symmetric group of equilateral triangle?
We can represent rotations of a equilateral triangle by matrices . Can we represent flips of a equilateral triangle by matrices ??
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$\begingroup$You can represent any lineal transformation with matrices, and flips are linear (assuming the center of the triangle is the origin).
Please note flips are usually called reflections.
$\endgroup$ 6 $\begingroup$If you wanted to use @J.W.Tanner's example of a reflection matrix, you need to position the vertices of the equilateral triangle accordingly.
Let the set of three vertices be $\{ \begin{bmatrix} 0 \\1 \end{bmatrix},\begin{bmatrix} \sqrt 3/2 \\ -1/2 \end{bmatrix},\begin{bmatrix} -\sqrt 3/2 \\ -1/2 \end{bmatrix}\}$. Then the matrix $M=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ acts as reflection in the $y$-axis (changing the sign of the $x$ coordinate). But we need six matrices to have a group of order six like the symmetry group of the equilateral triangle.
Note that the matrix that rotates counterclockwise about the origin by $2\pi/3$ (or $120^\circ$) is:
$$R=\begin{pmatrix} \cos (2\pi/3) & -\sin (2\pi/3) \\ \sin (2\pi/3) & \cos (2\pi/3 )\end{pmatrix}$$
This last matrix has multiplicative order three, $R^3 = I$, while a reflection matrix such as $M$ has multiplicative order two. Together the matrices $M$ and $R$ generate a group isomorphic to $S_3$, the symmetry group of the equilateral triangle.
In this representation of the symmetric group $S_3$ (every permutation of vertices corresponds to a placement of the equilateral triangle back onto itself) the rotations are $I,R,$ and $R^2$. The other three group elements are reflections, $M, MR,$ and $RM$. So both rotations and reflections can be realized as matrix transformations of the plane $\mathbb R^2$.
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