Proof for $X^3=I$ $\implies$ $X$ is diagonalizable.

Question

Let $X$ be a $3\times 3$ complex matrix, and suppose $X^3=I.$ Then, show that $X$ is diagonalizable.

I searched the solution, and this seems to work, but there is a part I don't understand.

Proof

Let me use this proposition :

Let $A$ be $3\times 3$ matrix. If a polynomial $P(x)$ satisfies $P(A)=O$, then $P(x)$ is devided by minimal polynomial of $A$ ; $\varphi_A(x)$.

Now, for $P(x):=x^3-1=(x-1)(x-\omega)(x-\omega^2)$, I have $P(X)=X^3-I=O,$ so by the proposition, $P(x)$ is devided by $\varphi_X(x).$

Thus, $\varphi_X(x)$ doesn't have repeated roots, and therefore $X$ is diagonalizable.

---

I don't understand the part " Thus, $\varphi_X(x)$ doesn't have repeated roots."

What I know is that $P(x)$ can be devided by $\varphi_X(x)$, so I have $P(x)=\varphi_X(x)\cdot B(x)$ for some polynomial $B(x)$.

Why this leads the fact that $\varphi_X(x)$ doesn't have repeated roots ?

Postscript

I reffered to Ben's answer and Stephen's answer in $A^3 = I$. Find the possible Jordan Forms???

Answer 1

$X^3=I$ implies $x^3-1$ annihilates $X$. Let $m$ be minimal polynomial of $X$. By proposition, $m|(x^3-1)$. Polynomial $x^3-1$ split into linear factors as $$\left(x-1\right)\left(x+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right) \left(x+\frac{1}{2}+\frac{\sqrt{3}}{2}i\right).$$ Since $m$ divides $x^3-1$, we have $m$ is of form $(x-c_1)\cdots (x-c_k)$, where $c_i\in \Bbb{C}$ and $c_i\neq c_j$, if $i\neq j$. Note $1\leq k\leq 3$.

Answer 2

Since $\varphi_X(x)\mid P(x) $ , set of zeros of $\varphi_X(x)$ is contained in the set of zeros of $P(x) $.

As you have noticed , $P(x)=\varphi_X(x)\cdot B(x)$ for some polynomial $B(x) $

Suppose $\alpha$ is zero of $\varphi_X(x)$ , then $P(\alpha) =0$.

Now suppose $\alpha$ is a simple zero ( zero of multiplicity $1$) of $P(x) $ then $(x-\alpha) \mid P(x) $ and $(x-\alpha) ^n\nmid P(x) $ for $n\ge 2$ .In other words, if $\varphi_X(x)\mid P(x) $ then the multiplicity of $\alpha$ as a zero of $\varphi_X(x)$ can be atmost $1$.