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

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Hint: Jordan decomposition. – Aaron Mazel-Gee Dec 9 '10 at 11:41
@Aaron Dear Aaron, you should post this as an answer. Then people will be less tempted to completely do the homework for Josh in another answer. And since this should actually suffice for Josh to do the exercise, he will be able to accept your answer or ask for further hints in the comments underneath it. – Alex B. Dec 9 '10 at 12:46
Alex, you're right. I've been hesitant to post answers like this because it feels like I'm whoring for points. But I guess comments can get you points too... – Aaron Mazel-Gee Dec 27 '10 at 9:03
up vote 9 down vote accepted

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.

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Thanks That is precisely my problem. I don't know how/what to conclude. – Josh Dec 9 '10 at 14:11
@Josh: Which are the $1\times 1$ diagonal matrices such that $A^5=I$? Make a list. When you are done: Which are the $2\times 2$ diagonal matrices such that $A^5=I$? Make a list ... If you keep going, I assure you that before you get to $10\times 10$ matrices you'll know the general answer. I, for one, will not tell you the answer to this :) – Mariano Suárez-Alvarez Dec 9 '10 at 14:37

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.

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You can format your formulas to a more readable form using dollars as in LaTeX. E.g. $3^{1+2}$ becomes $3^{1+2}$. See this guide for details: – Joonas Ilmavirta Apr 20 '15 at 14:10

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}

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