Let $ V=\mathbb{C}^n $ and let $ V^* $ be the dual space of $ V $.
Let $ M(n,\mathbb{C}) $ be the vector space of all $ n \times n $ complex matrices. Let $ U(n) $ be the group of unitary matrices. Let $ G $ be a subgroup of $ U(n) $.
There is a natural action of $ G $ on $ V $ by $ g \cdot v=gv $, and a natural action of $ G $ on $ V^* $ by $ g \cdot v=v^* g^* $ where $ g^*=g^{-1} $ is the conjugate transpose of $ g $. Let $ W $ be a subrepresentation of $ V $, call this representation $ \lambda: G \to GL(W) $. And let $ \lambda^*: G \to GL(W^*) $ be the corresponding subrepresentation of $ V^* $.
There is also a natural action of $ G $ on matrices by conjugation $ g \cdot T = gTg^{-1} $. Let $ R $ be a subrepresentation of $ M(n,\mathbb{C}) $ with respect to this conjugation action. Call this representation $ \omega: G \to GL(R) $.
Take two vectors $ w_1,w_2 \in W $ and a matrix $ T \in R $ and consider the product $ w_1^*Tw_2 $ where $ w_1^* $ denotes the dual vector of $ w_1 $ with respect to the standard Hermitian form.
Is it true that if the decomposition of $ \lambda^* \otimes \omega \otimes \lambda $ into irreducibles has no copies of the trivial representation then it must be the case that $$ w_1^* Tw_2=0 $$ for all $ w_1,w_2 \in W $ and $ T \in R $?
Context: This appears to be a form of/ related to/ generalization of/ what physicists call the Wigner-Eckart theorem. https://en.wikipedia.org/wiki/Wigner%E2%80%93Eckart_theorem
Consider the representation $ (W^* \otimes R \otimes W, \lambda^* \otimes \omega \otimes \lambda) $. The map $$ W^* \otimes R \otimes W \to \mathbb{C} $$ given by $$ w_1 \otimes T \otimes w_2 \mapsto w_1^*Tw_2 $$ is a homomorphism of $ G $ representations. In particular note that the action of $ g $ here is by $$ g \cdot (w_1^* \otimes T \otimes w_2) =(g \cdot w_1^*) \otimes (g \cdot T) \otimes (g \cdot w_2)=(w_1^* g^*) g T g^* (g w_2)=w_1^* T w_2 $$ so the action of $ g $ on $ \mathbb{C} $ is trivial. So the image of this irrep must be either the trivial rep $ \mathbb{C} $ or the zero rep $ 0 $. But a completely reducible representation can only have a quotient which is the identity irrep $ \mathbb{C} $ if it already includes a copy of the identity irrep. Thus if $ \lambda^* \otimes \omega \otimes \lambda $ does not include a copy of the trivial irrep then the quotient cannot be the trivial irrep and thus must be the $ 0 $ representation. In other words we have $$ w_1^*Tw_2=0 $$ for all $ w_1,w_2 \in W $ and $ T \in R $ if there are no copies of the trivial irrep in $ \lambda^* \otimes \omega \otimes \lambda $.