\lambda_i v)=\lambda_i^2 v\Rightarrow T(Tv)=\lambda_i^2 v\Rightarrow T^2v=\lambda_i^2 v.$$
This argument can be inductively continued to show that the eigenvalues of $T^k$ are $\lambda_1^k,\lambda_2^k,\cdots,\lambda_n^k$. Because $T^k$ is diagonalizable, these eigenvalues are distinct (a proof that I'm omitting for brevity here). Because the $\lambda_i^k$'s are distinct, it follows that the $\lambda_i$'s are distinct. Therefore, $T$ is diagonalizable (again based on the proof that I'm omitting).
Here are my questions:
(a) First and foremost, is this a valid proof? I'm skeptical because it is so much simpler than the proof I've linked (which is in terms of matrices, not linear operators, but that shouldn't make a significant difference). I also haven't (explicitly) used the fact that $T$ is invertible, so I feel something is missing.
(b) It seems like this problem can be approached using Jordan canonical form, but I'm struggling to do so. Any suggestions would be appreciated.
Best Answer
Since (a) is sorted out by the comments, let me only comment on (b). The point here is that $T$ is invertible if and only if all Jordan blocks have a non-zero eigenvalue and it is diagonalizable if and only if all Jordan blocks have size one. If you bring $T$ to Jordan canonical form, then a matrix representation of $T^k$ is obtained from just taking the $k$-th powers of the Jordan blocks of $T$. Hence you can prove your claim by showing that if $J$ is a Jordan-block of size at least two with non-zero eigenvalue, then for each $k\in\mathbb N$, the matrix $J^k$ is not diagonalizable. But this is easily verified directly, since $J^k$ is always upper triangular with (equal) non-zero entries on the main diagonal and also the entries in the diagonal right above the main one are non-zero.