[proofplan] We first classify an arbitrary irreducible finite-dimensional module by finding a genuine highest weight vector. The generalized eigenspace decomposition for the operator $h$ gives a maximal generalized $h$-weight, and the raising relation forces $e$ to vanish on that generalized eigenspace. From a genuine eigenvector in it, the lowering string under $f$ is computed using the $\mathfrak{sl}_2$ commutation relations; finite-dimensionality forces the highest weight to be a nonnegative integer and the string to span the whole irreducible module. Finally, for each nonnegative integer $m$, we check the defining relations on the displayed basis and prove irreducibility from the lowering string, giving existence and uniqueness. [/proofplan]

[step:Choose a highest weight vector in an arbitrary irreducible module]Let $V$ be a nonzero finite-dimensional irreducible $\mathfrak{sl}_2(\mathbb C)$-module. For $\lambda\in\mathbb C$, define the generalized $h$-eigenspace \begin{align*} V[\lambda]:=\{v\in V:(h-\lambda I_V)^N v=0\}, \end{align*} where $N:=\dim_{\mathbb C}V$ and $I_V:V\to V$ is the identity map. By the generalized eigenspace decomposition for a linear operator on a finite-dimensional complex [vector space](/page/Vector%20Space), the operator $h:V\to V$ has at least one eigenvalue, its set of eigenvalues is finite, and $V$ is the [direct sum](/page/Direct%20Sum) of its nonzero generalized $h$-eigenspaces. From \begin{align*} [h,e]=2e \end{align*} we get \begin{align*} (h-(\lambda+2)I_V)e=e(h-\lambda I_V) \end{align*} for every $\lambda\in\mathbb C$. Hence $e(V[\lambda])\subset V[\lambda+2]$. Choose $\lambda\in\mathbb C$ such that $V[\lambda]\ne 0$ and $V[\lambda+2]=0$, which is possible because only finitely many generalized eigenspaces are nonzero along every chain $\lambda+2\mathbb Z_{\ge 0}$. Then $e(V[\lambda])=0$. The restriction of $h-\lambda I_V$ to $V[\lambda]$ is nilpotent, so its kernel is nonzero. Choose $v_0\in V[\lambda]$ with $v_0\ne 0$ and \begin{align*} (h-\lambda I_V)v_0=0. \end{align*} Thus \begin{align*} hv_0=\lambda v_0. \end{align*} Also, \begin{align*} ev_0=0. \end{align*} So $v_0$ is a highest weight vector of highest weight $\lambda$.[/step]

[guided]We need a vector killed by $e$ and simultaneously an honest eigenvector for $h$. Since $h$ may not be diagonalizable at the start of the proof, we work first with generalized eigenspaces. Let $N:=\dim_{\mathbb C}V$, and for each $\lambda\in\mathbb C$ define \begin{align*} V[\lambda]:=\{v\in V:(h-\lambda I_V)^N v=0\}, \end{align*} where $I_V:V\to V$ is the identity map. By the generalized eigenspace decomposition for a finite-dimensional complex operator, $h:V\to V$ has at least one eigenvalue, its set of eigenvalues is finite, and $V$ is the direct sum of its nonzero generalized $h$-eigenspaces. This is the spectral input that lets us choose a top generalized weight later. The relation $[h,e]=2e$ means $he-eh=2e$, and rearranging gives \begin{align*} he=e(h+2I_V). \end{align*} Equivalently, for every $\lambda\in\mathbb C$, \begin{align*} (h-(\lambda+2)I_V)e=e(h-\lambda I_V). \end{align*} Raising this identity to the $N$-th power gives \begin{align*} (h-(\lambda+2)I_V)^N e=e(h-\lambda I_V)^N. \end{align*} Therefore, if $v\in V[\lambda]$, then \begin{align*} (h-(\lambda+2)I_V)^N ev=e(h-\lambda I_V)^N v=0, \end{align*} so $ev\in V[\lambda+2]$. This proves that $e$ shifts generalized $h$-weights upward by $2$. Because there are only finitely many nonzero generalized eigenspaces, we may choose a nonzero generalized weight space $V[\lambda]$ at the top of one of these upward chains, so that $V[\lambda]\ne 0$ and $V[\lambda+2]=0$. Then the shift property gives \begin{align*} e(V[\lambda])\subset V[\lambda+2]=0, \end{align*} so $e$ kills every vector in $V[\lambda]$. It remains to make the vector an actual eigenvector rather than merely a generalized eigenvector. On the nonzero finite-dimensional space $V[\lambda]$, the operator $h-\lambda I_V$ is nilpotent by definition of generalized eigenspace. A nilpotent operator on a nonzero finite-dimensional vector space has nonzero kernel. Hence there exists $v_0\in V[\lambda]$ such that $v_0\ne 0$ and \begin{align*} (h-\lambda I_V)v_0=0. \end{align*} Since $e$ kills all of $V[\lambda]$, the same vector also satisfies $ev_0=0$. Thus \begin{align*} hv_0=\lambda v_0. \end{align*} Also, \begin{align*} ev_0=0. \end{align*} This is the desired highest weight vector.[/guided]

