If A is an nxn matrix of complex numbers such that some power of A is identity, then i want to prove that A is diagonalizable over C

Question

If A is an nxn matrix of complex numbers such that some power of A is identity, then i want to prove that A is diagonalizable over C. How the proof will be followed from the jordan canonical forms?

Answer 1

Let $J$ be the JCF of $A$, and $S \in GL_n(\mathbb{C})$ such that $A=SJS^{−1}$. Then for some $k \geq 1$, we have

$$I_n=A^k=(SJS^{−1})^k=SJ^kS^{−1} \Rightarrow J^k = I_n.$$

Can you argue now that all of the Jordan blocks in $J$ are $1\times1$ blocks?

If we have

$$J=J_{r_1}(\lambda_1) \oplus \ldots \oplus J_{r_i}(\lambda_i)$$

Where $J_{r_{j}}(\lambda_{j})$ is the Jordan block of dimension $r_j$ corresponding to eigenvalue $\lambda_j$, then

$$J^k = J_{r_1}(\lambda_1)^k \oplus \ldots \oplus J_{r_i}(\lambda_i)^k$$

because of the way block matrices power up. Thus, we can study how a single Jordan block behaves when powered. You'll want to work out the details yourself, but a block $J_r(\lambda)$ of size $r$ for eigenvalue $\lambda$ satisfies

$$J_r(\lambda)^N = \begin{bmatrix} \lambda^N & {N \choose 1}\lambda^{N-1} & {N \choose 2}\lambda^{N-2} & \ldots & \ldots & \ldots & {N \choose r-1}\lambda^{N-r+1} \\ 0 & \lambda^N & {N \choose 1}\lambda^{N-1} & \ldots & \ldots & \ldots & {N \choose r-2}\lambda^{N-r+2}\\ \vdots & & & \ddots& & & \\ \vdots & & & & \ddots& & \vdots \\ \vdots & & & & & \ddots & \vdots\\ 0 & \ldots & \ldots & \ldots & \ldots & \ldots & \lambda^N\end{bmatrix}$$

You should be able to see that therefore the only way for us to have $J^k = I_n$ is if all Jordan blocks in $J$ are of dimension 1.