How to prove that this matrix is diagonalizable?

Question

I am trying assignment questions of Linear algebra and couldn't solve this particular question regarding diagonalizabilty.

Let $n \times n$ complex matrix $A$ satisfies $A^k = I$ the $n \times n $ identity matrix, where $k$ is a positive integer $>1$ and let $1$ not be an eigenvalue of $A$. Then how to prove that A is necessarity Diagonalizable?

As $A^k=I$ and 1 is not an eigenvalue so $(A-I ) (A^{k-1}+...+ I)=0$ implies that $(A^{k-1}+...+ I)=0$ but I am unable to move foreward .

Can you please help?

Answer 1

You could have assumed less. The condition "1 is not an eigenvalue of $A$" is unnecessary.

Recall that a complex matrix $A$ is diagonizable if and only if its minimal polynomial has no multiple roots. If $A^k = I$, then the minimal polynomial of $A$, which we denote by $f(x)$, necessarily divides $x^k - 1$. We conclude that $f(x)$ cannot have multiple roots since $x^k - 1$ has $k$ distinct roots in $\mathbb{C}$. Hence $A$ is diagonizable.

Alternatively, we may use Jordan Canonical Form to see the diagonizability. Suppose on the contrary $A$ is not diagonalizable, then there must exist some nontrivial Jordan block $B$ of the Jordan Canonical Form in the following form: $$\begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots& \ddots & \vdots \\ 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & \lambda \end{pmatrix}$$ which satisfies $B^k = I$. This is impossible by calculating, for example, the $(1,2)$-entry of $B^k$, since $\lambda \neq 0$.

Answer 2

There is a Theorem which says that If any square Matrix A is Diagonizable then any positive power of A ie A^k , k belong to Z+. A^k is also Diagonizable

BUT THE CONVERSE part is true Only if A is Invertible ie. If We are given that A^k is Diagonizable and A is Invertible Then A is Diagonizable . You can see the proof here If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.

Here We are given that A^k = I So A is Invertible and Identify is Always diagonizable So A is Diagonizable.

Hope This Would be Helpful for You

Answer 3

Let $A=SJS^{-1}$ be the Jordan normal form then $A^{k}=(SJS^{-1})^k=(SJ^{k}S^{-1})=I$. Then multiplying by $S^{-1}$ on the left and $S$ on the right yields $J^k=I$. If $J$ had a Jordan block $J_i$ of size $n>1$ corresponding to the eigenvalue $\lambda_{i}$, then $J_i^k$ would have super diagonal entries $(2\lambda_{i})^{(k-1)}\neq 0$.