$5$-dimensional irreducible representation of $\mathcal{A}_5$ I know that the character table of $\mathcal{A}_5$ is the following:
$$\begin{array}{c|c|c|c|c|c|c|c|c}
 & 1 & 15 & 20 & 12 & 12\\ \hline
\mathcal{A}_5 & id & (12)(34) & (123) & (12345) & (12354)\\ \hline
\chi_{\text{triv}} &  1 & 1 & 1& 1 & 1  \\
\chi_{\text{std}} & 4 & 0 & 1 & -1 & -1\\
\gamma & 5 & 1 & -1 & 0 & 0\\
\chi_{1} & 3 & -1 & 0 & \frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2}\\
\chi_{2} & 3 & -1 & 0 & \frac{1-\sqrt{5}}{2} & \frac{1+\sqrt{5}}{2} \end{array}$$
The standard representation is irreducible since the action of $\mathcal{A}_5$ on $\{1,\dots,5\}$ is doubly transitive. The last two characters can be computed by using the fact that $\mathcal{A}_5$ is isomorphic to the isometry group of the Icosahedron.
My question is about the character $\gamma$. I think this character can be found as follows: if $V$ denotes the standard representation, the symmetric square $S^2V$ has character $\chi_{S^2V}=(10,2,1,0,0)$ which satisfies $\Vert\chi_{S^2V}\Vert^2=3$, so $S^2V$ is the sum of $3$ irreducible representations. Finally because $\langle \chi_{S^2V},\chi_{triv}\rangle=\langle \chi_{S^2V},\chi_{std}\rangle=1$, we can define 
$$\gamma:=\chi_{S^2V}-\chi_{std}-\chi_{triv}.$$ 
I would like to construct $\gamma$ differently.
I know that there is an isomorphism $\varphi:\mathcal{A}_5\to\mathrm{PSL}_2(\mathbb{F}_5)$ and we can use this and the fact that $\mathrm{PSL}_2(\mathbb{F}_5)$ acts doubly-transitively on the set of of lines in $\mathbb F_5^2$, i.e. $\mathbb P^1(\mathbb F_5)$ which has $6$ elements, to create a $5$-dimensional irreducible representation on $\mathcal{A}_5$. Let's call this representation $(W,\rho_W)$ and $\gamma_W$ the character of this representation. If we already know the character table of $\mathcal A_5$ we can say that $\gamma_W$ is equal to $\gamma$.
My question is: is it possible to compute $\gamma_W$ without using this fact ? 
Since the action of $\mathrm{PSL}_2(\mathbb{F}_5)$ on $\mathbb P^1(\mathbb F_5)$ is faithful, the morphism $\rho_W:\mathcal A_5\to GL(W)$ is injective and the order of $\rho_W(g)$ and $g$ is the same, for $g\in\mathcal A_5$, but I can't say more.
I am a beginner in representation theory so please correct me if there is anything wrong in my reasoning.

I just recall the lemma that I used several times:
Lemma: If a finite group $G$ acts transitively on a finite set $X$, then the permutation representation decomposes into $\Bbb CX = \operatorname{Vect}(e)\oplus V$ with $e:=\sum_{x\in X}x$ and  $V:=\lbrace \sum_{x\in X}\lambda_x x~\vert~\sum_{x\in X}\lambda_x=0\rbrace$ and $\rho_V$ is irreducible if and only if the action of $G$ on $X$ is doubly-transitive.
 A: I am not sure exactly what you are asking, but computing the values of $\gamma_W$ is easy. It is the permutation character of the permutation representation of degree 6 minus the trivial character, so for $g \in A_5$, $\gamma_W(g)$ is equal to one less than the number of fixed points of $g$ in this permutation representation.
Now the image of the representation is doubly transitive of degree $6$, so its point stabilizer has order $60/6=10$.
So elements $g$ of order $3$ fix no points and hence $\gamma_W(g)=-1$.
Elements $g$ of order $5$ clearly fix a unique point, so $\gamma_W(g)=0$.
Finally elements $g$ of order $2$ fix two points, so $\gamma_W(g) = 1$. To see that, you could either argue that since $A_5$ is simple, $g$ must induce an even permutation of the six points, and hence it must consist of exactly two transpositions. Alternatively, you could calculate it directly. The diagonal matrix in ${\rm SL}_2(5)$ with entries $2$ and $3$ induces an element of order $2$ in ${\rm PSL}_2(5)$, and you can calculate its action on the projective line.
