Linear Algebra: If $A^3 = I$, does $A$ have to be $I$?

Question

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?

Answer 1

An example of a matrix $A$ such that $A^3 = I$ is $$ A = \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{matrix} \right) $$

Answer 2

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?

Answer 3

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}

Answer 4

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.