Invertible linear operator with diagonalizable power is diagonalizable

Question

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).

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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.

Answer 1

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.