Explicit decomposition of tensor product of a representation with its dual representation Consider a finite group $G$ with an irreducible representation $\rho$: $G \to \mathbf{GL}(W)$ where $W$ is a finite-dimensional vector space over $\mathbb{C}$. Let $\rho^\ast$ be the dual representation on $W^\ast$. Let $n$ = dim($W$). In S. Sternberg's book Group theory and physics, he makes the claim

Under $G$, the space $W \otimes W^\ast$ decomposes into a direct sum of $n$ copies of $W$.

Ultimately, I would like to prove this statement. But to build intuition I've been trying to see how this works explicitly for the case $n = 2$. However, I have not been able to see how the statement is true even in this case, so I would like help to finish this explicit example.
I begin with the matrix representation $R_{ij}(s)$ for an arbitrary $s \in G$, written with respect to a chosen orthonormal basis. Then, I think $s$ acts on a general element $u \otimes v \in W \otimes W^\ast$ as follows:
$$
(R \otimes R^\ast) 
\begin{bmatrix} u_1v_1 \\ u_1 v_2 \\ u_2 v_1 \\ u_2 v_2 \end{bmatrix}
=
\begin{bmatrix}
R_{11} R^\ast_{11}u_1 v_1 + R_{11} R^\ast_{12}u_1 v_2 + R_{12} R^\ast_{11}u_2 v_1 + R_{12} R^\ast_{12}u_2 v_2 \\
R_{11} R^\ast_{21}u_1 v_1 + R_{11} R^\ast_{22}u_1 v_2 + R_{12} R^\ast_{21}u_2 v_1 + R_{12} R^\ast_{22}u_2 v_2  \\
R_{21} R^\ast_{11}u_1 v_1 + R_{21} R^\ast_{12}u_1 v_2 + R_{22} R^\ast_{11}u_2 v_1 + R_{22} R^\ast_{12}u_2 v_2 \\
R_{21} R^\ast_{21}u_1 v_1 + R_{21} R^\ast_{22}u_1 v_2 + R_{22} R^\ast_{21}u_2 v_1 + R_{22} R^\ast_{22}u_2 v_2 
\end{bmatrix}
$$
I don't see how to find the stable subspaces here that are each isomorphic to $W$. I'm guessing the unitarity of $R$ will come into play, but even so I still don't see the stable subspaces, and would appreciate assistance.
Edited addition: I'm still getting used to notation, so I apologize if anything here is non-standard. In the above I used $R^\ast$ to mean the matrix representation of $s$ for its action on $W^\ast$. It does not mean the conjugate transpose of $R$. Incidentally, if I understand things correctly, I think $R^\ast$ will be simply the complex conjugate of $R$ (but not transposed), or possibly identical to $R$, depending on convention. But this might be part of my confusion.
 A: Ah okay I see what's going on:
So $W \otimes W^*$ carries a natural $G \times G$ action with $(g,h) \cdot v\otimes w = gv \otimes hw$.  The action of $G$ I mentioned in the comments corresponds to restricting this action to the "diagonal" copy of $G$ in $G \times G$ where the coordinates are equal.  If we do this then as I mentioned the theorem is very false, in fact $W \otimes W^*$ always has a natural copy of the trivial representation.
If instead we restrict to the first copy of $G$ inside $G \times G$ -- those pairs $(g,e)$ where $e$ is the identity, then what he says is true.  By restricting to this copy of $G$ really there is no action on the $W^*$ factor, so if we pick a basis $x_1, x_2, \dots, x_n$ for $W^*$ then $W\otimes span(x_i)$ is a subrepresention isomorphic to $W$ in the obvious way and $W \otimes W^* = (W\otimes span(x_1)) \oplus (W\otimes span(x_2)) \dots \oplus (W\otimes span(x_n))$, which is the form you want.
I think it might be worth doing an example where $G = GL_n(\mathbb{C})$ and $W$ is just the obvious representation of $GL_n(\mathbb{C})$ acting by matrix multiplication on $\mathbb{C}^n$ thought of as column vectors.  In this case $W \otimes W^*$ is the vector space of $n\times n$ matrices, and the action of $G \times G$ is given by $(g,h) \cdot A = gAh^{-1}$.
Here we see that the diagonal copy of $G$ is acting by conjugation and indeed the scalar matrices form the copy of the trivial representation I mentioned.  The action of the first factor that he is considering is just $(g,e) \cdot A = gA$,  but now if we think of how matrix multiplication works $G$ is just acting separately on the different columns of $A$ and each of the $n$ columns is just another copy of $\mathbb{C}^n$.
