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Let $\mathbf{\Sigma}$ be a $N \times N$ Hermitian and positive definite complex matrix. Let us define the matrix $\mathbf{A}$ as \begin{equation} \mathbf{A} \triangleq \mathbf{\Sigma}^{-T} \otimes \mathbf{\Sigma}^{-1} - \frac{1}{N} \mathrm{vec}(\mathbf{\Sigma}^{-1})\mathrm{vec}(\mathbf{\Sigma}^{-1})^H, \end{equation} where $(\cdot)^T$ and $(\cdot)^H$ are the transpose and the Hermitian operators, respectively. Moreover $\mathrm{vec}$ is the operator stacking the columns of a matrix on top of each other and $\otimes$ indicates the Kronecker product.

Is it possible to write the matrix $\mathbf{C}$ such that $\mathbf{A}=\mathbf{C}^H\mathbf{C}$, explicitly as function of $\mathbf{\Sigma}$ (or $\mathbf{\Sigma}^{-1}$)?

Thanks!

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  • $\begingroup$ By $\Sigma^{-T}$ you mean $(\Sigma^T)^{-1}$? $\endgroup$ – jobe Mar 9 '18 at 13:53
  • $\begingroup$ Yes, exactly. _ $\endgroup$ – Vuk Mar 9 '18 at 13:57
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Yes. Let $P=\Sigma^{-1/2}$. Then $P^T\otimes P$ is Hermitian (because $P$ is) and $\operatorname{vec}(\Sigma^{-1})=(P^T\otimes P)\operatorname{vec}(I)$. Therefore \begin{align} A&=(P^T\otimes P)^2-\frac1N(P^T\otimes P)\operatorname{vec}(I)\operatorname{vec}(I)^H(P^T\otimes P)\\ &=(P^T\otimes P)\left(I-\frac1N\operatorname{vec}(I)\operatorname{vec}(I)^T\right)(P^T\otimes P) \end{align} Note that $\Pi=I-\frac1N\operatorname{vec}(I)\operatorname{vec}(I)^T$ is the orthogonal projection on the linear space of (vectorised) traceless matrices. Therefore $\Pi^H\Pi=\Pi^2=\Pi$. Consequently, $A=C^HC$ when $C=\Pi(P^T\otimes P)$. As every positive definite matrix has a unique positive definite square root, $P$ and $C$ are indeed functions of $\Sigma$.

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  • $\begingroup$ Can we go a little further? For example, the matrix $\mathbf{I} - \frac{1}{N}\mathrm{vec}(\mathbf{I})\mathrm{vec}(\mathbf{I})^T$ can be interpreted as the orthogonal projection matrix on the orthogonal complement of the $\mathrm{span}(\mathrm{vec}(\mathbf{I}))$, i.e. $\Pi^{\perp}_{\mathrm{vec}(\mathbf{I})}$, right? Can this help us to simplify the expression? $\endgroup$ – Vuk Mar 14 '18 at 9:08
  • $\begingroup$ @Vuk Yes, as you said, it is an orthogonal projection, so you may take itself as its positive semidefinite square root. $\endgroup$ – user1551 Mar 14 '18 at 10:41

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