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Give an example of a 4x4 matrix where $A \neq I$, $A^2 \neq I$, and $A^3 = I$.

I found a 2x2 matrix where $A \neq I$ and $A^2 = I$, but this problem is more complex and has me completely stumped.

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Sure! Will certainly do. –  LearningPython Oct 3 '12 at 6:08
Fixed, accepted previous answers from those who have helped me. –  LearningPython Oct 3 '12 at 6:10
Does it have to be a real matrix? Have you heard of rotation matrices? –  Kevin Carlson Oct 3 '12 at 6:12
I think it can be any type of matrix. –  LearningPython Oct 3 '12 at 6:12
If $A\neq I$ and $A^3=I$, then it's not possible for $A^2$ to equal $I$. So it's not necessary to include the condition $A^2\neq I$. –  alex.jordan Oct 3 '12 at 22:29
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6 Answers

Here is a $2\times2$ example $$ \begin{bmatrix} -\frac12&\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&-\frac12 \end{bmatrix}\tag{1} $$ This works because it is a matrix representation of $e^{2\pi i/3}=-\frac12+i\frac{\sqrt{3}}{2}$; that is, a rotation by $\frac{2\pi}{3}$. Thus, squaring it gives another $2\times2$ example $$ \begin{bmatrix} -\frac12&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&-\frac12 \end{bmatrix}\tag{2} $$ $(1)$ and $(2)$ can easily be extended to $4\times4$ examples in many ways. Here is one using $(1)$: $$ \begin{bmatrix} -\frac12&\frac{\sqrt{3}}{2}&0&0\\ -\frac{\sqrt{3}}{2}&-\frac12&0&0\\ 0&0&\hspace{7pt}1\hspace{7pt}\vphantom{-\frac12}&0\\ 0&0&0&\hspace{5pt}1\hspace{5pt}\vphantom{-\frac12} \end{bmatrix}\tag{3} $$

Yet another $2\times2$ matrix:

As Hagen von Eitzen points out, we can consider the unit vectors $u,v,w$, which are separated by $\frac{2\pi}{3}$

$\hspace{5cm}$enter image description here

and note that $u+v+w=0$ to get that $w=-u-v$. Let $\begin{bmatrix}a\\b\end{bmatrix}$ be a coordinate vector using the basis vectors $\{u,v\}$. Then rotation by $\frac{2\pi}{3}$ has the following action: $$ \begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\mapsto\begin{bmatrix}v&w\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}0&-1\\1&-1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\tag{4} $$ Thus, the matrix for rotating by $\frac{2\pi}{3}$ under the basis $\{u,v\}$ is $$ \begin{bmatrix}0&-1\\1&-1\end{bmatrix}\tag{5} $$ which, as Gerry Myerson points out, is the companion matrix for $x^2+x+1$. A companion matrix is annihilated by its polynomial, so $(5)$ is annihilated by $x^3-1=(x-1)(x^2+x+1)$.

The square of $(5)$ also satisfies the specified conditions: $$ \begin{bmatrix}-1&1\\-1&0\end{bmatrix}\tag{6} $$ The answer given by Marc van Leeuwen uses $(6)$.

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"Unfortunately" this works only if we are allowed to divide by 2 and take the square root of 3. Thus this is not a solution for matrices with entries in $\mathbb Z$ or in $\mathbb F_2$. –  Hagen von Eitzen Oct 3 '12 at 12:37
"Fortunately" these restrictions were not imposed. As I have noted elsewhere, Gerry Myerson and Marc van Leeuwen have nice answers using only $0$ and $\pm1$. As fgp notes, the $2\times2$ real matrices can be extended to exponents other than $3$. –  robjohn Oct 3 '12 at 13:28
Well, since the problem statement did not specify that the characteristic is $\ne 2$, assuming that characteristic is $\ne 2$ seems to be invalid. Unless one takes the "matrices are always real by default" approach. Which may be valid, but uneducational. –  Hagen von Eitzen Oct 3 '12 at 13:33
@HagenvonEitzen: Your definition of "educational" strikes me as very different from the standard one. This answer is the most educational of the bunch, if you ask me (which is not a knock against the other examples). Universal and educational are not synonyms. If the requester had wanted an instance of this phenomena that worked over all commutative rings with unity, he could have specified as such. Without such a specification, the request to me implicitly leaves it up to the responder to choose the most appropriate setting. –  Cam McLeman Oct 3 '12 at 22:08
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Nobody has really highlighted the geometric interpretation yet, so I will. You're looking for a linear mapping (in 4 dimensions) that will take a vector $x$ to itself if applied three times in a row, yet will yield distinct intermediate vectors $Ax$, $A^2x$.

