Consider the square with vertices $(\pm 1, \pm 1)$. The symmetry group $D_4$ of this square acts by permuting the vertices
- Show that each permutation comes from a linear transformation. Compute the trace of this representation.
- Show that it is irreducible.
By comes from a linear transformation, does the question only demand that I argue that any action of $D_4$ is a combination of rotations and reflections and hence linear? Or, is there something else that the phrase comes from is supposed to mean?
Also, how do we compute the representation? What is the vector space corresponding to which we need to find the representation?
Disclaimer 1: In the mathematical field of representation theory, group representations describe abstract groups in terms of bijective linear transformations of a vector space to itself (i.e. vector space automorphisms); in particular, they can be used to represent group elements as invertible matrices so that the group operation can be represented by matrix multiplication.
Disclaimer 2: A representation of a group G on a vector space V over a field K is a group homomorphism from G to GL(V), (called the general linear group on V). That is, a representation is a map
$\rho \colon G\to \mathrm {GL} \left(V\right)$ such that $\rho (g_{1}g_{2})=\rho (g_{1})\rho (g_{2}),\qquad {\text{for all }}g_{1},g_{2}\in G.$
In the case at hand, where V is of finite dimension $n$, it is common to choose a basis for $V$ and identify $GL(V)$ with $GL(n, K)$, the so called group of n-by-n invertible matrices on the field K. Thus
$\rho \colon G\to \mathrm {GL} \left(n,K\right)$, i e. the matrix representation.
One can show there exist hom(e)omorphisms $\phi:G\to\mathrm{GL}_2(\mathbb R)$ which map $s$ to each of your matrices $s_s$ and $r$ to each of your matrices $r_r$, and one can verify that the eight matrices $s_s$ and $r_r$ (see below) satisfy the same relations as the generators $s$ and $r$.
Invertible matrices for rotation and reflection of regular $n$-gons:
Let $r_{k}$ be the rotation around $(0,0)$ of angle $\alpha_{k}:=\frac{k\cdot 2\pi}{n}$
and $s_{k}$ the reflection across the line throught $(0,0)$ angled $\frac{\alpha_{k}}{2}=\frac{k\cdot \pi}{n}$ against the positive x-axis.
You then have, for $n$-gon in general:
$$r_{k}=\begin{pmatrix}\cos(\alpha _{k})&-\sin(\alpha _{k})\\\sin(\alpha _{k})&\cos(\alpha _{k})\end{pmatrix}$$
$$s_{k}=\begin{pmatrix}\cos(\alpha _{k})&\sin(\alpha _{k})\\\sin(\alpha _{k})&-\cos(\alpha _{k})\end{pmatrix}$$
The group (re)presentation of such a Dihedral group $D_n$ is:
$D_n=⟨α,β:α^n=β^2=e,βαβ=α^{−1}⟩$
The group presentation of the Dihedral group $D_4$ is thus given by:
$D_4=⟨a,b:a^4=b^2=e,bab=a^{−1}⟩$
Alternatively, with $r,s$ from above, our Dihedral group $D_4$ has the group presentation:
$D_4=⟨r,s:r^4=s^2=e,srs=r^{−1}⟩$
That is, the dihedral group $D_4$ is generated by two elements $r$ (rotation) and $s$ (reflection) such that:
$(1):r^4=e$
$(2):s^2=e$
$(3):sr=r^{3}s$ which is $srs=r^{-1}$.
The eight elements (read: linear transformations) of the Dihedral group $D_{4}$ in matrix form then are:
\begin{aligned}r_{0}&={\bigl (}{\begin{smallmatrix}1&0\\0&1\end{smallmatrix}}{\bigr )},&r_{1}&={\bigl (}{\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}}{\bigr )},&r_{2}&={\bigl (}{\begin{smallmatrix}-1&0\\0&-1\end{smallmatrix}}{\bigr )},&r_{3}&={\bigl (}{\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}}{\bigr )},\\s_{0}&={\bigl (}{\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}}{\bigr )},&s_{1}&={\bigl (}{\begin{smallmatrix}0&1\\1&0\end{smallmatrix}}{\bigr )},&s_{2}&={\bigl (}{\begin{smallmatrix}-1&0\\0&1\end{smallmatrix}}{\bigr )},&s_{3}&={\bigl (}{\begin{smallmatrix}0&-1\\-1&0\end{smallmatrix}}{\bigr )}.\end{aligned}
These eight matrices are computed by inserting multiples of $\frac{\pi}{2}$ in the above formulae for $n$-gon, and they are representative of the eight group operations.
In two-dimensional space $R^2$, linear maps are generally described by 2 × 2 matrices, and above eight matrices are the eight mappings you are looking for. And you are thus quite right argueing that any action of D4 is a combination of rotations and reflections and hence linear.
What basis to choose?
Note: The set R$^2$ of the ordered pairs of real numbers is a vector space under the operations of component-wise addition and scalar multiplication; that is
$(a,b)+(c,d)=(a+c,b+d)$
$\lambda (a,b)=(\lambda a,\lambda b)$, where $\lambda$ is any real number.
A simple basis of this vector space consists of the two vectors $e_1 = (1, 0)$ and $e_2 = (0, 1)$.
These vectors form a basis (called the standard basis) because any vector $v = (a, b)$ of R$^2$ may be uniquely written as $\displaystyle {v} =a {e}_{1}+b {e}_{2}$.
However, any other pair of linearly independent vectors of R$^2$, such as $(1, 1)$ and $(−1, 2)$, forms a basis of R$^2$ as well. So long as you choose a pair of linearly independent vectors, you're good to go.
As for their traces:
The trace of a square matrix $A$, denoted $tr(A)$, is defined to be the sum of elements on the main diagonal (from the upper left to the lower right) of A.
In your example, $tr(r_0)=2$, $tr(r_2)=-2$, and all other traces are $0$.