[proofplan] Let $d=|\lambda|$ and realize $S_\lambda(E)$ as the image of a Young symmetrizer on $E^{\otimes d}$. The case $\ell(\lambda)>n$ follows directly from column antisymmetry, because a column of length greater than $\dim E$ forces an exterior power of degree greater than $n$. For $\ell(\lambda)\leq n$, we invoke the precise Schur-Weyl and Young symmetrizer theorem: tensor space decomposes into irreducible polynomial $GL(E)$-modules tensored with Specht modules, and the Young symmetrizer image is exactly one copy of the $GL(E)$-factor indexed by $\lambda$. The same theorem supplies the highest weight, and connectedness of $GL_n(\mathbb C)$ transfers irreducibility from the group representation to the differentiated [Lie algebra](/page/Lie%20Algebra) representation. [/proofplan] [step:Realize the Schur functor as a Young symmetrizer image in tensor space] Let \begin{align*} d:=|\lambda|=\sum_{i\geq 1}\lambda_i. \end{align*} If $d=0$, then $\lambda=\varnothing$, the Young symmetrizer construction gives $S_\lambda(E)=\mathbb C$, and the diagonal torus acts by the constant character $1$. Differentiating this character gives highest weight $0$, which equals $\sum_{i=1}^n\lambda_i\varepsilon_i$. Hence assume $d\geq 1$ for the remainder of the proof. Choose a Young tableau $T$ of shape $\lambda$ whose boxes are filled by $\{1,\dots,d\}$. Let $R_T\leq S_d$ denote the subgroup preserving every row of $T$, and let $C_T\leq S_d$ denote the subgroup preserving every column of $T$. Define elements of the group algebra $\mathbb C[S_d]$ by \begin{align*} a_T:=\sum_{\sigma\in R_T}\sigma. \end{align*} Define also \begin{align*} b_T:=\sum_{\tau\in C_T}\operatorname{sgn}(\tau)\tau. \end{align*} Finally put \begin{align*} c_T:=a_Tb_T\in\mathbb C[S_d]. \end{align*} The [symmetric group](/page/Symmetric%20Group) $S_d$ acts on $E^{\otimes d}$ by permuting tensor factors, and $GL(E)$ acts diagonally by \begin{align*} g\cdot(v_1\otimes\cdots\otimes v_d):=(gv_1)\otimes\cdots\otimes(gv_d). \end{align*} These two actions commute, since applying the same [linear map](/page/Linear%20Map) to each tensor factor is independent of the order in which the tensor factors are permuted. With this convention for Young symmetrizers, define \begin{align*} S_\lambda(E):=c_T(E^{\otimes d})\subseteq E^{\otimes d}. \end{align*} Because the diagonal action of $GL(E)$ on $E^{\otimes d}$ has matrix entries that are [homogeneous polynomial](/page/Homogeneous%20Polynomial) functions of degree $d$ in the matrix entries of $g\in GL(E)$, every $GL(E)$-submodule of $E^{\otimes d}$ is a polynomial $GL(E)$-module. Thus $S_\lambda(E)$ is polynomial whenever it is nonzero. [/step] [step:Use column antisymmetry to prove vanishing when the partition has too many parts] Assume $\ell(\lambda)>n$. The first column of the Young diagram of $\lambda$ has length $\ell(\lambda)$, which is greater than $\dim_{\mathbb C}E=n$. Let $I\subset\{1,\dots,d\}$ be the set of tensor positions lying in this first column. The operator $b_T$ antisymmetrizes the tensor positions indexed by $I$, so the $I$-part of every tensor in $b_T(E^{\otimes d})$ lies in the alternating quotient corresponding to \begin{align*} \bigwedge^{\ell(\lambda)}E. \end{align*} Since $\ell(\lambda)>\dim_{\mathbb C}E$, the exterior power $\bigwedge^{\ell(\lambda)}E$ is zero. Therefore \begin{align*} b_T(E^{\otimes d})=0. \end{align*} It follows from $c_T=a_Tb_T$ that \begin{align*} S_\lambda(E)=a_Tb_T(E^{\otimes d})=0. \end{align*} This proves the asserted vanishing case. [/step] [step:Apply Schur-Weyl duality and the Young symmetrizer theorem in the nonzero range] Assume $\ell(\lambda)\leq n$. We use the following external standard form of Schur-Weyl duality together with the Young symmetrizer theorem. For the commuting actions of $GL(E)$ and $S_d$ on $E^{\otimes d}$ over $\mathbb C$, there is a $GL(E)\times S_d$-module decomposition \begin{align*} E^{\otimes d}\cong\bigoplus_{\substack{\mu\vdash d, \, \ell(\mu)\leq n}} L_\mu(E)\otimes \operatorname{Sp}_\mu. \end{align*} Here $\operatorname{Sp}_\mu$ is the Specht module of shape $\mu$, and $L_\mu(E)$ is a nonzero irreducible polynomial $GL(E)$-module whose highest weight with respect to the diagonal torus and upper triangular Borel is \begin{align*} \mu_1\varepsilon_1+\cdots+\mu_n\varepsilon_n. \end{align*} The Young symmetrizer theorem further states that, for a tableau $T$ of shape $\lambda$, the operator $c_T$ annihilates the summands with Specht shape not isomorphic to $\lambda$, and on the $\lambda$-summand its image is \begin{align*} L_\lambda(E)\otimes c_T\operatorname{Sp}_\lambda. \end{align*} Moreover $c_T\operatorname{Sp}_\lambda$ is one-dimensional and nonzero. The hypotheses of this theorem are satisfied: $E=\mathbb C^n$ is a complex [vector space](/page/Vector%20Space) of dimension $n$, $d=|\lambda|$ is a nonnegative integer, and the actions above are exactly the commuting diagonal $GL(E)$-action and tensor-place permutation action of $S_d$. Since $\ell(\lambda)\leq n$, the partition $\lambda$ occurs among the indices in the [Schur-Weyl decomposition](/theorems/8454). Therefore \begin{align*} S_\lambda(E)=c_T(E^{\otimes d})\cong L_\lambda(E)\otimes c_T\operatorname{Sp}_\lambda. \end{align*} Since $c_T\operatorname{Sp}_\lambda$ is a one-dimensional nonzero complex vector space on which $GL(E)$ acts only through the first tensor factor, this is isomorphic to $L_\lambda(E)$ as a $GL(E)$-module. Hence $S_\lambda(E)$ is nonzero and irreducible as a polynomial $GL(E)$-module. The same Schur-Weyl statement identifies the highest weight of $L_\lambda(E)$, and hence of $S_\lambda(E)$, as \begin{align*} \lambda_1\varepsilon_1+\lambda_2\varepsilon_2+\cdots+\lambda_n\varepsilon_n. \end{align*} [guided] The point of Schur-Weyl duality is that tensor space has two commuting symmetries. The group $GL(E)$ changes the vector placed in each tensor factor, while $S_d$ changes the positions of the tensor factors. Because these operations commute, tensor space can be decomposed into simultaneous building blocks for both actions. We invoke the following external standard theorem in its precise form. If $E$ is a complex vector space of dimension $n$ and $d\geq 0$, then the commuting $GL(E)$- and $S_d$-actions on $E^{\otimes d}$ satisfy \begin{align*} E^{\otimes d}\cong\bigoplus_{\substack{\mu\vdash d, \, \ell(\mu)\leq n}} L_\mu(E)\otimes \operatorname{Sp}_\mu. \end{align*} In this decomposition, $\operatorname{Sp}_\mu$ is the Specht module of shape $\mu$, and $L_\mu(E)$ is a nonzero irreducible polynomial $GL(E)$-module. The theorem also records the highest weight of $L_\mu(E)$: with respect to the diagonal torus and the upper triangular Borel, it is \begin{align*} \mu_1\varepsilon_1+\cdots+\mu_n\varepsilon_n. \end{align*} The condition $\ell(\mu)\leq n$ is part of the theorem and reflects the fact that