Let's start with the 2-dimensional case. There, it's easy to see that the mapping you want is a rotation. If $A$ represents the rotation by an $n$-th of a full circle (i.e, $360/n$ degree, or $2\pi/n$ red), then $A^n$ rotates by a full circle or equivalently not at all.

The rotation by $\varphi$ rad is given by $\begin{pmatrix} \cos\varphi & \sin\varphi \\ -\sin\varphi & \cos \varphi\end{pmatrix}$. Thus, $$A = \begin{pmatrix} \cos\frac{2\pi}{n} & \sin\frac{2\pi}{n} \\ -\sin\frac{2\pi}{n} & \cos\frac{2\pi}{n}\end{pmatrix}$$ is a mapping with $A^0 \ldots A^{n-1} \neq I$ and $A^n = I$.

To extend this to more than two dimensions, you can simply use the identity mapping for the other dimensions. Or, in the 4-dimensional case, you could also combine two rotation matrixes. The resulting matrix then has block-diagonal form with two $2x2$ blocks.

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This was essentially the idea of my answer. I just gave the example of $n=3$. –  robjohn Oct 3 '12 at 13:26
@robjohn Yeah, there are lots of good answeres here. I added mine because I wanted to demonstrate that this particular problem can be solved purely by using geometric intuition - it doesn't require any knowledge of matrix algebra (Well, apart from being able to write down a $2x2$ rotation matrix that is). –  fgp Oct 4 '12 at 18:56
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Very nice; another $2\times2$ example! –  robjohn Oct 3 '12 at 12:00
This is (in two dimensions) the rotation of the plane by 120° with respect to a basis with 120° between the basis vectors (of same length). –  Hagen von Eitzen Oct 3 '12 at 12:41
@HagenvonEitzen: It (the upper $2\times2$ block) is also the endomorphism induced by the cyclic permutation matrix given by Gerry Meyerson on the quotient by the obviously invariant subspace spanned by $(1,1,1)^\top$, with respect to the images in the quotient of the first two standard basis vectors. Or, what amounts to the same, the (inverse) companion matrix for $1+X+X^2$. –  Marc van Leeuwen Oct 4 '12 at 6:18
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Let $b_1,b_2,b_3,b_4$ be a basis. Define $A$ such that $$Ab_1=b_2, Ab_2=b_3, Ab_3=b_1$$ and $Ab_4=b_4$ and extend $A$ linearly. Then for scalars $\alpha_k$, $k=1,2,3,4$ $$A(\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4)=\alpha_1b_2+\alpha_2b_3+\alpha_3b_1+\alpha_4b_4 \ne\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4$$ and $$A^2(\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4)=\alpha_1b_3+\alpha_2b_1+\alpha_3b_2+\alpha_4b_4 \ne\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4$$ while $$A^3(\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4)=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4.$$

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This generalizes the answer given by Gerry Myerson. (+1) –  robjohn Oct 3 '12 at 7:20
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Since powers of diagonal matrices just raise the elements to those powers individually, we can simply pick a diagonal matrix such that all the diagonal elements are cube roots of $1$. If you're allowed complex matrices, the solution is easy.

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You can do the same thing with real matrices. –  robjohn Oct 3 '12 at 6:33
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Here's a $3\times3$: $$\pmatrix{0&1&0\cr0&0&1\cr1&0&0\cr}$$ You should be able to get a $4\times4$ out of this.

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It looks prettier than mine :-) (+1) –  robjohn Oct 3 '12 at 6:41
Maybe so, but you made the important point that it can be done with a $2\times2$. –  Gerry Myerson Oct 3 '12 at 7:37
Marc van Leeuwen has an aesthetic $2\times2$ example that uses $0$ and $\pm1$! –  robjohn Oct 3 '12 at 12:04
Yes, I should have thought of it - just take the companion matrix for $x^2+x+1$. –  Gerry Myerson Oct 3 '12 at 13:06
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