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I'm working on some Random Matrix Theory related stuff for my thesis, and i've come across the following problem:

Consider a (normalized) spiked Wigner matrix $\mathbf{A}$ $$ \mathbf{A} = \frac{\beta}{n}\mathbf{xx}^T + \mathbf{W}, $$ where $\mathbf{x}$ is an n-dimensional vector (in our case it's on the Boolean hypercube) such that $\|x\|_2 = \sqrt{n}$, $\beta < 1$, and $$ \mathbf{W}_{ij} \sim \begin{cases} \mathcal{N}(0,1/n), \quad i\neq j \\ \mathcal{N}(0,2/n), \quad i = j. \end{cases} $$

I'm trying to show that it is asymptotically unitarily invariant (or not, perhaps). My intuition is that by the BBP phase transition, the spectrum of $\mathbf{A}$ remains unfazed by the spike (i.e. it's the Wigner semicircle distribution).

By unitary invariance, I mean that I want to show that for a Haar distributed matrix $\mathbf{V}$, we can write out $$ \mathbf{A} = \mathbf{V\Lambda V}^T $$ in the large system limit $n \rightarrow \infty$, where $\boldsymbol{\Lambda}$ is the diagonal containing the eigenvalues of $\mathbf{W}$.

Any ideas on how to show this property?

Thanks

Edit: So far i've looked into this BBP phase transition, but i can't formalize an argument. I've also attempted some matrix perturbation, but i'm unfortunately a bit novice & it doesn't seem to help me achieve the goal. Also, i've tried showing that $$ \mathbf{A} \overset{d}{=} \mathbf{UWU}^T + \frac{\beta}{n}\mathbf{xx}^T \\ = \mathbf{UWU}^T + \frac{\beta}{n}\mathbf{U(U^Tx)(U^Tx)}^T\mathbf{U}^T \\ = \mathbf{U}(\mathbf{W} + \frac{\beta}{n}\mathbf{(U^Tx)(U^Tx)^T})\mathbf{U}^T$$ where $\mathbf{U^Tx}$ would thus asymptotically be unitarily invariant (i think?), and I could somehow show that the outer product goes to zero by $\frac{1}{n}$ factor? I'm definitely open to ideas!

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  • $\begingroup$ doesn't your $A$ have an "outlier" eigenvalue with an eigenvector that depends on $x$ ? that would break the unitary invariance of the spectrum of $A$ $\endgroup$ Commented Oct 24 at 19:24
  • $\begingroup$ @CarloBeenakker that's the BBP part i was talking about. In the case where $\beta < 1$, the spike is "absorbed" in the noise and the outlier is still within the bulk of the eigenvalues $\endgroup$
    – Kawa
    Commented Oct 25 at 5:36

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