$D_{2n} = \{ r, s: r^n = s^2 = e, srs = r^{-1} \}$ is the dihedral group with order $2n$.
I'm trying to classify the $2$-dimensional irreducible $\mathbb{C}$-representations of $D_{10}$ up to isomorphism. This page on the subwiki Groupprops says that there should be two such representations of the form
$$r = \begin{pmatrix} \cos(2 \pi k / 5) & -\sin(2 \pi k / 5) \\ \sin(2 \pi k / 5) & \cos(2 \pi k / 5) \end{pmatrix}, s=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad k = 1,2$$
which intuitively makes sense considering $D_{10}$ as the symmetries of a pentagon. However, I'm trying to derive this a priori with no knowledge of how many representations there are or what form they may be in. How do I derive a priori all possibilities for what $s$ could be, so I know that I have all the representations? I'd want to show that I've got all possibilities for $r$ and $s$, and then can work from there on showing there are two isomorphism classes.
I've asked a similar question here and gotten an answer, I just don't understand how we can derive this and show there are no more possibilities.
For $K$ a field if you have matrix representation
$\rho : G\to GL_n(K)$
then for any matrix $M\in GL_n(K)$
$\psi : G \to GL_n(K)$
given by $\psi(g) = M\rho(g)M^{-1}$ is also a representation which is said to be isomorphic to the earlier.
For a degree two representation suppose $s\to S$ and $r\to R$. Since $S^2 = I$ eigenvalues are $1$ and/or $-1$. If both eigenvalues of $S$ were $1$(or $-1$) then for some basis $S$ will be $I$(or $-I$). In this case $S$ commutes with $R$ $\implies$ $SRS = R = R^{-1}$ $\implies$ $R^2 \ I$ which together with $R^5 = I$ implies $R = I$ hence we are getting degree two reducible representation. Which means for getting irreducible representation two eigenvalues be $1$ and $-1$. Then for some basis of $K^2$ you can always have $S = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
So no matter which degree $2$ representation you are having, you can have $S = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ in both. Obviously the representation of $r$ will be different for different non-isomorphic representations. To find representations of $r$ you can once again use rotations, this time by $72°$ as we are working with $D_{10}$. Now let's denote this anticlockwise rotation by $R$ which in matrix form looks like
$R = \begin{pmatrix} \cos(2 \pi / 5) & -\sin(2 \pi / 5) \\ \sin(2 \pi / 5) & \cos(2 \pi / 5)\end{pmatrix}$
Let's call
$ S = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
As before, one obvious representation will be $s \to S$ and $r \to R$. Another will be $s \to S$ and $r \to R^2$ which is to say that applying $r$ rotates $\mathbb R^2$ twice by $72°$ or equivalently by $144°$. To see that these two are non isomorphic suppose contrary they were isomorphic. Which means there is $P\in GL_2(K)$ such that $ S = PSP^{-1}$ and $R^2 = PRP^{-1}$. But you can check that these two equations are not compatible with $R^5 = S^2 = I$ and $SRS = R^{-1}$.
For knowing how many and what type of irreducible representations are there priori you need to know that if there are $r$ non-isomorphic irreducible representations of $G$ with degree $d_i$ then $d_1^2 +\ldots+d_r^2 = |G|$. Moreover $r$ is same as the number of conjugacy classes of $G$. For small groups using these results you know beforehand the number and degrees of irreducible representations. To actually find them is a difficult task in general. For groups defined by using some geometry that geometry itself provides a tool to find some of the representations. For other groups there are other numerous techniques. But I don't know if there is any full proof technique that always works.