antisymmetrizing more than $n$ vectors in an $n$-dimensional space gives zero. We also invoke the corresponding Young symmetrizer theorem. For a tableau $T$ of shape $\lambda$, the Young symmetrizer $c_T=a_Tb_T$ selects the Specht shape $\lambda$: it annihilates all Schur-Weyl summands whose Specht factor is not isomorphic to $\operatorname{Sp}_\lambda$, and on the $\lambda$-summand its image is \begin{align*} L_\lambda(E)\otimes c_T\operatorname{Sp}_\lambda. \end{align*} The same theorem states that $c_T\operatorname{Sp}_\lambda$ is one-dimensional and nonzero. This is exactly the missing non-cancellation input in a direct tensor calculation: it says that the row symmetrization and column antisymmetrization do not cancel when the shape has at most $n$ rows. We now check the hypotheses. The vector space in the theorem is $E=\mathbb C^n$, so it is a complex vector space of dimension $n$. The integer in the theorem is $d=|\lambda|$. The $GL(E)$-action is the diagonal action on $E^{\otimes d}$, and the $S_d$-action is the tensor-place permutation action; these are the actions used in the construction of $S_\lambda(E)$. Since we are in the case $\ell(\lambda)\leq n$, the partition $\lambda$ appears in the Schur-Weyl [direct sum](/page/Direct%20Sum). Applying the Young symmetrizer theorem to the Schur-Weyl decomposition gives \begin{align*} S_\lambda(E)=c_T(E^{\otimes d})\cong L_\lambda(E)\otimes c_T\operatorname{Sp}_\lambda. \end{align*} The factor $c_T\operatorname{Sp}_\lambda$ is a one-dimensional nonzero complex vector space, and $GL(E)$ acts only on the $L_\lambda(E)$ factor. Thus, as a $GL(E)$-module, $S_\lambda(E)$ is isomorphic to $L_\lambda(E)$. Since $L_\lambda(E)$ is irreducible and nonzero by Schur-Weyl duality, $S_\lambda(E)$ is a nonzero irreducible polynomial $GL(E)$-module. Finally, the highest weight is not being inferred from an unproved candidate vector. It is part of the Schur-Weyl highest-weight identification of the summand $L_\lambda(E)$. Therefore the highest weight of $S_\lambda(E)$ is \begin{align*} \lambda_1\varepsilon_1+\lambda_2\varepsilon_2+\cdots+\lambda_n\varepsilon_n. \end{align*} [/guided] [/step] [step:Differentiate the irreducible group representation to the Lie algebra] It remains to justify that the differentiated $\mathfrak{gl}_n(\mathbb C)$-module is irreducible. Let $W\subseteq S_\lambda(E)$ be a complex linear subspace stable under the differentiated action of $\mathfrak{gl}_n(\mathbb C)$. For every $X\in\mathfrak{gl}_n(\mathbb C)$, the operator $X$ preserves $W$, so the finite-dimensional exponential series $\exp(X)$ also preserves $W$. The connected complex Lie group $GL_n(\mathbb C)$ is generated by exponentials from its Lie algebra. Hence $W$ is stable under the whole $GL_n(\mathbb C)$-action. Since $S_\lambda(E)$ is irreducible as a $GL_n(\mathbb C)$-module in the case $\ell(\lambda)\leq n$, the only $GL_n(\mathbb C)$-stable subspaces are $0$ and $S_\lambda(E)$. Therefore the only $\mathfrak{gl}_n(\mathbb C)$-stable subspaces are $0$ and $S_\lambda(E)$, so the differentiated module is irreducible. The highest weight for the diagonal torus differentiates to the same linear functional on the diagonal Cartan subalgebra $\mathfrak h$, namely \begin{align*} \lambda_1\varepsilon_1+\lambda_2\varepsilon_2+\cdots+\lambda_n\varepsilon_n. \end{align*} Combining this with the vanishing argument for $\ell(\lambda)>n$ proves the theorem. [/step]