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Let $\Bbb{S}_{++}^n$ denote the space of symmetric positive definite (SPD) $n\times n$ real matrices, and let $A,B\in\Bbb{S}_{++}^n$.

Is it possible to express the logarithm of $A^{-1}B$ as a difference of the form $P-Q$, where $P,Q$ are $n\times n$ real matrices, not necessarily SPD, but P must depend solely on $A$ and $Q$ must depend solely on $B$, i.e., $$ \log(A^{-1}B)=P-Q. $$

Note that, in genral, $AB\neq BA$.

EDIT: This question is a part of a more general question I asked on MathSE. I am not sure that this is the right approach for attacking my original problem, but I am interested in this one in any case.

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  • $\begingroup$ Have you done some numerical experiments? In particular, testing longer chains such as $(P-Q)+(Q-R)+(R-S)=(P-S)$. $\endgroup$
    – Federico Poloni
    Commented Jun 23, 2015 at 8:26
  • $\begingroup$ Thanks for your reply @FedericoPoloni. No, the truth is that I did not. But how could do that. I cannot figure out how the logarithm should be treated. $\endgroup$
    – nullgeppetto
    Commented Jun 23, 2015 at 8:29
  • $\begingroup$ What do you mean by "treated"? Matlab, Mathematica and Python (sympy) all have library functions to compute matrix logarithms without trouble. $\endgroup$
    – Federico Poloni
    Commented Jun 23, 2015 at 8:33
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    $\begingroup$ The Riemannian distance is not negative definite; that is, $\exp(-\gamma d^2(X,Y))$ is not a positive definite function. $\endgroup$
    – Suvrit
    Commented Jun 25, 2015 at 21:38
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    $\begingroup$ Here is a link to a paper that answers your "kernel" question in complete detail (the log-Euclidean transform indeed leads to a kernel but at the expense of "killing" the curvature): www2.compute.dtu.dk/~sohau/papers/cvpr2015/feragen_cvpr2015.pdf $\endgroup$
    – Suvrit
    Commented Jun 25, 2015 at 22:40

1 Answer 1

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Solution is inspired by the comment of Federico Poloni. Suppose there exist such functions, i.e. $$log(A^{-1}B)=P(A)-Q(B).$$ for all SPD-matrices $A,B$. Putting $A=B$ yields $P(A)=Q(A)$ for all SPD-matrix $A$. Now if the equation above holds we would have $$log(A^{-1}B)+log(B^{-1}C)+log(C^{-1}A)=0.$$ However simply choosing random matrices shows that this indentity does not hold. Here is a matlab code: d=2;

A=gen_rand_spd(d);

B=gen_rand_spd(d);

C=gen_rand_spd(d);

logm(A^(-1)*B)+logm(B^(-1)*C)+logm(C^(-1)*A)

function [A] = gen_rand_spd(d)

A=rand(d);

A=A+A';

A=expm(A);

end

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  • $\begingroup$ Thanks for your answer. Well, that's clear. Would you mind take a look at my original question at MathSE? I'm interested in showing that the squared distance $d^2$ is negative definite. $d$ in my case is a geodesic metric between SPD matrices in $\Bbb{S}_{++}^n$. Is this something possible to be shown? Thanks once again! math.stackexchange.com/questions/1334670/… $\endgroup$
    – nullgeppetto
    Commented Jun 23, 2015 at 8:44
  • $\begingroup$ I read the question on MathSE but to be honest I didn't understand it. What do you mean by negative definite? By the way you gave the definition of the metric induced by the Riemannian metric and not the Riemannian metric itself. Also I dont understand why the metric could also be defined as $||log(A^{-1}B)||_F$. $\endgroup$
    – user35593
    Commented Jun 23, 2015 at 9:02
  • $\begingroup$ To tell the truth, I am not familiarized with this theory, that's why I am here! I gave that metric because that was proposed by Förstner and Moonen. If you want take a look to the paper I give below (Eq.13). I need to use a distance metric to measure the similarity between two SPD matrices (but not the Log-Euclidean one). So, I came up with this one, which also seems to work properly in my case. However, I would like to show that I could use this distance to construct a PD kernel $k(\cdot,\cdot)=\exp(-\gamma d^2(\cdot,\cdot))$, and that's why I need $d^2$ to be negative-definite, ...> $\endgroup$
    – nullgeppetto
    Commented Jun 23, 2015 at 9:21
  • $\begingroup$ ... just like the square Euclidean distance (which is negative-definite). Would you like to suggest my some other approach maybe? What would be the metric you mention above? $\endgroup$
    – nullgeppetto
    Commented Jun 23, 2015 at 9:22
  • $\begingroup$ Well, I forgot to give the link of the paper: ipb.uni-bonn.de/fileadmin/publication/pdf/… $\endgroup$
    – nullgeppetto
    Commented Jun 23, 2015 at 9:24

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