how to prove that two representations are irreducible $L_1:S_3 \to GL_2(F)$ that do this :
$$
L_1((12)) =\begin{pmatrix}0&1\\1&0\end{pmatrix} ,\\
 L_1(1,2,3)=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}
$$
$L_2:S_3 \to GL_2(F)$
$L_2(1,2)= \begin{pmatrix}0&1\\1&0\end{pmatrix}$
$L_2(1,2,3)= \begin{pmatrix}0&1\\-1&-1\end{pmatrix}$
I have to prove that L1 & L2 are irreducible in any field .
I'm already proved that $L_1$ isomorphic to $L_2$  , I know that in field C if there are two isomorphic representations then they are irreducible but how can we prove that in any field other than C these two representations are irreducible?
 A: We want to show that $L_1$ is not reducible. $L_1$ is a representation over $F^2$, In order for $L_1$ to be reducible, $L$ would have to have non-trivial (and hence one-dimensional) sub-representation. In other words, if $L_1$ is reducible, then there must be a non-zero vector $v \in F^2$ such that $L_1(g)(v) \in \operatorname{span}(v)$ holds for every $g \in S_3$.
So, suppose for the purpose of contradiction that $v$ is such a vector. There must exist a $\lambda$ for which
$$
L_1((1\ \ 2)) v = \lambda v.
$$
However, the only eigenvalues of $L_1((1\ \ 2))$ are $\lambda = 1, -1$.
Suppose first that $v$ is an eigenvector of $L_1((1\ \ 2))$ associated with $\lambda = 1$. This means that $v$ has the form $v = t \cdot (1,1)$ for some non-zero $t \in F$. We then compute
$$
L_1((1\ \ 2\ \ 3)) v = \pmatrix{-t\\ 0}.
$$
However, there is no $\lambda$ such that $\lambda \cdot v = (-t,0)$, which contradicts the requirement that $v$ is an eigenvector of $L_1((1\ \ 2\ \ 3))$.
Similarly, if $v$ is an eigenvector of $L_1((1\ \ 2))$ associated with $\lambda = -1$, then we have $v = t \cdot (1,-1)$ for some $t \in F$. We then compute
$$
L_1((1\ \ 2\ \ 3)) v = \pmatrix{t\\ 2t}.
$$
However, there is no $\lambda$ such that $\lambda\cdot v = (t,2t)$, which contradicts the same condition again.
Thus, $L_1$ has no non-trivial sub-representation, which is to say that $L_1$ is irreducible.
