# Factorisation of Gaussian random matrix into random Hermitian and correction factor

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.

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 $$Q$$.
• 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.
• What about $B=I_n$, $Q$ independent of $\Gamma$ and $\Delta=\Gamma$? I don't see any requirement on $\Delta$. Jun 28, 2021 at 11:08
• The question specifies $B \neq I_n$. Otherwise the factorisation would be trivial.
• $B=2I_n$ and $\Delta=\Gamma/2$ is allowed. Extra constraints are required to prevent trivial solutions. Jun 28, 2021 at 13:14