Matrix $A$ with $A^{3}=I$ and Canonical Forms

Question

Let $A \in M_{2}(\mathbb{Q})$ be a matrix which satisfies $A^{3}=I$, for $I$ being the identity matrix and $A$ different from $I$. Express $A$ in rational canonical form and in Jordan form as a matrix in $\mathbb{C}$.

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What would be a way to get started with this problem? I know that for the minimum polynomial of $A$, then one has $m_{A}(x)|x^{3}-1$ (as the minimum polynomial divides any other polynomial that makes $A$ vanish), and this polynomial factors into distinct factors in $\mathbb{C}$ (I think).

How could one use this information to write both canonical forms for a matrix satisfying this property? Not sure how to approach this.

Answer 1

$x^{3}-1=(x-1)(x^{2}+x+1)$, is the factorisation over $\mathbb{Q}[x]$. $m_{A}(x)$ has to divide $x^{3}-1$ and has degree at most $2$. Now, since $A$ is not the indentity, the minimal polynomial over $\mathbb{Q}$ is $x^{2}+x+1$. Now, you can factor $x^{2}+x+1$ over $\mathbb{C}$.

Answer 2

$$m_A(x) \mid x^3 - 1 = (x-1)(x^2+x+1)$$

The latter is the irreducible factorization of $x^3-1$ over $\mathbf{Q}$.

Since $A$ is $2 \times 2$ matrix, then $\deg m_A (x) \le 2$ by Cayley-Hamilton Theorem. Therefore, $m_A(x) = x-1$ or $m_A(x) = x^2 + x + 1$. The former would imply $A = I$ which by assumption cannot be, so the latter must be true. Then since the minimal polynomial is already degree 2, this completely determines the rational canonical form.

Over $\mathbf{C}$, which is algebraically closed, $x^3 - 1$ splits. Namely, if $\zeta_3$ is a primitive root of unity, then $$x^3 - 1 = (x - \zeta _ 3) (x - \zeta_3 ^2 ) (x-1)$$

In this form, it is clear that $x^3 - 1$ has no repeated roots and so the minimal polynomial also cannot have repeated roots. This is equivalent to saying that the Jordan Canonical Form is actually a diagonal matrix with such a minimal polynomial. Then its third power must be zero so all of its entries must be third roots of unity.