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Working in machine learning, I try to find a way to compare time series, which can be considered as semi-continuous matrices belonging to $\mathbb R^{n \times \mathbb R}$ (a column corresponds to n-dimensional the value of the time series at a given time). A first step would be to consider the Gramian of the time series $u \in \mathbb R^{n \times \mathbb R}$: $M(u) = u.u' \in \mathbb R^{n \times n}$ (where $'$ is the transpose operator and $.$ is the integration along time). The good point is that $M(u)$ is semi-definite positive and that there is a natural Riemannian distance on the space of positive definite matrices (see the first pages of this paper for a short intro): $$d(M,N) = \|Log(M^{-1}.N) \|_{\mbox{Froebenius}}$$ This is an idea to measure the similarity between the two time series.

For some (convoluted) reason, I am now interested in the matrices of the form $P(u) = u.A.u' \in \mathbb R^{n \times n}$, where $A$ can be seen as a skew-symmetric matrix of $\mathbb R^{\mathbb R \times \mathbb R}$. So I have two questions:

  1. Is there a natural distance on the set of skew symmetric (or skew hermitian) matrices? I guess the answer is "not in such a general framework". Thus my second question is:
  2. In the more particular case presented above: $P(u)$ is a Gram-like skew symmetric matrix, do we have such a distance? Note that $A$ is the same for all the time series we want to compare.
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  • $\begingroup$ Actually the distance for posdef matrices is: $\|\log(M^{-1/2}NM^{-1/2})\|_F$... $\endgroup$
    – Suvrit
    Sep 27, 2012 at 10:12
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    $\begingroup$ I don't know if that is a natural distance, but a non-boring choice is given by mapping skew-Hermitian matrices to unitary matrices by using the exponential map, and then just using the Riemannian distance in the mapped space. $\endgroup$
    – Suvrit
    Sep 27, 2012 at 10:22

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Since skew-Hermitian matrices form the Lie algebra of the unitary group $U(n)$, the space has a natural positive definite inner product given by $$ (A, B) = - tr (AB). $$ This inner product, of course, defines a distance. The inner product is invariant under conjugation by elements of $U(n)$.

Real skew-symmetric matrices form the Lie algebra of the orthogonal group. It also has a conjugation invariant positive definite inner product, in fact only one up to a scalar multiple. It is also $-tr(AB)$.

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    $\begingroup$ Isn't it the Frobenius norm? $\endgroup$ Sep 27, 2012 at 11:29
  • $\begingroup$ Up to a scaler the inner product is the Killing form. I don't know that the corresponding norm is called. $\endgroup$ Sep 27, 2012 at 13:23
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    $\begingroup$ I mean, this construction should give the "trivial" norm $\left(\sum_{i,j} A_{ij}^2\right)^{1/2}$, if I am not missing something. $\endgroup$ Sep 27, 2012 at 16:18

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