Irreducible invariant tensor spaces of the general linear group $GL_n(\mathbb{C})$

I am learning the representation theory of the general linear group $GL_n(\mathbb{C})$.

As far as I understand, the way to decompose the $\nu$-fold tensor product space $V^{\otimes \nu}=V \otimes V \otimes \cdots \otimes V$, where $V$ is an $n$-dimensional vector space over the complex field, to irreducible invariant subspaces with respect to the action of the group $GL_n(\mathbb{C})$ is to use the fact that the action of $GL_n(\mathbb{C})$ commutes with the action of the permutation group $S_\nu$.

We first decompose the full space $V^{\otimes \nu}$ into irreducible invariant subspaces. For each irreducible representation $D_i$ (of dimension $d_i$) of $S_\nu$, we have $m_i \times d_i$ basis functions $\{ f^{i}_{pq}| 1\leq p \leq m_i, 1\leq q \leq d_i \}$, such that in each row, the $d_i$ basis functions span an irreducible invariant subspace of the representation $D_i$, moreover, the representation matrices of $S_\nu$ with respect to them are independent of the row index $p$.

Now the problem is, I can understand that the operator $a\in GL_n(\mathbb{C})$ does not couple two basis functions belonging to two different (nonequivalent) representations of $S_\nu$, but I cannot see why each column of $\{ f^{i}_{pq}| 1\leq p \leq m_i, 1\leq q \leq d_i \}$ constructs an irreducible invariant subspace for $GL_n(\mathbb{C})$.

• This is not obvious. If I understand what you're asking for, this is the statement of Schur-Weyl duality. – Qiaochu Yuan Dec 16 '15 at 17:26