A symplectic form $\beta$ on $V$, a finite-dimensional real vector space, is a nondegenerate skew-symmetric bilinear map $\beta:V\times V\longrightarrow\Bbb R$. We define the symplectic group of $V$ as the following subgroup of $\mathbf{GL}(V)$:
$$\mathbf{Sp}(V)\overset{\rm def}{=}\{g\in\mathbf{GL}(V)\,\colon\,\beta(gu,gv)=\beta(u,v)\,\forall\,u,v\in V \}$$
Clearly $\mathbf{Sp}(V)$ acts on $V$ by left multiplication. I am trying to prove that this representation is an irreducible representation of $\mathbf{Sp}(V)$.
I am not sure how to approach this. I am new to Lie groups and their representations. Do I need to use, somehow, the existence of a symplectic basis (a basis of $V$ of the following form)? $$\{e_i\}_{i=1}^n\cup\{f_i\}_{j=1}^n\quad \beta(e_i,e_j)=\beta(f_i,f_j)=0\quad\beta(e_i,f_j)=\delta_{ij}.$$
I think using this basis it is enough to prove that the standard action of $\mathbf{Sp}(2n,\Bbb R)$ which is when $V=\Bbb R^{2n}$ and $\beta$ is the symplectic form given by $$\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix}.$$
Any help would be appreciated. Thanks!
You should try to prove something a little bit stronger: for any non-zero $x\in V$, there is a symplectic basis with $e_1=x$.
Then if $x,y\in V$ are non-zero, you can choose symplectic bases $\mathcal{B}=(e_1,\dots,e_n,f_1,\dots,f_n)$ and $\mathcal{B}'=(e'_1,\dots,e'_n,f'_1,\dots,f'_n)$ with $e_1=x$ and $e'_1=y$.
Then show that if some $g\in GL(V)$ sends $\mathcal{B}$ to $\mathcal{B'}$, then actually $g\in Sp(V)$, and that gives you a $g\in Sp(V)$ such that $g\cdot x=y$.
Use that to deduce that $V$ is irreducible. (The point of this proof is the important fact that a symplectic space, unlike a quadratic space, is homogeneous: all points are equivalent.)