TITLE: Square root-related calculations with matrices
QUESTION [6 upvotes]: If $\mathbf A$ is an $n \times n$ matrix such that $\mathbf A^6 = \mathbf I_n$ (the identity matrix), is it true that either $\mathbf A^3 = \mathbf I_n$ or $\mathbf A^3 = \mathbf -I_n$?
I'm struggling to solve this question.
I re-wrote $\mathbf A^3$ as $\mathbf B$, which gives $\mathbf B^2 = \mathbf I_n$.
I do know that, if $\mathbf B^2 = \mathbf I_n$, then it is not necessarily true that $\mathbf B = \pm\mathbf I$ (I had a look at the following link, though I haven't fully understood the explanations given: If $A^2 = I$ (Identity Matrix) then $A = \pm I$).
Can I (and if so, how do I) apply this fact to the case where $\mathbf B$ is the cube of another matrix $\mathbf A$?

REPLY [2 votes]: The condition that $A^n=I$ is equivalent to saying that all the eigenvalues of $A$ are $n^{th}$ roots of $1$. So, in two dimensions, the only matrices squaring to $I$ but not equal to $I$ or $-I$ are those having eigenvalues of $1$ and $-1$ - hence a characteristic matrix of $x^2-1$. This means that
$$\pmatrix{a&&b\\c&&d}$$
squares to $I$ and is not $I$ or $-I$ if and only if:
$$(a-x)(d-x)-bc=x^2-1$$
meaning
$$ad-bc=-1$$
$$a+d=0.$$
An instance of this is Git Gud's example, but there are infinitely many such examples - in particular, any matrix of the form
$$\pmatrix{\lambda && \alpha \\ \frac{1-\lambda^2}{\alpha} &&-\lambda}$$
suffices for any chose of $\lambda$ and $\alpha$.