Prove that if the dual representation is irreducible then so is the representation itself. Prove that if the dual representation is irreducible then so is the representation itself.
I know how to prove the opposite direction, but this direction I can not, could anyone give me a hint for this??  
 A: Suppose $V$ is your representation and $V^*$ is irreducible. 
Let $W\subset V$ be an invariant subspace. Let $W^\bot = \{f\in V^* \mid \forall x\in W, f(x) =0\}$. Then if $f\in W^\bot$ and $x\in W$, $g\in G$, $g^{-1}\cdot x\in W$ so that $f(g^{-1}\cdot x) = 0$, and so $(g\cdot f)(x) = 0$ by definition of the action on $V^*$. 
Thus $g\cdot f \in W^\bot$ : $W^\bot$ is an invariant subspace of $V^*$. By irreducibility, this means $W^\bot = 0$ or $V^*$. 
If $W^\bot =0$, then this means that $W=V$ : indeed let $S$ be a complement of $W$ in $V$, then $W^\bot =0$ means that every linear form on $S$ is $0$, which implies that $S=0$.
If $W^\bot = V^*$, then that means $W=0$, because it means that there is no nonzero linear form on $W$. 
So any invariant subspace of $V$ is $0$ or $V$, in other words $V$ is irreducible
A: This is entirely symmetric. Suppose the dual representation $W^*$ is irreducible. Then by what you can prove, $W^{**}$ is irreducible. If $v\in W$, we define $v^{**}\in W^{**}$ to be the unique vector such that for all $f\in W^*$ we have $v^{**} (f) = f(v)$. 
If $g\in G$, then $(g\cdot v)$ maps to the unique vector $(g\cdot v)^{**}$ such that $(g\cdot v)^{**}(f) = f(g\cdot v)= (g\cdot f)(v) = (g\cdot v^{**})(f)$, so $(g\cdot v)^{**} = g\cdot v^{**}$. This proves that it is a homomorphism of representations. To prove injectivity and surjectivity, if $(v-w)^{**}=0$, then $f(v)-f(w) = 0$ for all $f$, hence $v=w$. The kernel is thus trivial, and it is surjective because it has the same dimension.
