Let $(ρ,V)$ be a representation of a two-dimensional Lie algebra $g.$ The bracket on $g$ is defined for the basis vectors by $[e_1, e_2] = e_2.$ If $\rho(e_1)$ is a nilpotent operetor, then $\rho(e_2)$ is the zero operator.
We have that all the eigenvalues of $\rho(e_1)$ are zero, and $ρ(e_1) ρ(e_2)- ρ(e_2)ρ(e_1)= ρ(e_2).$ Can you give me some hints? Thanks.
Let us define in every Lie algebra $[x^{\{k\}},y]$, $k\ge 1$, by induction: $$[x^{\{1\}},y]=[x,y], [x^{\{k+1\}},y]=[x,[x^{\{k\}},y]].$$.
Then in any ring with bracket operation $[x,y]=xy-yx$ we have $$[x^{\{2\}},y]=x^2y-2xyx+yx^2, [x^{\{k\}},y]=\sum_{i=0}^k l_i x^{i}yx^{k-i}, k\ge 1\qquad (1) $$ for some integers $l_i$.
Now in your case $[e_1,e_2]=e_2$, so $[e_1^{\{k\}},e_2]=e_2$. If you denote $x=\rho(e_1), y=\rho(e_2)$, then in the matrix ring you should have $[x^{\{k\}},y]=y$ for every $k\ge 1$. We know that for some $k$, $x^k=0$. Then by $(1)$ $y=[x^{\{2k\}},y]=0$ because every term in $(1)$ is $0$. Q.E.D.