4
$\begingroup$

Suppose $(x_i(t))$ is a $n$-dimensional time-series, where $t$ is an integer between $1$ and $T$ (time is discrete) and $i$ an integer between $1$ and $n$, and I assume $n<T$. From this time-series, one can construct two interesting covariance matrices:

The spatial covariance matrix : $A_{ij} = \sum_{t} x_i(t)x_j(t)$

The temporal covariance matrix : $B_{st} = \sum_{i} x_i(s)x_i(t)$

If one puts $x_i(t)$ in matrix form $X$, with $X_{it}=x_i(t)$, then $A=X.X'$ and $B=X'.X$.

So from the singular value decomposition of $X$ we know that $A$ and $B$ share exactly the same eigenvalues (the only difference is that, in addition, $B$ has $T-n$ zero eigenvalues).

Therefore, my question is : what does it MEAN ? Is there an explanation of this fact somewhere ? Do you have some physical interpretation ?

$\endgroup$
1
$\begingroup$

For a meaningful answer to your question about "physical interpretation", I need to work with empirical covariance matrices, so averaged over many trials. (You consider a single trial in your question, but that somehow obscures the interpretation.) My conclusion is:

If you find that the empirical spatial and temporal covariance matrices share the same positive eigenvalues, then you know that your data is self-averaging in both space and time.

Consider a data set $x_{ij}^{(k)}$ of measured values at position $i$, time sample $j$ in the $k$-th trial. The empirical spatio-temporal covariance matrix is defined by

$$C_{ii',jj'}=\frac{1}{N_{\rm trials}}\sum_{k=1}^{N_{\rm trials}}x_{ij}^{(k)}x_{i'j'}^{(k)}.$$

Empirical spatial and temporal covariance matrices are constructed as partial traces,

$$S_{ii'}=\sum_{j}C_{ii',jj},\qquad T_{jj'}=\sum_{i}C_{ii,jj'}.$$

In general, these two matrices $S$ and $T$ will have different sets of positive eigenvalues.

However, if the data is self-averaging in time, then the empirical spatial covariance matrix is $\bar{S}_{ii'}=\sum_{j}x_{ij}^{(k)}x_{i'j}^{(k)}$ independent of the sample index $k$; if moreover the data is self-averaging in space, then the empirical temporal covariance matrix is $\bar{T}_{jj'}=\sum_{i}x_{ij}^{(k)}x_{ij'}^{(k)}$ independent of $k$. These two matrices $\bar{S}$ and $\bar{T}$ (corresponding to your $A$ and $B$) have the same set of positive eigenvalues.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.