TITLE: If a real $2\times2$ matrix satisfies $A^4=I$, does it follow that $A^2=\pm I$?
QUESTION [11 upvotes]: I would like to know if a  $2 \times 2$ matrix $A$ satisfies $A^4=I$, also satisfies $A^2=I$ or $A^2=-I$ if all elements of $A$ are real. 
Also, I would really appreciate your help on further generalizations. 
i.e) Properties of $A$ such that $A^n=I$.

REPLY [2 votes]: If $A^{2}\ne I$, then there is a non-zero vector $X$ such that $Y=(A^{2}-I)X \ne 0$. Then $(A^{2}+I)Y=0$. If $A^{2}\ne -I$, then $Y=(A^{2}+I)X \ne 0$ for some $X$, which gives $(A^{2}-I)Y=0$. Combining these, you get $Y_1,Y_2$ such that $A^{2}Y_1=Y_1$ and $A^{2}Y_2=-Y_2$. Using this as a basis,
$$
   A^{2}=P^{-1}\left[\begin{array}{cc}1 & 0\\0 & -1\end{array}\right]P.
$$
That's not possible because $\det(A^{2})=\det(A)^{2} \ne -1$. So, either $A^{2}=I$ or $A^{2}=-I$.