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I have only learned about calculus and linear algebra, so I don't know about differential algebra.

I got to know about the concept of "geodesic" recently. What I need to know is this:

Suppose I have a $10\times 10$ matrix $Z$ and $Z = XY$, where $X$ is $10\times 3$, $Y$ is $3\times 10$. I have two points in a manifold, $(X_1, Y_1, Z_1)$ and $(X_2, Y_2, Z_2)$. I want to calculate the "geodesic path" between this two point in this manifold which satisfy $Z=XY$. How can I do that?

At present, I think the idea may be similar with how to calculate the geodesic path between two point $(x_1, y_1, z_2)$ and $(x_2, y_2, z_2)$ which satisfy $z=xy$ in a $3$-dimensional space. But I don't know how to calculate the simplified case easier, as I know nothing about differential geometry.

I'm looking forward to your response!

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  • $\begingroup$ Hi @bangliu. Welcome! You might find this MathJax tutorial helpful for this site in future :) $\endgroup$ – Shaun May 23 '14 at 20:45
  • $\begingroup$ Interesting problem. It's weird enough to have to associate matrices with coordinates, but even weirder for two coordinates to be matrices of different sizes. What is the context of such a problem? $\endgroup$ – Muphrid May 23 '14 at 20:52
  • $\begingroup$ @Muphrid The context is: consider the Z matrix is the some kind of "distance" matrix, and Z(i,j) is the distance between iten i and j. Maybe you may ask that Z should be symmetric, but the thing is Z(i,j) and Z(j, i) maybe a little different. Now I want to know how this matrix changes during a time period. I know Z1 and Z2, and I want to inference Z between time t1~t2 using the "geodesic path". If you haven't met the case of matrix, could you help me with the simplified case: z=xy in the 3-dimentional space? Thanks! $\endgroup$ – Excalibur May 23 '14 at 21:03
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    $\begingroup$ @Shaun Hi, thanks, the link is helpful! $\endgroup$ – Excalibur May 23 '14 at 21:03
  • $\begingroup$ @Muphrid Maybe more detailed: for example, in a network, we have N nodes, the latency between node i, j is Z(i,j). The latency matrix can be factorized as two matrices multiplication: Z = XY, where X is Nr, and Y is rN. $\endgroup$ – Excalibur May 24 '14 at 0:10

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