Let $A$ be a real $2×2$ matrix such that $A^6=I$. The total number of possibilities for the characteristic polynomial of $A$ is: Let $A$ be a real $2×2$ matrix such that $A^6 = I$ (where $I$ denote the identity $2×2$ matrix). The total number
of possibilities for the characteristic polynomial of $A$ is:
Annihilating polynomial is $x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1).$ Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial.
So, Possible CHaracteristic polynomials are $(x-1)(x+1),(x^2+x+1),(x^2-x+1)$.We can not take any 2-degree polynomial factor from the annihilating polynomial. Since complex roots occur in pairs. But the answer given is $5$. Can you help me where is my mistake?
 A: This statement is false:

Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial. 

$$f(t) = (t-1)(t^4+7t^3+105t-999)$$
Annihilates $A=I$. The characteristic polynomial of $A$ does not divide $f(t)$.
It seems like you may be confusing characteristic polynomial with minimal polynomial.
Without loss of generality, you may assume $A$ is diagonal (it is diagonalizable, and similar matrices have the same characteristic polynomial). Your matrix, up to similarity will be
$$\pmatrix{1 & 0 \\ 0 & 1}, \pmatrix{1 & 0 \\ 0 & -1}, \pmatrix{-1 & 0 \\ 0 & -1},$$
$$ \pmatrix{-\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i & 0 \\ 0 & -\dfrac{1}{2} - \dfrac{ \sqrt{3}}{2}i}, \pmatrix{\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i & 0 \\ 0 & \dfrac{1}{2} - \dfrac{ \sqrt{3}}{2}i}$$
And so it's characteristic polynomial (respectively) will be one of
$$(x-1)^2, (x-1)(x+1), (x+1)^2, x^2+x+1, x^2-x+1$$
A: To find the possibilities for the characteristic polynomial of an $n\times n$ matrix $A$, it helps to know some facts about the minimal polynomial of $A$. Let $c(x)$ be the characteristic polynomial of $A$, and let $m(x)$ be the minimal polynomial. Here are some facts about $m(x)$:


*

*The degree of $m(x)$ is at least 1, and the lead coefficient of $m(x)$ is 1.

*For any polynomial $p(x)$, if $p(x)$ annihilates $A$, then $m(x)$ divides $p(x)$. Hence, if a polynomial $p(x)$ has smaller degree than $m(x)$, then $p(x)$ won't annihilate $A$. This is why $m(x)$ is called the minimal polynomial.

*The Cayley-Hamilton theorem says that $c(x)$ annihilates $A$. So $m(x)$ divides $c(x)$. Hence if $A$ is $n\times n$, then the degree of $m(x)$ is at most $n$.

*$m(x)$ and $c(x)$ have the same roots. The roots of $m(x)$ are the eigenvalues of $A$. Note that even though $m(x)$ and $c(x)$ have the same roots, $m(x)$ and $c(x)$ may be different since the multiplicities of the roots may be different.
Now let's let $A$ be $2\times2$, and suppose $A^6=I$, so that $x^6-1$ annihilates $A$.
Then $m(x)$ is of degree at most 2, and $m(x)$ divides $x^6-1$. Hence $m(x)$ is either $x-1$, $x+1$, $(x-1)(x+1)$, $x^2-x+1$, or $x^2+x+1$.
$m(x)$ divides $c(x)$, and $m(x)$ and $c(x)$ have the same roots, so $c(x)$ must respectively be either $(x-1)^2$, $(x+1)^2$, $(x-1)(x+1)$, $x^2-x+1$, or $x^2+x+1$.
