Let $V$ be an irreducible representation of $G = GL(2, \mathbb{C})$. Is it true that for every $\mathbf{v} \in V$ and every two-dimensional subspace $W \subset V$ the $G$-orbit $G\mathbb{v}$ intersects $W$?
From irreducibility we know that the orbit spans all of $V$, but by itself that is not enough to conclude that it meets all two-dimensional subspaces as is illustrated by the case that $G$ is a finite group. So somehow the topology of $G$ or the fundamental theorem of algebra should be involved.
Let $n = \dim V$. For $n = 1, 2$ the statement is trivial. In the case $n = 3$ I was also able to show that the statement is true (see the proof below), but only by exploiting the fact that this representation has a very special structure: it is the space of traceless 2-by-2-matrices on which $G$ acts by conjugation. This is very helpful in understanding what the orbits look like: for any $\lambda \in \mathbb{C}$ the set of non-zero matrices with eigenvalues $\{-\lambda, \lambda\}$ is a $G$-orbit and so is the singleton set $\{0\}$.
For higher dimensional $G$-irreps I don't have a similarly simple description of the orbits - any pointer to the literature to where I could find such a description of orbits in $GL(2, \mathbb{C})$ or more generally $GL(m, \mathbb{C})$-irreps would already be welcome!
My argument for $n = 3$: Let $\lambda \in \mathbb{C}$ and linearly independent matrices $X = \begin{pmatrix} x_1 & x_2 \\ x_3 & -x_1 \end{pmatrix}, Y = \begin{pmatrix} y_1 & y_2 \\ y_3 & -y_1 \end{pmatrix} \in Mat(2, \mathbb{C}) \cong V$ be given. Our goal is to show that the space of matrices spanned by $X$ and $Y$ contains a matrix $Z = \alpha X + \beta Y$ with eigenvalues $\lambda$ and $-\lambda$.
We distinguish two case. When $X$ has a non-zero eigenvalue $\mu$ then $(\lambda/\mu) X$ has eigenvalue $\lambda$ with the same eigenvector. Since $(\lambda/\mu) X$ is moreover traceless its other eigenvalue must be $-\lambda$.
When on the other hand both eigenvalues of $X$ are zero, the $G$-orbit of $X$ contains the matrix $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ and hence we may assume without loss of generality that $X = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$.
Look at the polynomial $f(\beta) = \det(X + \beta Y - \lambda I)$. If $f$ has any zeroes $\beta_0$ then $X + \beta_0Y$ is a traceless two-by-two matrix with at least one eigenvalue equal to $\lambda$. As it is traceless it follows that its other eigenvalue equals $-\lambda$.
We distinguish two further cases: the case where $f$ is constant and the case where it is not. In the latter case $f$ has at least one zero by the Fundamental Theorem of Algebra and we are done. In order to rule out the other case, we take a look at what it means for $f$ to constant in $\beta$.
Expanding out $f$ we find $f(\beta) = (x_1 + \beta y_1 - \lambda)(-x_1 - \beta y_1 - \lambda) - (x_2 + \beta y_2)(x_3 + \beta x_3)$ $= \det(X - \lambda I) - (2x_1y_1 + x_2y_3 + x_3y_2)\beta + (\det Y)\beta^2$
so that $f$ is constant if and only if both $\det Y = 0$ and $2x_1y_1 + x_2y_3 + x_3y_2 = 0$. However, since we assumed wlog that $x_1 = x_3 = 0$ and $x_2 = 1$ the latter equality reduces to $y_3 = 0$. Together with the condition that $\det Y = tr Y = 0$ this implies that $Y$ is a scalar multiple of $X$, contradicting the assumption of linear independence.
So that settles that, but as I said it relies heavily on the structure of the three dimensional irreducible representation as the space of traceless 2-by-2-matrices and so I don't see how to generalize this to the higher dimensional cases.
The answer is no.
Take $Sym^n(V)$ with $n \ge 4$ where $V$ is the standard $2$-dimensional representation. Let's say $x$ and $y$ are basis elements for $V$, then this space is just the space of homogeneous degree $n$ polynomials in $x$ and $y$ with the natural action of $GL_2$.
If we look at the orbit of $x^n$ under this action it is all polynomials of the form $(ax+by)^n$ with $a$ and $b$ not both zero. This orbit doesn't intersect the codimension 2 subspace where the coefficients of both $x^n$ and $y^n$ are zero. Since $n \ge 4$ this space has dimension at least $2$ so in particular the orbit misses all $2$-dimensional subspaces of this.