[step:Compute the lowering string generated by the highest weight vector] For each $i\in\mathbb Z_{\ge 0}$, define \begin{align*} w_i:=f^i v_0\in V. \end{align*} Using $[h,f]=-2f$, induction on $i$ gives \begin{align*} hw_i=(\lambda-2i)w_i. \end{align*} We also claim that for every $i\ge 1$, \begin{align*} [e,f^i]=i f^{i-1}(h-i+1). \end{align*} For $i=1$, this is $[e,f]=h$. If it holds for $i$, then using $[A,BC]=[A,B]C+B[A,C]$ gives \begin{align*} [e,f^{i+1}]=[e,f^i]f+f^i[e,f]. \end{align*} Substituting the induction hypothesis and using $hf=f(h-2)$ gives \begin{align*} [e,f^{i+1}]=i f^{i-1}(h-i+1)f+f^i h. \end{align*} Thus \begin{align*} [e,f^{i+1}]=i f^i(h-i-1)+f^i h=(i+1)f^i(h-i). \end{align*} This proves the identity. Since $ev_0=0$ and $hv_0=\lambda v_0$, the identity gives, for every $i\ge 1$, \begin{align*} ew_i=e f^i v_0=i f^{i-1}(h-i+1)v_0=i(\lambda-i+1)w_{i-1}. \end{align*} [/step]

[step:Use finite dimensionality to force the highest weight to be a nonnegative integer] The nonzero vectors among the $w_i$ are $h$-eigenvectors with pairwise distinct eigenvalues $\lambda-2i$. Since $V$ is finite-dimensional, not all $w_i$ can be nonzero. Let $r\in\mathbb Z_{\ge 0}$ be the largest integer such that $w_r\ne 0$. Then \begin{align*} w_{r+1}=0. \end{align*} Applying $e$ to this equality and using the formula from the previous step gives \begin{align*} 0=ew_{r+1}=(r+1)(\lambda-r)w_r. \end{align*} Because $r+1\ne 0$ in $\mathbb C$ and $w_r\ne 0$, we obtain \begin{align*} \lambda=r. \end{align*} Therefore the highest weight is a nonnegative integer. Write $m:=r$. The subspace \begin{align*} W:=\operatorname{span}_{\mathbb C}\{w_i:0\le i\le m\} \end{align*} is nonzero and is stable under $h$, $f$, and $e$ by the formulas above and by $w_{m+1}=0$. Hence $W$ is a nonzero $\mathfrak{sl}_2(\mathbb C)$-submodule of $V$. Since $V$ is irreducible, $W=V$. Thus $V$ has basis $\{w_i:0\le i\le m\}$ and action \begin{align*} hw_i=(m-2i)w_i \end{align*} for $0\le i\le m$, \begin{align*} fw_i=w_{i+1} \end{align*} for $0\le i<m$, \begin{align*} fw_m=0, \end{align*} \begin{align*} ew_i=i(m-i+1)w_{i-1} \end{align*} for $1\le i\le m$, and \begin{align*} ew_0=0. \end{align*} [/step]

