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Let $A$ be an $n\times n$ random matrix with i.i.d entries (say standard Gaussian) $A_{ij}$. I know that there is a CLT type result known for the determinant of $A$. More precisely there is a CLT for $\log(|\det(A)|)$. For any fixed $(i, j)$, If we look at the minor $M_{ij}$ of $A$, it follows that $\log(|M_{ij}|)$ also satisfies the same CLT.

I want to know if anything is known about the joint distribution of $(M_{11}, M_{22})$? I would be more generally interested in understanding the joint law of $(M_{ij}: 1\leq i, j\leq k)$ for an arbitrary (but fixed) $k$. In particular, can one expect any kind of asymptotic independence between $( M_{11}, M_{22})$ as $n\to \infty$?

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There is certainly no asymptotic independence between $\det M_{11}, \det M_{22}$. From the base times height formula for parallelepipeds we see that \begin{align*} \frac{|\det M_{12}|}{|\det M_{22}|} &= \frac{|v_1 \wedge v_3 \wedge \dots \wedge v_n|}{|v_2 \wedge v_3 \wedge \dots \wedge v_n|}\\ &= \frac{\mathrm{dist}(v_1, V)}{\mathrm{dist}(v_2,V)}\\ &= \frac{|v_1 \cdot u|}{|v_2 \cdot u|} \end{align*} where $v_i = (A_{i1}, A_{i3}, \dots, A_{in})$, $V \subset {\bf R}^{n-1}$ is the span of $v_3,\dots,v_n$, and $u$ is a unit normal to $V$. For fixed $V$, $v_1 \cdot u, v_2 \cdot u$ are independent gaussians, and so we conclude that $\log |\det M_{12}| - \log |\det M_{22}|$ is bounded in probability, and similarly for $\log |\det M_{12}| - \log |\det M_{11}|$, hence $\log |\det M_{11}| - \log |\det M_{22}|$ is also bounded in probability. On the other hand, as you observed it is known that $\log |\det M_{11}|$, $\log |\det M_{22}|$ are asymptotically gaussian with variance comparable to $\log n$, so there is no asymptotic independence here.

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  • $\begingroup$ Aah! Thank you so much. That's pretty neat argument. I was not hoping asymptotic independence, but could not formulate this simple neat argument. Just as a quick follow up, has there been any attempt to understand the joint distribution of minors? Let's say even for principle minors $(M_{11}, M_{22})$? $\endgroup$
    – Raghav
    Commented Jan 14, 2023 at 18:31
  • $\begingroup$ I don't know of any paper explicitly dealing with this, but perhaps the singular value decomposition $A = \sum_i \sigma_i u_i^* v_i$ will give good results, since the joint distribution of the singular values is known (see e.g., mathscinet.ams.org/mathscinet-getitem?mr=964668) and $u_i,v_i$ are iid orthonormal frames which have roughly Gaussian components (see e.g., arxiv.org/abs/math/0601457) independently of the $\sigma_i$. Note from Cramer's rule that $M_{jj} = (\prod_i \sigma_i) \sum_i \frac{u_{i,j} v_{i,j}}{\sigma_i}$. $\endgroup$
    – Terry Tao
    Commented Jan 14, 2023 at 20:28
  • $\begingroup$ @TerryTao I don't see how you apply Cramer's rule to obtain this relation for $M_{jj}$? What is the linear system you relate this too in order to apply Cramer? $\endgroup$
    – Salini Mendisi
    Commented Jan 16, 2023 at 1:18
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    $\begingroup$ A small typo: there is also an unspecified sign $\pm$ in the formula. I am using Cramer's rule in the form that computes components of the inverse: en.wikipedia.org/wiki/Cramer%27s_rule#Finding_inverse_matrix . In this application, we have $(A^{-1})_{jj} = \frac{1}{\det A} M_{jj}$. From the SVD we have $(A^{-1})_{jj} = \sum_i \frac{u_{i,j} v_{i,j}}{\sigma_i}$ and also $\det A = \pm \prod_i \sigma_i$. $\endgroup$
    – Terry Tao
    Commented Jan 16, 2023 at 13:32

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