Padron my ignorance, I am a physicist who never really fully understood group theory but just used it when necessary.
Let's consider a $C_5$ group corresponding to $5$-fold rotation by $\frac{2\pi}{5}$ radians.
I need a fourth dimensional representation of this group. The way I would do this, is to find the irreps and then just stack them up in block diagonal form to form higher dimensional representations.
However, I know for a fact that the following is a valid representaion for the rotation $r$:
$$\mathcal{R}(r) = \left (\begin{array}{ccc} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ \end{array} \right ).$$
Which is the companion matrix to the polynomial $1 + t + t^2 + t^3 + t^4.$
If I take $r$ as the 5th root of unity, $r = e^{i\frac{2\pi}{5}}$, I can see that satisfies the same equation to the one for $t$ above (set to = 0). Which, from what I understand, would provide the algebra of the group?
Could someone please help in clearing my confusion about the relation, if any, between:
group element representation
companion matrix
the group algebra
The rotation $\tau$ satisfies the relation $\tau^5=1$. In fact, the group $C_5$ is generated by $\theta$ with a defining relation $\theta^5=1$. So to get a representation of $C_5$, it is equivalent to getting the linear map $T$ corresponding to $\tau$. This linear map must also satisfy $T^5=1$. Now, since $T^5-1=(T-1)(T^4+T^3+T^2+T+1)$, if $T$ is such that the second bracket is $0$, then certainly $T^5=1$ ( of course, you can also say take $T-1=0$, and that would give $T$ identity matrix, not that interesting.
Now, to get $T$ that satisfies $T^4+T^3+T^2+T+1=0$, you can do this: $T(e_i)= e_{i+1}$, for $i=1,2,3$. Now $T(e_4) = T(T^3 e_1)=T^4(e_1)$, so make this $-(T^3+T^2+T+1)(e_1)=-e_1-e_2-e_3-e_4$. Now you have a $T$ that satisfies $T^4+T^3+T^2+T+1(e_1)=0$, and since $e_i=T^{i-1}(e_1)$, it also satisfies $T^4+T^3+T^2+T+1(e_i)=0$, for $i=1,\ldots,4$, and so $T^4+T^3+T^2+T+1=0$.
The group $C_5$ is $\langle \tau \mid \tau^5=1\rangle$. So its group algebra over a field $k$ is $k[X]/(X^5-1)=k[X]/((X-1)(X^4+X^3+X^2+X+1))$. If the characteristic of the field is $0$ ( say $\mathbb{R}$ or $\mathbb{C}$) then we can write the group algebra as $k[X]/(X-1)\times k[X]/(X^4+X^3+X^2+X+1)$. If $k=\mathbb{C}$, the second factor further decomposes as $\prod_{l=1}^4 \mathbb{C}[X]/(X-\omega_l)$, where $\omega= e^{2\pi i/5}$.