I am asked to prove that a $3 \times 3$ complex matrix satisfying $A^3 = I$ can be diagonalized.
I tried to study the eigenvalues of $A$, however this does not provide much information about if $A$ is diagnoalizable or is not,
Does anyone have another idea?
On
Let $\alpha = e^{2\pi i/3}$ so that $\alpha^{3}=1$. You have $$ (A-\alpha I)(A-\alpha^{2} I)(A-\alpha^{3}I)=0. $$ You can show that $$ 1 = \frac{(x-\alpha^{2})(x-\alpha^{3})}{(\alpha-\alpha^{2})(\alpha-\alpha^{3})}+\frac{(x-\alpha)(x-\alpha^{3})}{(\alpha^{2}-\alpha)(\alpha^{2}-\alpha^{3})}+\frac{(x-\alpha)(x-\alpha^{2})}{(\alpha^{3}-\alpha)(\alpha^{3}-\alpha^{2})}. $$ To see why this is true, notice that the first quadratic $p_1$ has values $\{1,0,0\}$ at $\{\alpha,\alpha^{2},\alpha^{3}\}$, the second $p_2$ has values $\{0,1,0\}$ and the third $p_3$ has values $\{0,0,1\}$ so that the sum of the three has values $\{1,1,1\}$. A quadratic that takes the value $1$ in three distinct places must be identically $1$. Therefore, $$ I = p_1(A)+p_2(A)+p_3(A). $$ Also notice that $A^{3}-I=0$ gives $$(A-\alpha I)p_1(A)=0 \\ (A-\alpha^{2}I)p_2(A)=0 \\ (A-\alpha^{3}I)p_3(A)=0. $$ Every vector $x$ can be written as $x=x_1+x_2+x_3$ where $Ax_1=\alpha x_1$, $Ax_2=\alpha^{2}x_2$ and $Ax_3=\alpha^{3}x_3$. You can then argue that there is a basis of eigenvectors of $A$, which makes $A$ diagonalizable.
In line with Noah Olander's hint, we will look at minimal polynomials.
Since $A^3 = I$, then we know that the minimal polynomial of $A$ divides $$x^3 - 1 = (x^2+x+1)(x-1)$$
Note that the roots of $x^3 -1$ are all distinct (They are actually the third roots of unity). That is, the minimal polynomial of $A$ can have no repeated roots.
A matrix is diagonalizable over the field $F = \mathbb{C}$ if and only if its minimal polynomial is a product of distinct linear factors over $\mathbb{C}$. Since the minimal polynomial of $A$ must have distinct linear factors over $\mathbb{C}$, $A$ must be diagonalizable.