Consider a hermitian form $F$ on $V=\mathbb{C}^n$ which is invariant under a subgroup $H \subset GL(n,\mathbb{C})$. Such a form satisfies $\forall h \in H$: $$h^t F \overline{h}=F,\quad F = \overline{F}^t.$$
My question now is the following:
If we have such an non-trivial Form $F$, invariant under $H$ and additionally assume, that $H$ acts irreducible on $\mathbb{C}^n$, does this then already imply that $F$ has to be non-degenerate?
My idea goes as follows. If we consider the kernels $${}^{\perp}V=\{v \mid \forall w \in V, F(v,w)=0\} \subseteq V$$ $$V^{\perp}=\{w \mid \forall v \in V, F(v,w)=0\} \subseteq V$$ then one verifies that those are $H$-invariant. Let $v \in {}^{\perp}V$ and $h \in H$, then we have $\forall w \in V$: $$(hv)^t F w = v^t h^t F w = v^t F \overline{h}^{-1}w = 0$$ as $v \in {}^{\perp}V$, $F$ is $H$-invariant and $\overline{h}^{-1}w \in V$. We conclude $hv \in {}^{\perp}V$, i.e. those kernels are $H$-invariant subspaces of $V$. Since $H$ acts irreducible on $V$ we get ${}^{\perp}V = \mathbb{C}^n$ or ${}^{\perp}V = \{0\}$. Only the latter case remains, as $F$ is non-trivial by assumption. Since both kernel are trivial, $F$ is per definition non-degenerate.