content_image/307736.jpg content_image/307737.jpg content_image/307733.jpg
## 5.3 Proof of Proposition 4 

To save space, we only derive the asymptotic power envelope for the relatively more difficult case of real-valued data and µ -based tests. According to the Neyman- Pearson lemma, the most powerful test of h = 0 against the simple alternative h = ( h$_{1}$, ..., h$_{r}$ ) is the test which rejects the null when L$_{p}$ ( h ; µ$_{p}$ ) is larger than a critical value C. It follows from Theorem 1 that, for such a test to have asymptotic size α , C must be 

where 

<img src='content_image/307736.jpg'>

<img src='content_image/307737.jpg'>

<img src='content_image/307733.jpg'>

Now, according to Le Cam’s third lemma and Theorem 1, under h = ( h$_{1}$, ..., h$_{r}$ ) , ln L$_{p}$ ( h ; µ$_{p}$ ) d → N ( m ( h ) + W ( h ) , W ( h )) . The asymptotic power (15) follows. □ 

## 5.4 Invariance issues and Proof of Proposition 5 

Before turning to the proof of Proposition 5, let us clarify the invariance issues in the problem under study. For basic definitions (invariant, maximal invariant, etc.), we refer to Chapter 6 of Lehmann and Romano (2005). 

Suppose that X is a p × n random matrix with vec ( X ) ∼ N (0 , I$_{n}$ ⊗ Σ). This model is clearly invariant under the group G $_{p}$, acting on R p × $^{n}$, of left-hand multi- plications by a p × p orthogonal matrix x ↦→ Qx, x ∈ R p × $^{n}$, Q ∈ O ( p ) ; so are the null hypothesis H$_{0}$ and the alternative H$_{1}$. Letting m = min ( n, p ) , the m -tuple λ ( X ) = ( λ$_{1}$, ..., λ$_{m}$ ) of non-zero eigenvalues of 1 $_{n}$XX ′ is clearly invariant under that group, since 1 $_{n}$xx ′ and 1 n ( Qx ) ( Qx $_{)}$′ = 1 $_{n}$Qxx ′ Q ′ share the same eigenvalues λ ( x ) for any orthogonal matrix Q and any matrix x ∈ R p × $^{n}$. However, λ ( X ) is 

