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

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

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    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 $A^2\neq I$.2012-10-03

6 Answers 6

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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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    @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.2012-10-03
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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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    Yes, I should have thought of it - just take the companion matrix for $x^2+x+1$.2012-10-03
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$\begin{pmatrix}-1&1&0&0\\-1&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$

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    @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 $1+X+X^2$.2012-10-04
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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.2012-10-03
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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)2012-10-03
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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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    @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).2012-10-04