I found this table on here:
$a$ is the first element of a dihedral group $D_n$, i.e. the rotation by an angle $2\pi/n$.
Few questions on that:
1) Non real, non orthogonal representation matrices.
Let's take a vector, say $\left ( \begin{array} & 1 \\ 0 \end{array} \right )$.
For a real, orthonogal matrix, we are in $\mathbb{R}^2$: $\left ( \begin{array} & \cos(2\pi k/n) & -\sin(2\pi k/n) \\ \sin(2\pi k/n) & \cos(2\pi k/n) \end{array} \right ) \cdot \left ( \begin{array} & 1 \\ 0 \end{array} \right ) = \left ( \begin{array} & \cos(2\pi k/n) \\ \sin(2\pi k/n) \end{array} \right ), $ which is fine ✅.
For a complex unitary matrix, we are in $\mathbb{C}^2$: $\left ( \begin{array} & e^{2\pi ik/n} & 0 \\ 0 & e^{2\pi ik/n} \end{array} \right ) \cdot \left ( \begin{array} & 1 \\ 0 \end{array} \right ) = \left ( \begin{array} & e^{2\pi ik/n} \\ 0 \end{array} \right ), $ so in $\mathbb{C}$ I get $\cos(2\pi k/n) + i\sin(2\pi k/n)$ which is the same as above so it's fine ✅.
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The 4th column is the issue.
What is $\mathbb{Q}(\cos(2\pi/n))$ and what does it mean?
If I take a vecor $\left ( \begin{array} & 1 \\ 0 \end{array} \right )$ (in what space??), and use the matrix from the table: $\left ( \begin{array} & 0 & -1 \\ 1 & 2\cos{2\pi k/n} \end{array} \right ) \cdot \left ( \begin{array} & 1 \\ 0 \end{array} \right ) = \left ( \begin{array} & 0 \\ 1 \end{array} \right )$, what does this mean ?
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2) Also, how does the minimal polynomial help/fit in findig the representation?
So you know that in this representation, a generator of the rotations part of the group is meant to act by rotating the plane by the angle $2 \pi k /n$. Writing that transformation down as a matrix with real coefficients requires a choice of basis of the plane $\mathbb{R}^2$.
If the standard basis $e_1, e_2$ is chosen, i.e., a unit vector along the $x$-axis and a unit vector along the $y$-axis, then we get the "real orthogonal" column of that table, which is the expected rotation matrix.
To get the other column, draw a regular unit $n$-gon in the plane, centred at the origin. Each vertex of the $n$-gon is a unit vector: choose two adjacent ones and call those $v_1$, $v_2$. Now, how is a rotation through $2 \pi / n$ written in this basis? Write $R$ for the rotation: we must have $R v_1 = v_2$ by our choice of basis. Some coordinate geometry should then tell you that, in terms of $v_1$ and $v_2$, $R v_2 = -v_1 + 2 \cos(2 \pi / n) v_2$. So in the basis $(v_1, v_2)$, the matrix of $R$ is $$ \begin{pmatrix} 0 & - 1 \\ 1 & 2 \cos(2 \pi / n) \end{pmatrix}$$ and taking this matrix to the power $k$ will give a rotation by $2 \pi k / n$.
So there is no difference in the representations: they are just the same representation written down using different bases. This is what it means to be isomorphic as representations. The characteristic polynomial of a matrix is invariant under change of basis, which could be why the table lists it as well.
Finally, $\mathbb{Q}(a)$ is notation which means the field extension of $\mathbb{Q}$ by the element $a$, where $a$ could be real or complex. Here the table is just noting what $\mathbb{Q}$ needs to be extended by to make the representation work, which here is just cosine of a particular angle. (Notably, we do not need the full power of the complex numbers).
Later addition: How can we derive the matrix $R$? Let $$M = \begin{pmatrix} \cos(2 \pi / n) & -\sin(2 \pi / n) \\ \sin(2 \pi / n) & \cos(2 \pi / n) \end{pmatrix}$$ be the rotation matrix. We essentially want to find a basis in which $M$ will only have rational numbers and $\cos(2 \pi / n)$. Let $v_1$ be any nonzero vector in the plane, and let $v_2 = Mv_1$ be the rotation applied to $v_1$. Then we want to compute $Mv_2 = M^2 v_1$ in the $(v_1, v_2)$ basis. $M$ satisfies its characteristic polynomial $x^2 - 2 \cos(2 \pi / n) x + 1 = 0$, and so we have $$M^2 v_1 - 2 \cos(2 \pi / n) Mv_1 + v_1 = 0$$ which we rearrange to see $$Mv_2 = -v_1 + 2 \cos(2 \pi / n) v_2$$ which gives the second column of the matrix $R$.