TITLE: Linear Algebra: If $A^3 = I$, does $A$ have to be $I$?
QUESTION [5 upvotes]: So it's been a while since I've taken Linear Algebra, but my friend asked me a question, that I couldn't answer.
If a matrix $A$ exists such that $A^3 = I$, does $A$ have to equal the identity matrix $I$?
My first instinct was to say no, but...
(edited out my incorrect math)
EDIT: thanks guys for the awesome examples
EDIT2: Followup question: Is there a way to solve for all possibilities of A if given A^3 = I?

REPLY [5 votes]: Why, the general answer is just as simple: consider a $120^\circ$ rotation around any axis.
In 2D, the possibilities are limited to Jacky Chong's answer, its transposition, and $I$.
In 3D, the general answer is $$\left(\begin{array}{c|c|c}
{3\over2}x^2-{1\over2} & {3\over2}xy+{\sqrt3\over2}z& {3\over2}xz-{\sqrt3\over2}y\\ \hline
{3\over2}xy-{\sqrt3\over2}z& {3\over2}y^2-{1\over2} & {3\over2}yz+{\sqrt3\over2}x\\ \hline
{3\over2}xz+{\sqrt3\over2}y& {3\over2}yz-{\sqrt3\over2}x & {3\over2}z^2-{1\over2} \\
\end{array}\right)$$
where $x,y,z$ are any numbers such that $x^2+y^2+z^2=1$. Note that when $(x,y,z)=({1\over\sqrt3},{1\over\sqrt3},{1\over\sqrt3})$, this produces Bolton Bailey's answer.
In 4D and beyond, things get a bit hairy.