Euclidean matrix distance
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From: Anton on 14 Jun 2010 14:09 Walter Roberson <roberson(a)hushmail.com> wrote in message <wStRn.72751$Gx2.29908(a)newsfe20.iad>... > Anton wrote: > > > Woops! Is there perhaps another distance metric commonly used for > > matrices (tensors)? > > Looking at the wikipedia entry about distance metrics, it appears that > in differential geometry, metrics are only meaningful for vector spaces, > not for tensors. > > You might, of course, not be restricting yourself to differential > calculus, but if not then "distance metric" might have a rather > different meaning than Euclidean distance -- e.g., you might be > concerned about the geodesic rather than Euclidean distance. I think you > might have to give a bit more context. Sure thing. I'm working with a special type of MRI data. This data has been specifically processed to tell you something about the diffusion behavior of water. So I have a decent number of "images" composed of 3x3 tensors - and these tensors are what describe the way in which water is diffusing. My goal is to find the difference between these images. For various reasons, any two images are _almost_, but not quite the same. I'm not sure geodesic distance is applicable, because that's more like graph theory.
From: Philip on 15 Jun 2010 09:49 "Anton " <aorliche(a)gmail.com> wrote in message <hv5r81$q4m$1(a)fred.mathworks.com>... > Walter Roberson <roberson(a)hushmail.com> wrote in message <wStRn.72751$Gx2.29908(a)newsfe20.iad>... > > Anton wrote: > > > > > Woops! Is there perhaps another distance metric commonly used for > > > matrices (tensors)? > > > > Looking at the wikipedia entry about distance metrics, it appears that > > in differential geometry, metrics are only meaningful for vector spaces, > > not for tensors. > > > > You might, of course, not be restricting yourself to differential > > calculus, but if not then "distance metric" might have a rather > > different meaning than Euclidean distance -- e.g., you might be > > concerned about the geodesic rather than Euclidean distance. I think you > > might have to give a bit more context. > > Sure thing. I'm working with a special type of MRI data. This data has been specifically processed to tell you something about the diffusion behavior of water. So I have a decent number of "images" composed of 3x3 tensors - and these tensors are what describe the way in which water is diffusing. My goal is to find the difference between these images. For various reasons, any two images are _almost_, but not quite the same. > > I'm not sure geodesic distance is applicable, because that's more like graph theory. There are plenty of metrics applicable on tensor spaces, which generalise or not the Euclidean metrics for vector spaces. In the specific case of diffusion tensors in MRI, they can be interpreted as the space of positive definite matrices, which has a natural structure as a 'symmetric space' in differential geometry (the standard book for this is Helgason, Differential geometry, Lie groups, and symmetric spaces). This natural structure induces a natural differential geometric metric. I have to do a bit of self-publicity here, but I think is the easiest is to refer to our paper 'A rigorous framework for diffusion tensor calculus', Magnetic Resonance in Medicine Volume 53, Issue 1, Date: January 2005, Pages: 221-225 http://www3.interscience.wiley.com/journal/109860141/abstract To summarise: the (geodesic) distance between two diffusion tensors A and B is the square root of the sum-of-squared logarithms of eigenvalues of A^(-1/2)*B*A^(-1/2). Simplifications of this formula have been proposed, such as log-Euclidean, which is also simple and fast. Which is most appropriate to use is still an open question, and depends on requirements on invariance etc, so the simple Frobenius (arithmetic, Euclidean) distance is still possible (cf Pasternak).
From: Anton on 15 Jun 2010 13:14 Hi guys. First off, thanks for the help. Thank you to Philip for the paper reference. Your answers led me to learn a little bit about the math involved in this whole endeavor (like it or not). It turns out that the quantity I was looking to calculate is d(D1 ,D2 ) = sqrt( trace[(D1 − D2 )^2] ) where trace(D1) = λ1 + λ2 + λ3 Paper reference: "Similarity Measures for Matching Diffusion Tensor Images" Alexander et al. NOW my question is, is this the same as the Frobenious distance previously mentioned? I ask since it would be nice to call just one functions :-)
From: Bruno Luong on 15 Jun 2010 14:01
> NOW my question is, is this the same as the Frobenious distance previously mentioned? I ask since it would be nice to call just one functions :-) The answer is yes if D1 and D2 are symmetric. No otherwise. Bruno |