= Solution
{c}
\b[There is a sign error in the printed assertion under the standard <sl2 triple> convention.] Work first with the operators exactly as printed. Since the $v_i$ form a <Hamel basis>, their prescribed images extend to <linear maps> on $V$; each vector is a finite <linear combination> of <basis vectors>.
For $i\geq1$, the <operator commutators> with $h$ are
$$
\begin{aligned}
[h,e]v_i&=\bigl(-2(i-1)(i-1)+2i(i-1)\bigr)v_{i-1}=2ev_i,\\
[h,f]v_i&=\bigl(-2(i+1)^2+2i(i+1)\bigr)v_{i+1}=-2fv_i.
\end{aligned}
$$
At $i=0$, the first identity holds because both sides vanish; the second holds because $hf(v_0)=-2v_1$ and $fh(v_0)=0$. On the other hand, for $i\geq1$,
$$
[e,f]v_i=i(i+1)v_i-i(i-1)v_i=2iv_i=-hv_i.
$$
At $i=0$, both $ef(v_0)=e(v_1)=0$ and $fe(v_0)=0$, so this identity also holds. Thus the actual relations are
$$
\boxed{[h,e]=2e,\qquad[h,f]=-2f,\qquad[e,f]=-h.}
$$
In particular, $[e,f]v_1=2v_1\ne-2v_1=hv_1$ in <characteristic zero>; the requested equality cannot hold as printed. Setting
$$
\boxed{E=-e,\qquad H=h,\qquad F=f.}
$$
corrects the sign and gives $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. These are the standard <sl2 triple> relations, so the standard <basis> of the <sl2 Lie algebra> acts on $V$ by $E,H,F$, giving a <Lie algebra representation>.
For completeness, these operators are <linearly independent>: applying $aE+bH+cF=0$ to $v_0$ gives $c=0$; applying it to $v_1$ then gives $b=0$; applying it to $v_2$ gives $a=0$. Hence the printed operators span a <Lie subalgebra> isomorphic to the <sl2 Lie algebra>, even though their printed normalization fails. Replacing the coefficient $i-1$ in the printed definition of $e(v_i)$ by $1-i$ is one equivalent way to repair the statement.
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