Using spectral decomposition, we can write any symmetric matrix as

$$\Sigma = Q \Lambda Q^{\top}$$

where $Q$ is orthonormal, and

$$\Lambda = \text{diag}(\lambda_1, ..., \lambda_p)$$

with $\lambda_1 \geq ... \geq \lambda_p \geq 0$.

An alternative parameterization can be made for the covariance matrix in terms of eigenvalues $\lambda_1,...,\lambda_p$ and $Q$ can be expressed using Euler angles in terms of $p(p-1)/2$ angles, $\theta_{ij}$, where $i = 1,2,...,p-1$ and $j = i, ..., p-1$.[1]

Can someone elaborate on this method such that given a function with p eigenvalues and $p(p-1)/2$ angles I can build a valid $\Sigma$

[1]: Hoffman, Raffenetti, Ruedenberg. "Generalization of Euler Angles to N‐Dimensional Orthogonal Matrices". J. Math. Phys. 13, 528 (1972)

  • $\begingroup$ yes. It is a angles. edited $\endgroup$
    – Kumar
    Jul 8 '15 at 13:59
  • $\begingroup$ You have to specify a convention for what those angles mean. Even in $\mathbb{R}^3$ there are multiple conventions. If you want to use the convention in the paper you reference, then you will have to explain it here in your post. $\endgroup$
    – whuber
    Jul 8 '15 at 17:57

Those angles are sufficient to define the eigenvectors, to within normalization of the eigenvectors. Therefore you can use them to determine the eigenvectors, which after normalization, can be used to populate Q (there are two possibilities differing by a factor of -1 for the signs of elements for each column of Q, , but they cancel out due to multiplying Q and Q'). I presume the details are provided in the paper.

  • 1
    $\begingroup$ can you show your idea with an example? $\endgroup$
    – TPArrow
    Jul 8 '15 at 14:39
  • $\begingroup$ If p =2, one angle is sufficient to define orientation of the 1st eigenvector relative to the axes, and 2nd eigenvector is orthogonal to that. Similarly, if p = 3, 3 angles plus orthogonality of the 3 eigenvectors is sufficient to defene the orientation of all 3 eigenvectors. Etc. As I wrote, I presume the details are in the paper, which I don't have access to. $\endgroup$ Jul 8 '15 at 14:52

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