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?

• The equation $(A-I)(A^2+A+I)=0$ does not imply that one of the factors is the zero matrix.
– daw
Sep 21 '16 at 6:13
• A real matrix can have only real entries, but can certainly have complex eigenvalues. That is the reason why although $x^2-x+1$ has only complex roots, it certainly can be the characteristic polynomial of a real matrix. Sep 21 '16 at 6:16
• One can take linear map $T:\mathbb{R}^3 \to \mathbb{R}^3$ defined by T(x,y,z)=(z,x,y). It satifies $T^3=I$ but $T\neq I$. Sep 21 '16 at 6:34
• "The equation (A−I)(A2+A+I)=0(A−I)(A2+A+I)=0 does not imply that one of the factors is the zero matrix." Sorry, I had no idea what I was doing XD. Is there any way to solve for all possibilities of A if given A^3 = I? Sep 21 '16 at 6:57
• @GeoffHuang Next time use the search feature first. There were already many solutions containing an example, like this. Sep 21 '16 at 10:52

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}

• Man you guys are all awesome. Sep 21 '16 at 6:52

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

• Man you guys are all awesome. Sep 21 '16 at 6:52

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:

• Yes, thanks. I'm a bit rusty, but thinking about it this way makes perfect sense now. Sep 21 '16 at 6:53
• Is that how Jacky Chong got his example? Sep 21 '16 at 7:01
• I imagine so, yes.
– MJD
Sep 21 '16 at 7:01

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.