Prove that $T^n$ is diagonalizable.

Question

Prove or give a counterexample: If $V$ is a complex vector space and $\text{dim V} = n$ and $T \in L(V)$, then $T^n$ is diagonalizable.

In order to show that $T$ is diagonalizable I need to show that I have $n$ distinct eigenvalues.

If I use the theorem that states that if $V$ is a complex vector space and $T \in L(V)$ then there is a basis consisting of the generalized eigenvectors of $T$, then is this sufficient, because I feel like this is more complicated than that? Any tips or help? Thank you!

Answer 1

A generalization of the previous answer:

for every $a$:

$$\left(\begin{array}{cc}1&a\\0&1\end{array}\right)^2=\left(\begin{array}{cc}1&2a\\0&1\end{array}\right).$$

RHS is usually not diagonalizable.

Answer 2

Possible counterexample, there may be a mistake: consider $A:=\displaystyle\left(\begin{array}{cc} i& 1 \\ 0& i \end{array}\right)$ which is in its Jordan Canonical Form and hence, non-diagonalizable. Notice that $A^2=\displaystyle\left(\begin{array}{cc} -1& 2i \\ 0& -1 \end{array}\right)$ which has Jordan Canonical Form $\displaystyle\left(\begin{array}{cc} 1& -1 \\ 0& 1 \end{array}\right)$, which isn't diagonalizable.

Let me know if there is a mistake.

Answer 3

Hint : Consider any non diagonal nilpotent matrix or non diagonal matrix with all the eigen values are same in general.