Let $V$ be a vector space of dimension $n$. 
Let $S^k V$ be a representation of $GL(n)$.
I would like to know if there exists some characterization of finite dimensional $GL(n)$ modules $V\_1,V\_2$ such that exists $GL(n)$ mapping $T:S^k V \rightarrow V\_1\otimes V\_2$ such that for some $x$ matrix $T(x)$ is invertible .


Added:
Just to clarify: The mapping $T$ itself may not be invertible I am asking that $T(x)$ will be invertible in $V\_1\otimes V\_2$.
For example the mapping $S^2 V \rightarrow V\otimes V$ is not invariable, but $T(x)$ is invartable for most $x$'s.


I am interested for a base field $\mathbb{C}$. And of course this may happen only in case when $S^kV$ is a sub-representation of $V\_1 \otimes V\_2$. In fact I do not know for which irreps $V\_1, V\_2$ representation $V\_1 \otimes V\_2$ have $S^kV$ as a component. I will be happy if you can give me a reference for this question. 



Representations of $GL(n)$ containing $S^kV$
Regarding the question of pairs, here is the answer (hope, I'm not mistaken).


Let me first change the notation a little bit. Suppose, we are looking for irreps $V\_1=\Sigma^{\lambda}V$ and $V\_2=\Sigma^{\mu}V$ where $\lambda = (\lambda\_1,\ldots,\lambda\_n)$ and $\mu = (\mu\_1,\ldots,\mu\_n)$ resp., such that there exists an embedding $S^kV\to V\_1^\*\otimes V\_2$.


It's immediate that such an embedding exists iff $V\_2=\Sigma^{\mu}V$ is an irreducible factor in $V\_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum\_i\lambda\_i$. Secondly, one has the following inequalities:


* $\mu\_1\geq \lambda\_1$,
* $\lambda\_1\geq \mu\_2 \geq \lambda\_2$,
* $\ldots$
* $\lambda\_{n-1}\geq \mu\_n \geq \lambda\_n$.


Finally, if such an embedding exists, it's unique by the very same L-R rule.


UPD: Forgot to mention that the main question still seems to be very hard as even checking that $\dim V\_1=\dim V\_2$ is a huge problem.


UPD 2: Let me answer your main question (despite UPD). Let's forget about representation theory and do some linear algebra. What we have is a surjective map $V\_1\otimes U\to V\_2$, where $\dim V\_1 = \dim V\_2$. Let the corresponding bilinear map be $h:V\_1\times U\to V\_2$. We want to find such $u\in U$ that $h\_u = h(-, u)$ is invertible, equivalently, of maximal rank.


Well, this seems to be quite obvious: let $r$ be the maximal rank of $h\_u$ among all $u\in U$. Suppose $r<\dim V\_2$. Let $u\_0\in U$ be such that $\mathrm{rk}\ h\_{u\_0}=r$. Take any $v\in V\_2\setminus\mathrm{Im}\ h\_{u\_0}$. Then there exists such $u\_1\in U$ that $v\in \mathrm{Im}\ h\_{u\_1}$. Now it's easy to see that the rank of a general linear combination $\alpha h\_{u\_0} + \beta h\_{u\_1} = h\_{\alpha u\_0+\beta u\_1}$ will be greater than $r$.


