Let $SU(2)= {\{\left( \begin{array}{cc} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{array} \right)\in\mathcal{M}_{2\times2}(\mathbb{C}):(\alpha,\beta)\in\mathbb{C}^{2},\ \alpha\bar{\alpha}+\beta\bar{\beta}=1 }\}$
$su(2)$ Lie algebra of SU(2).
$B(H_s)={\{f:H_s\longrightarrow H_s: \text{linear and continuous}}\}$
$H_s=\text{span}{\{v_k^s:k=0,1,...,s}\}$
$H:=\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)$$\ $$E:=\left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right)$$\ $$F:=\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)$
We will agree to $s\in\mathbb{N}$ the function $\psi_s:su(2)\longrightarrow B(H_s)$ defined by the actions:
$\psi_s(H)v_k^s=(s-k)v_k^s$, $\psi_s(E)v_k^s=\lambda_k v_{k-1}^s$, $\psi_s(F)v_k^s=\lambda_{k+1} v_{k+1}^s$, where $\lambda_k=\sqrt{k(s-k+1)}$.
Have many doubts, but the most important is, how can I show that $\psi_s$ is a homomorphism?
In other words what you want is to check that the map defined above is a lie algebra representation. Now from your basis, $[E,F]=H$,$[H,E]=2E$ and $[H,F]=-2F$. So you want to check the following three sets of equations for every $s$ and for every $k$:
$(\psi_s(E)\psi_s(F)-\psi_s(F)\psi(E))v_k^s=\psi_s(H)(v_k^s)$
$(\psi_s(H)\psi_s(E)-\psi_s(E)\psi(H))v_k^s=2\psi_s(E)(v_k^s)$
$(\psi_s(H)\psi_s(F)-\psi_s(F)\psi(H))v_k^s=-2\psi_s(H)(v_k^s)$
Hope you can do it now. It's just a matter of writing things down.