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By the Bartlett decomposition, one has that for $k \leq n$ and $\mathbf{\Gamma}_{n\times k} \in \mathbb{R}^{n\times k}$ a standard Gaussian matrix with independent entries

$$\mathbf{\Gamma}_{n\times k} \sim \mathbf{Q}_{n\times k}\mathbf{R}_{k\times k}$$

for $\mathbf{Q}_{n\times k}, \: \mathbf{R}_{k\times k}$ statistically independent, $\mathbf{Q}_{n\times k}$ uniformly distributed on the Stiefel manifold $\mathbb{V}_{k}^n$ and $\mathbf{R}_{k\times k}$ random upper-diagonal, such that the $(i,i)$-th diagonal entry is distributed as a $\chi^2_{n-i+1}$ random variable, whilst the $(i,j)$-th super-diagonal entry is a standard Gaussian and all entries are mutually independent.

Question: suppose $n = k$. Is there any known factorisation of a square Gaussian matrix into the product of a random Hermitian and a correction factor as below?

$$\mathbf{\Gamma}_{n\times n} \sim \mathbf{Q}_{n\times n}\mathbf{B}_{n\times n}\mathbf{Q}_{n\times n}^T\mathbf{\Delta}_{n\times n}$$

for $\mathbf{Q}_{n\times n}$ as before, an arbitrary but fixed diagonal $\mathbf{B}_{n\times n} \neq \text{Id}$ and random $\mathbf{\Delta}_{n\times n}$, independent of $\mathbf{Q}_{n\times n}$ (which may depend on $\mathbf{B}_{n\times n}$). Thanks in advance.

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Write the SVD of $\Gamma$, say $\Gamma = \sum_i q_i s_i v_i^T$. with $s_1,...,s_n>0$ the singular values and $q_i, v_i$ are the left and right singular vectors. If $Q=[q_1|...|q_n]$, $B=diag(s_1,...,s_n)$ and $V=[v_1|...|v_n]$ then $\Gamma=QBV^T$. The crux of the matter is that $(Q,B,V)$ are mutually independent with $Q,V\in O(n)$.

From here, $\Gamma=QBQ^T QV^T$ to obtain the desired form. It remains to show that $\Delta=QV^T$ is independent of $Q$.

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  • $\begingroup$ Thanks a lot for your answer. This helps. Just a remark: in the required decomposition of $\mathbf{\Gamma}_{n\times n}$, the diagonal matrix $\mathbf{B}_{n\times n}$, can be arbitrary but should be non-random. So the above argument requires a small fix to work as desired. $\endgroup$
    – user274127
    Jun 26, 2021 at 10:22
  • $\begingroup$ What about $B=I_n$, $Q$ independent of $\Gamma$ and $\Delta=\Gamma$? I don't see any requirement on $\Delta$. $\endgroup$
    – jlewk
    Jun 28, 2021 at 11:08
  • $\begingroup$ The question specifies $B \neq I_n$. Otherwise the factorisation would be trivial. $\endgroup$
    – user274127
    Jun 28, 2021 at 12:30
  • $\begingroup$ $B=2I_n$ and $\Delta=\Gamma/2$ is allowed. Extra constraints are required to prevent trivial solutions. $\endgroup$
    – jlewk
    Jun 28, 2021 at 13:14

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