Linear Algebra: If $A^3 = I$, does $A$ have to be $I$? 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?
 A: Do you know that 2×2 matrices correspond to linear transformations of the plane, and that composing the transformations multiplies the corresponding matrices?
Can you think of a linear transformation of the plane which, repeated three times, is the identity transformation?
Mouse over for hint:

 How about a rotation?

A: 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.
A: Consider the matrix
\begin{align}
A=
\begin{pmatrix}
\cos \frac{2\pi}{3} & -\sin\frac{2\pi}{3}\\
\sin\frac{2\pi}{3} & \cos \frac{2\pi}{3}
\end{pmatrix}
\end{align}
A: An example of a matrix $A$ such that $A^3 = I$ is
$$
A =
\left(
\begin{matrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{matrix}
\right)
$$
