# Random matrix determinant problem

Suppose we have a a set of random matrices in the complex field of the form $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ where $a_i$ are constant positive real scalars and $v_i$ are random complex valued matrices with all elements i.i.d. circular symmetric Gaussian for all $i$. All the $n$ matrices $v_i$ are of same dimension. And $I$ is the identity matrix and $v_i^H$ is the conjugate transpose of $v_i$. Thus an element of the matrices $v_iv_i^H$ for all $i$, is the sum of two squared i.i.d normal random variables which makes it exponential, and $v_iv_i^H$ are symmetric.

Now we want to maximize the following determinant over $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ $$\mathbb{E}\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right).$$

Here $\mathbb E$ is the expectation. Essentially we pick one matrix for the numerator and all the rest go in the denominator. Since all matrices are i.i.d, can I claim that the matrix which should go on the numerator is the one with the highest $a_i$ ?

Edit: With regards to first comments, it seems asymptotic analysis is the appropriate for such a question. So we consider the solution when the dimensions of all $v_i$ grow.

I ran a simulation with the $a_i=i$, where $i\in \{1,\dots,10\}$ and matrices of $20 \times 20$. With randomly generated normals with mean zero and variance 1. The results seems to confirm the claim. The function is maximum when the matrix with $a_{10}=10$ goes on the numerator. P.S.: Random version of my previous question A determinant problem with symmetric PSD matrices

May be related to Expected determinant of a random NxN matrix.

• are you interested in asymptotics when n is fixed and the dimension of the v_i's is large, or in exact expressions? The former is doable, I don't know about the latter. – ofer zeitouni Jan 22 '14 at 7:22
• @oferzeitouni I was reading other problems about random matrices and most seem to be interested in asymptotics. I am very much happy to know the asymptotic approach to this question. Would be very grateful if you could please explain how that can be done. – MLT Jan 22 '14 at 12:19
• @oferzeitouni I did a simulation which for matrices of $20 \times 20$ and it seems to hold. – MLT Jan 22 '14 at 16:02

First, I understand the question as one asking about $$det(I+a_i v_iv_i^*(I+\sum_{j\neq i} a_jv_jv_j^*)^{-1}) =det(I+\sum_ia_iv_iv_i^*)/det(I+\sum_{j\neq i} a_j v_j v_j^*)=:A/B$$ Since $A$ does not depend on $i$, the question is about the asymptotics of $B$. Letting $\lambda_i$ denote the eigenvalues of the matrix $X=\sum_{j\neq i} a_j v_j v_j^*$, with empirical measure $L_N=N^{-1}\sum \delta_{\lambda_i}$, ($N$ is the dimension of the $v_i$s), you have $$N^{-1}\log B=\int \log(1+x) dL_N(x)\to \int\log(1+x)d\mu(x)$$ where $\mu$ is the limit of $L_N$. So the issue is the evaluation of $\mu$. But because $X$ is a sum of freely independent matrices, the law of $\mu$ can be computed: it is the free convolution of (rescaled, by $a_i$) Marchenko-Pastur laws.
• Could you please explain why you took the $\log$ of $B$? – MLT Jan 22 '14 at 16:55
• Thanks. I would like to clarify if my understanding is correct please. The distribution of the eigenvalues of $a_iv_iv_i^H$ as dimension tends to $+\infty$, is given by the Marchenko–Pastur distribution. Then we need the distribution of sum of these Marchenko–Pastur distributions, which are independent. For some reason these distributions do not commute therefore we cannot do normal convolution and instead have to do something called free convolution? – MLT Jan 22 '14 at 18:31
• Thank you. Then we proceed. To find the free convolution we need the R-transform. The R-transform of the M-P distribution (say \mu_i)of the eigenvalues of $a_iv_iv_i^H$, with the dimension ratio $\beta$ is $R_i(x)=\frac{a_i}{1-\beta x}$. Then the R-transform of the distribution of eigenvalues $\lambda_i$ of $X$ is the sum of R-transforms $R_{\lambda_i}=\sum_{j \neq i}R_j=\sum_{j\neq i} \frac{a_j}{1-\beta x}$ ? – MLT Jan 22 '14 at 19:06