The original question (that can be seen at the bottom of this post) was replaced by this first edit (below)
EDIT I
I give more details about my problem.
First of all let suppose to have K vectors $\boldsymbol{\omega}_k = \{\omega_{i,k}\}_{i=1}^n$, $k=1,\dots ,K$ of length $n$, distributed as multivariate normals: $$ \boldsymbol{\omega}_k \sim N_n(\mathbf{0}_n, \mathbf{C}_k), k=1,\dots ,K. $$ Now let suppose that i want to introduce dependence between the $\boldsymbol{\omega}_k$'s assuming that $$ \boldsymbol{\omega}_{i,\cdot}=(\omega_{i,1},\omega_{i,2},\dots , \omega_{i,K})' \sim N_K(\mathbf{0}_k, \boldsymbol{\Sigma}), i=1,\dots , n $$ notice that the $\omega$'s have all the same index $i$. Now we can ask what is the covariance between $\boldsymbol{\omega}_{i,\cdot}$ and $\boldsymbol{\omega}_{j,\cdot}$, with $i\neq j$.
There is not a unique answer. We have to define a matrix $\mathbf{A}$ such that $\boldsymbol{\Sigma}=\mathbf{A}\mathbf{A}'$, and to find the matrices $$ \mathbf{T}_k = [\mathbf{A}]_{\cdot,k} [\mathbf{A}]_{\cdot,k}', k=1,\dots ,K $$ where $[\mathbf{A}]_{\cdot,k}$ is the $k^{th}$ column of $\mathbf{A}$. Then the covariance between $\boldsymbol{\omega}_{i,\cdot}$ and $\boldsymbol{\omega}_{j,\cdot}$ is given by $$ \sum_{k=1}^K \mathbf{T}_k [\mathbf{C}_k]_{ij} $$ As you can see, the choice of $\mathbf{A}$ influences the covariance structure of my problem, and different $\mathbf{A}$'s give different models.
For example we can define $\mathbf{A}$ to be the Cholesky factorization, but then it is easy to see that we are saying that, marginally, the covariance matrix of $\boldsymbol{\omega}_1$ depends only on $\mathbf{C}_1$, the one of $\boldsymbol{\omega}_2$ depends on $\mathbf{C}_1$ and $\mathbf{C}_2$ and so on.
What I'm looking for is a $\mathbf{A}$ that let me able to have marginal distributions that do not depend on the ordering of the $\boldsymbol{\omega}$'s, i.e. that potentially the marginal density of $\boldsymbol{\omega}_1$ can be equal to the one of $\boldsymbol{\omega}_K$, that is not true for the Cholesky factorization and the spectral decomposition.
Old Question
Let suppose to have a covariance matrix $\boldsymbol{\Sigma}$. The spectral decomposition of a positive definite matrix tells us that we can write $$ \boldsymbol{\Sigma} = \boldsymbol{\Psi}\boldsymbol{\Lambda}\boldsymbol{\Lambda}\boldsymbol{\Psi}' $$ where the column vectors of $\boldsymbol{\Psi}$ are the normalized eigenvectors and $\boldsymbol{\Lambda}$ is a diagonal matrix where the $i^{th}$ element is the square root of the eigenvalue associated to the $i^{th}$ normalized eigenvector.
What I am interested in is the matrix $$ \mathbf{A} = \boldsymbol{\Psi}\boldsymbol{\Lambda} $$ where of course $\boldsymbol{\Sigma} =\mathbf{A}\mathbf{A}'$, assuming that the diagonal elements of $\boldsymbol{\Lambda}$ are placed in ascending order. It seems to me obvious that the space of $\mathbf{A}$ is constrained by the fact that I am imposing an ascending order of the eigenvalues, but what are precisely the constraints?