The original question (that can be seen at the bottom of this post) was replaced by this first edit (below)


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?

  • 3
    $\begingroup$ I think you are based on the misconception that the eigenvalues are ordered. This actually is guaranteed only with some particular eigen-decomposition algorithms (eg. Lanczos-based algorithms). Any permutation of the columns of $\Lambda$ alongside the corresponding permutation of $\Psi$ would be a perfectly valid eigen-decomposition. Moreover, notice that usually eigen-decompositions entail a balancing pre-processing step; it improves the conditioning of $\Sigma$ to produce more accurate results but it can affect your final results too (among others: their ordering!). (Nice question! +1) $\endgroup$
    – usεr11852
    Commented Apr 2, 2017 at 17:10
  • 1
    $\begingroup$ I am revising my upvote to a downvote because the new question is a completely new and different question which should be given it's own thread. Your adding it in diminishes the utility of my answer. $\endgroup$
    – Taylor
    Commented Apr 6, 2017 at 4:30
  • 1
    $\begingroup$ There can be no solution is some cases, which is intuitively clear. Assume that $K=2$ and that (0) all $\omega_{i,k}$ have unit variance (1) $\omega_{i,1}$ and $\omega_{i+1,1}$ have a strong correlation say $\rho_1 \approx 1$, while (2) $\omega_{i,2}$ and $\omega_{i+1,2}$ have a weak correlation say $\rho_2 \approx 0$. Then no value of $\Sigma_{1,2}$ can be exist because both $\omega_{i,2}$ and $\omega_{i+1,2}$ should then be close to $\Sigma_{1,2} \omega_{i,1}$, which conflicts with a small value for $\rho_2$. $\endgroup$
    – Yves
    Commented Apr 7, 2017 at 6:08

1 Answer 1


A covariance matrix has ${n \choose 2} + n = \frac{n(n+1)}{2}$ free elements. The constraints for the spectral decomposition are:

  1. The eigenvalues are positive
  2. The eigenvectors are orthogonal
  3. The eigenvectors are unit length.

I recently asked a question that had to do with this. @amoeba had a good comment that helped visualize these constraints and proved why the number of free elements in $\Psi$ was ${n \choose 2}$, and the number of free elements in $\Lambda^2$ was $n$.

Regarding the ordering of the eigenvalues, that may or may not be important to you. The spectral decomposition is only unique up to re-orderings of your $\Lambda^2$ diagonal matrix. If your task is to estimate the parameters of this matrix, each of these re-orderings will have corresponding estimates. In this case the model will not be identifiable.

However, if you are just theorizing about the covariance matrix, people sometimes just restrict their attention to one ordering of the eigenvalues, without losing any generality. Proving something is true for one ordering will usually guarantee truth for the other orderings.

Edit to address comment:

1,2 and 3 are constraints that every covariance matrix has, so it is as "free" as possible. Sampling from some distribution of $\Sigma$ is possible as long as long as the distribution exists, but it is also common to restrict the columns of $\Psi$ further, which is the same as fixing the ordering of your eigenvalues. What I mean is that sometimes they will restrict it to have $\psi_{i,j} = 0$ when $j > i$ and $\psi_{j,j} = 1$. In other words, it will be lower-diagonal with $1$s on the diagonal. The are other ways to restrict this space, but it would look something like this: $$ \Psi = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ \psi_{2,1} & 1 & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \psi_{p,1} & \psi_{p,2} & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ \psi_{n,1} & \cdots & \psi_{n,p-1} & \psi_{n,p} \end{array} \right]. $$

Why is this representation unique? If you re-order the diagonal matrix $\Lambda^2$, then you have to re-order the columns of $\Psi$, but then the columns of $\Psi$ won't follow this pattern. Why do this? Well now your posterior is not multi-modal.

  • 1
    $\begingroup$ Among other things, I'm trying to estimate the elements of $\boldsymbol{\Sigma}$, but in my application, to model the coregionalization, I need a matrix $\mathbf{A}$ such that $\boldsymbol{\Sigma}=\mathbf{A}\mathbf{A}'$. And I also need $\mathbf{A}$ to be as free as possible, also overparametrized, since, at the end, I'm just looking to the posterior samples of $\boldsymbol{\Sigma}$ (it's a Bayesian model). Maybe I can just introduce a variable that decides the ordering of the eigenvalues/eigenvectors $\endgroup$
    – niandra82
    Commented Apr 2, 2017 at 17:43
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    $\begingroup$ @niandra82 see edit $\endgroup$
    – Taylor
    Commented Apr 2, 2017 at 18:18
  • 1
    $\begingroup$ I edited my question, adding more details on my problem. Thank you $\endgroup$
    – niandra82
    Commented Apr 2, 2017 at 18:46
  • $\begingroup$ Minor: Covariance matrix has to be positive semi-definite, so its eigen values are non-negative. Not necessarily all positive. $\endgroup$
    – Zhubarb
    Commented Apr 11, 2017 at 9:05
  • $\begingroup$ @Berkmeister positive definite usually. Unless you're looking at singular random vectors $\endgroup$
    – Taylor
    Commented Apr 11, 2017 at 12:06

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