Let $G$ be a group and let $V$ be a complex vector space which is a representation of $G$. Let's write the (left) action of $g\in G$ on $v\in V$ as $gv$.
The dual vector space of $V$ is the set of linear maps from $V$ to $\mathbb{C}$, and is written as $V^*$. I'll use the notation $(\phi,v)$ to mean $\phi(v)$ if $\phi\in V^*$ and $v\in V$.
The dual representation of $V$ is defined to be the vector space $V^*$ with the (right) $G$-action $(\phi g,v):=(\phi,gv)$ for $\phi\in V^*$, $v\in V$ and $g\in G$.
A representation of $G$ is irreducible if the only $G$-invariant subspaces are the zero subspace or the whole vector space.
I can show that if $V^*$ is an irreducible representation, then so is $V$. Indeed, if $W$ is a $G$-invariant subspace of $V$ then its annihilator $W^\perp=\{\phi\in V^*\colon w\in W\implies (\phi,w)=0\}$ is a $G$-invariant subspace of $V^*$, so either $W^\perp=V^*$, which clearly implies $W=0$, or $W^\perp=0$, which (by fiddling around with a Hamel basis) yields $W=V$.
Does the converse hold? That is, if $V$ is irreducible, must $V^*$ also be irreducible?
If $\dim V<\infty$ then $V^{**}$ is equivalent to $V$, so the answer is yes in this case. But I don't know what happens if $\dim V=\infty$.