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!
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 $\Sigma$.