= Solution
On $T=V^{\otimes n}$, define the <permutation> and diagonal actions by
$$
\sigma(v_1\otimes\cdots\otimes v_n)
=v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)},\qquad
g(v_1\otimes\cdots\otimes v_n)=gv_1\otimes\cdots\otimes gv_n.
$$
The inverse in the first formula gives a left action. Applying $g$ to every tensor factor and then reordering produces the same tensor as reordering and then applying $g$. Thus the two actions commute. The <Schur algebra> is
$$
\boxed{S_{\mathbb C}(m,n)=\operatorname{End}_{\mathbb CS_n}(T).}
$$
It is the <commutant of an operator algebra> of the <permutation> action.
To identify this commutant, use the multilinear isomorphism
$$
\operatorname{End}(T)\cong\operatorname{End}(V)^{\otimes n}.
$$
Conjugation by a <permutation> reorders the factors on the right. Therefore the invariant subspace is the space of <symmetric tensors> of degree $n$ in $\operatorname{End}(V)$. It is spanned by $a^{\otimes n}$. Indeed, the polarization identity
$$
\sum_{J\subseteq\{1,\ldots,n\}}(-1)^{n-|J|}
\left(\sum_{j\in J}a_j\right)^{\otimes n}
=\sum_{\sigma\in S_n}a_{\sigma(1)}\otimes\cdots\otimes a_{\sigma(n)}
$$
expresses every symmetrized elementary tensor as a <linear combination> of pure powers. Those symmetrized elementary tensors span the invariant subspace in <characteristic zero>.
One may restrict $a$ to invertible <endomorphisms> without changing the span. For fixed $a$, the vector-valued <polynomial> $(a+tI)^{\otimes n}$ has degree at most $n$. Choose $n+1$ distinct values of $t$ away from the finitely many roots of $\det(a+tI)$. <Polynomial interpolation> expresses its constant term $a^{\otimes n}$ as a <linear combination> of those invertible <tensor powers>. Consequently
$$
S_{\mathbb C}(m,n)=\operatorname{span}_{\mathbb C}\{g^{\otimes n}:g\in\operatorname{GL}(V)\}.
$$
The span is an algebra, since products are $(gh)^{\otimes n}$.
The image $A_T$ of $\mathbb CS_n$ is semisimple, as a quotient of a <semisimple algebra>. The <double-centralizer theorem for semisimple operator algebras> gives $A_T''=A_T$. Since the preceding computation says that $A_T'$ is the span of the general linear action, the two actions are \b[mutual commutants]. More explicitly, <semisimple module> decomposition gives
$$
\boxed{V^{\otimes n}\cong
\bigoplus_{\substack{\lambda\vdash n\\\ell(\lambda)\leq m}}
S^\lambda\otimes D_\lambda(V).}
$$
Here $D_\lambda(V)=\operatorname{Hom}_{S_n}(S^\lambda,T)$ is the multiplicity space, with its natural general linear action. The commutant algebra is the product of the full <endomorphism> algebras of these multiplicity spaces. Hence each nonzero $D_\lambda$ is irreducible, and distinct spaces have nonisomorphic general linear representations, because the operators $g^{\otimes n}$ span that commutant. A primitive $e_t$ picks a one-dimensional factor from $S^\lambda$, so $e_tT$ is an equivalent realization of $D_\lambda$ as a <Schur module>.
The range of shapes is exact. If $\lambda$ has more than $m$ rows, a column antisymmetrizes more than $m$ vectors and its action vanishes by $\Lambda^{m+1}V=0$. Conversely, for at most $m$ rows place the $i$th basis vector in every tensor position belonging to row $i$ of $t$. Row symmetrization multiplies this tensor by $\prod_i\lambda_i!$. Column antisymmetrization is nonzero, because all vectors within each column are distinct, and its different <permutations> give distinct basis tensors. Thus $h_tT\ne0$. This proves the <length bound for a Schur module> and the stated <Schur–Weyl duality>. For $n=0$, the empty partition and $V^{\otimes0}=\mathbb C$ give the trivial version.
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