[step:Construct the module with the displayed action for every nonnegative highest weight] Fix $m\in\mathbb Z_{\ge 0}$. Let $L(m)$ be the complex vector space with basis \begin{align*} \{v_i:0\le i\le m\}. \end{align*} Define linear maps $e,f,h:L(m)\to L(m)$ on this basis by \begin{align*} hv_i=(m-2i)v_i \end{align*} for $0\le i\le m$, \begin{align*} fv_i=v_{i+1} \end{align*} for $0\le i<m$, \begin{align*} fv_m=0, \end{align*} \begin{align*} ev_i=i(m-i+1)v_{i-1} \end{align*} for $1\le i\le m$, and \begin{align*} ev_0=0. \end{align*} We check the defining relations on the basis $\{v_i:0\le i\le m\}$. Since equality of linear maps is determined by equality on a basis, it is enough to compare the two sides on each $v_i$. For $0\le i\le m$, \begin{align*} [h,e]v_i=2ev_i, \end{align*} where both sides are zero when $i=0$, and for $i\ge 1$ this follows from comparing the $h$-weights of $v_i$ and $v_{i-1}$. Similarly, \begin{align*} [h,f]v_i=-2fv_i \end{align*} for all $0\le i\le m$, with both sides zero when $i=m$. Finally, for $1\le i<m$, \begin{align*} [e,f]v_i=e v_{i+1}-f e v_i=(i+1)(m-i)v_i-i(m-i+1)v_i=(m-2i)v_i=hv_i. \end{align*} For $i=0$, \begin{align*} [e,f]v_0=ev_1=m v_0=hv_0. \end{align*} For $i=m$, \begin{align*} [e,f]v_m=-f e v_m=-m v_m=hv_m. \end{align*} Thus the formulas define an $\mathfrak{sl}_2(\mathbb C)$-module. The vector $v_0$ satisfies \begin{align*} ev_0=0. \end{align*} Also, \begin{align*} hv_0=mv_0. \end{align*} Since $v_0$ is a basis vector, $v_0\ne 0$, so $L(m)$ has highest weight $m$. [/step]

[step:Prove the constructed module is irreducible] Let $0\ne U\subset L(m)$ be an $\mathfrak{sl}_2(\mathbb C)$-submodule. Choose a nonzero vector \begin{align*} u=\sum_{i=0}^m a_i v_i\in U, \end{align*} where $a_i\in\mathbb C$. Let $j$ be the largest index such that $a_j\ne 0$. Since $U$ is stable under $e$, the vector $e^j u$ lies in $U$. For $0\le i<j$, we have $e^j v_i=0$, while \begin{align*} e^j v_j=j!m(m-1)\cdots(m-j+1)v_0\ne 0. \end{align*} Hence $e^j u$ is a nonzero scalar multiple of $v_0$, so $v_0\in U$. Since $U$ is stable under $f$ and \begin{align*} v_i=f^i v_0 \end{align*} for every $0\le i\le m$, all basis vectors $v_i$ belong to $U$. Therefore $U=L(m)$, proving irreducibility. [/step]

[step:Identify all irreducible finite-dimensional modules uniquely] The classification argument above shows that every nonzero irreducible finite-dimensional $\mathfrak{sl}_2(\mathbb C)$-module is isomorphic to one of the constructed modules $L(m)$ with $m\in\mathbb Z_{\ge 0}$. It remains to prove uniqueness of $m$. In $L(m)$, the highest weight vector has $h$-eigenvalue $m$, and the set of $h$-eigenvalues is \begin{align*} \{m,m-2,\dots,-m\}. \end{align*} The largest eigenvalue in this string is $m$. Therefore, if $L(m)\cong L(n)$ as $\mathfrak{sl}_2(\mathbb C)$-modules, the corresponding linear isomorphism intertwines the action of $h$ and preserves the set of $h$-eigenvalues. Hence $m=n$. This proves that every irreducible finite-dimensional $\mathfrak{sl}_2(\mathbb C)$-module is isomorphic to exactly one $L(m)$. [/step]