Dimension1 Icosahedral
Group $G=A_5$ does not have 2-dimensional representation, its binary-extension $SL(2,5)$ has 2 2-dimensional representation, one representation is as follows:
\[\begin{align*} G=\langle \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}, \frac{1}{\sqrt{5}} \begin{pmatrix} -\zeta_5+\zeta_5^4 & \zeta_5^2-\zeta_5^3\\ \zeta_5^2-\zeta_5^3 & \zeta_5-\zeta_5^4 \end{pmatrix}, \begin{pmatrix} \zeta_5^3 & 0\\ 0 & \zeta_5^2 \end{pmatrix} \rangle. \end{align*}\]where $\zeta_5$ is a 5-th root of unity.
The Magma code is as follows:
FScale:=CyclotomicField(60); F:=FScale; Z5:=RootOfUnity(5); sq5:=2*Z5^3 + 2*Z5^2 + 1; s:=Z5-Z5^4; t:=Z5^2-Z5^3; G:=MatrixGroup<2,F| [0,1, -1,0], [-s/sq5,t/sq5, t/sq5,s/sq5], [Z5^3,0, 0,Z5^2]>;
The Burnside symbol is
\[\begin{align*} [\mathbb{P}^1 \circlearrowleft G]=&(1,G \circlearrowright k(x),())\\ &+(C_2,1 \circlearrowright k,(1))\\ &+(C_3,1 \circlearrowright k,(2))+(C_5,1 \circlearrowright k,(3)) \end{align*}\]where
\[\begin{align*} C_2=\langle \begin{pmatrix} 0 & \zeta_{10}^2\\ \zeta_{10}^3 & 0 \end{pmatrix} \rangle, C_3=\langle \frac{1}{5} \begin{pmatrix} -\zeta_{60}^{14}+\zeta_{60}^{12}-2\zeta_{60}^{6}+\zeta_{60}^{4}-2 & -3\zeta_{60}^{14}-2\zeta_{60}^{12}+4\zeta_{60}^{6}+3\zeta_{60}^{4}-1\\ 2\zeta_{60}^{14}-2\zeta_{60}^{12}-\zeta_{60}^{6}-2\zeta_{60}^{4}-1 & \zeta_{60}^{14}-\zeta_{60}^{12}+2\zeta_{60}^{6}-\zeta_{60}^{4}-3 \end{pmatrix} \rangle, C_5=\langle \frac{1}{5} \begin{pmatrix} 4\zeta_{60}^{14}+\zeta_{60}^{12}-2\zeta_{60}^{6}-4\zeta_{60}^{4}+3 & 2\zeta_{60}^{14}+3\zeta_{60}^{12}-\zeta_{60}^{6}-2\zeta_{60}^{4}+4\\ 2\zeta_{60}^{14}+3\zeta_{60}^{12}-\zeta_{60}^{6}-2\zeta_{60}^{4}-1 & \zeta_{60}^{14}-\zeta_{60}^{12}-3\zeta_{60}^{6}-\zeta_{60}^{4}+2 \end{pmatrix} \rangle, \end{align*}\]There are conjugation relations:
\[\begin{align*} (C_3,1 \circlearrowright k,(1))=(C_3,1 \circlearrowright k,(2))\\ (C_5,1 \circlearrowright k,(2))=(C_5,1 \circlearrowright k,(3)) \end{align*}\]Another projective linear representation of $A_5$ is given by
\[\begin{align*} G=\langle \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}, \frac{1}{\sqrt{5}} \begin{pmatrix} -\zeta_5+\zeta_5^4 & \zeta_5^2-\zeta_5^3\\ \zeta_5^2-\zeta_5^3 & \zeta_5-\zeta_5^4 \end{pmatrix}, \begin{pmatrix} \zeta_5^1 & 0\\ 0 & \zeta_5^4 \end{pmatrix} \rangle. \end{align*}\]where $\zeta_5$ is a 5-th root of unity.
The Magma code is as follows:
FScale:=CyclotomicField(60); F:=FScale; Z5:=RootOfUnity(5); sq5:=2*Z5^3 + 2*Z5^2 + 1; s:=Z5-Z5^4; t:=Z5^2-Z5^3; G:=MatrixGroup<2,F| [0,1, -1,0], [-s/sq5,t/sq5, t/sq5,s/sq5], [Z5^1,0, 0,Z5^4],>;
The Burnside symbol is
\[\begin{align*} [\mathbb{P}^1 \circlearrowleft G]=&(1,G \circlearrowright k(x),())\\ &+(C_2,1 \circlearrowright k,(1))\\ &+(C_3,1 \circlearrowright k,(2))+(C_5,1 \circlearrowright k,(4)) \end{align*}\]where
\[\begin{align*} C_2=\langle \begin{pmatrix} 0 & -\zeta_{10}^3\\ -\zeta_{10}^2 & 0 \end{pmatrix} \rangle, C_3=\langle \frac{1}{5} \begin{pmatrix} -\zeta_{60}^{14}+\zeta_{60}^{12}-2\zeta_{60}^{6}+\zeta_{60}^{4}-2 & 2\zeta_{60}^{14}-2\zeta_{60}^{12}-\zeta_{60}^{6}-2\zeta_{60}^{4}-1\\ -3\zeta_{60}^{14}-2\zeta_{60}^{12}+4\zeta_{60}^{6}+3\zeta_{60}^{4}-1 & \zeta_{60}^{14}-\zeta_{60}^{12}+2\zeta_{60}^{6}-\zeta_{60}^{4}-3 \end{pmatrix} \rangle, C_5=\langle \frac{1}{5} \begin{pmatrix} -2\zeta_{60}^{14}+2\zeta_{60}^{12}+\zeta_{60}^{6}+2\zeta_{60}^{4}+1 & -\zeta_{60}^{14}+\zeta_{60}^{12}+3\zeta_{60}^{6}+\zeta_{60}^{4}-2\\ 4\zeta_{60}^{14}+\zeta_{60}^{12}-2\zeta_{60}^{6}-4\zeta_{60}^{4}+3 & -3\zeta_{60}^{14}-2\zeta_{60}^{12}+4\zeta_{60}^{6}+3\zeta_{60}^{4}-1 \end{pmatrix} \rangle, \end{align*}\]There are conjugation relations:
\[\begin{align*} (C_3,1 \circlearrowright k,(1))=(C_3,1 \circlearrowright k,(2))\\ (C_5,1 \circlearrowright k,(1))=(C_5,1 \circlearrowright k,(4)) \end{align*}\]