classification up to similarity of complex n-by-n matrices

Question

Classify up to similarity all 3 x 3 complex matrices $A$ such that $A^n$ = $I$.

Answer 1

If you define $p=X^n-1\in\mathbb C[X]$, then $p(A)=0$. This tells you that the minimal polinomial $m_A$ of $A$ divides $p$ and, in particular, that $m_A$ has all its roots simple, because the same is true of $p$.

It follows that $A$ is diagonalizable, so, up to similarity, you can suppose that it is diagonal.

Can you see which are the diagonal matrices $A$ which satisfy the condition $A^n=I$?

NB: This argument does not depend on your knowing about Jordan canonical forms.

Answer 2

In fact, one does not need to know the characteristic polynomial in this case. Let the minimal polynomial be $p$, then $p\mid (x^3-1)$. It is important to see that $x^3-1$ has three distinct roots in $\mathbb{C}$. Hence $p$ cannot have repeated roots in $\mathbb{C}$. Thus $A$ must be diagonalizable over $\mathbb{C}$, with each diagonal entry a root of $x^3-1$. Hence $A$ is similar to \begin{equation} \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}, \qquad a,b,c \in \{1,e^{i2\pi/3},e^{-i2\pi/3}\} \end{equation}

Answer 3

It is an Hoffman Kunze exercise problem. It will be for $3\times 3$ matrices $A$, $A^3=I$.

My answer is, the minimal polynomial of $A$ will divide $X^3-1=0$. Now $x^3-1=(x-1)(x-\omega)(x-\omega^2)$ where $\omega^3=1$. So the minimal polynomial can be of the forms

1. $m=x-a$
2. $m=(x-a)(x-b)$
3. $m=(x-a)(x-b)(x-c)$

Now if $m=(x-a)$,then characteristic polynomial of $A$ will be equal to $(x-a)^3$. Hence $A$ is similar to a diagonal matrix with all entries equal.

If $m=(x-a)(x-b)$, then characteristic polynomial of $A$ will be equal to either $(x-a)^2(x-b)$ or $(x-a)(x-b)^2$. Hence $A$ is similar to a diagonal matrix with two entries equal.

If $m=(x-a)(x-b)(x-c)$, then characteristic polynomial of $A$ will be equal to either $(x-a)(x-b)(x-c)$. Hence $A$ is similar to a diagonal matrix with all entries are unequal.