# Invertible linear operator with diagonalizable power is diagonalizable

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.

• Could cite precisely what result are you omitting? You should be using invertibility somewhere. For example, nilpotent matrices such as Jordan blocks of eigenvalue zero are nilpotent (and so for a sufficiently high power, diagonalizable) and not diagonzalizable. Dec 6, 2018 at 0:53
• @GuidoA. mathworld.wolfram.com/DiagonalizableMatrix.html The diagonalization theorem gives that an $n\times n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors. These must correspond to $n$ distinct eigenvalues. And since an operator can be represented as a matrix... Dec 6, 2018 at 0:56
• Could you clarify your statement? The identity matrix $I_n \in M_n(\mathbb{k})$ seems pretty diagonalizable to me and has a unique eigenvalue. Dec 6, 2018 at 0:59
• @GuidoA. Hmmm. It's probably my mistake...but this seems to me that MathWorld has an error. Certainly $\mbox{dim }V$ eigenvalues implies diagonalizability, but the converse doesn't appear to be true, as suggested by MathWorld Dec 6, 2018 at 1:03
• Are you aware of the following fact: A linear operator $T$ on a finite-dimensional $\mathbb{C}$-vector space $V$ is diagonalizable if and only if there exists a polynomial $P$ satisfying $P\left(T\right) = 0$ and $P^\prime \neq 0$ (this means that the derivative of $P$ is not identically $0$ as a polynomial)? If you know this, I suggest you think about how to get such a polynomial for $T$ from such a polynomial for $T^k$. Dec 6, 2018 at 4:37

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.