I have a matrix which is change with time. Let me denote it as A(t). I know t=0 it is A(0) and I know t=1 it is A(1). A is symmetric positive semi-definite matrix. What I want to do is find the "geodesic path" for the matrix A(t) give A(0) and A(1). As far as I know, it seems need some Riemannian Geometry knowledge, but I haven't learned about Riemannian Geometry. Does anyone knows about the idea about how to get the geodesic path? Or recommend some papers or other materials which consider about this problem. Thanks very much!

Update: What if A have the manifold structure that A = B*C, where A is n*n, B is n*r and C is r*n? That means A is determined by B and C, and we can factorize B(0), C(0) from A(0), and B(1) C(1) from A(1). What I did is: as A = B*C, then A(i,j) = \sum_{k}B_{ik} * C_{kj} Denote B_{ik}*C_{kj} ad A(i,j,k), then A(i,j) = \sum_{k}A(i,j,k) To calculate the geodesic path of matrix A, what I did is calculate the geodesic path of A(i,j,k). This is solved by calculate the geodesic path on the 3-D surface of z = xy. After I calculated every A(i,j,k), I get the matrix of A(t). What I want to know is: does the matrix calculated in this way is really the "geodesic" matrix of A(t)? If it is, how to prove that my calculation is the right method? If not, how to get it?


One natural metric on the space of positive semi-definite matrices is just the Euclidean metric on this cone in $\Bbb R^{n(n+1)/2}$. That is, the set of symmetric $n\times n$ matrices is naturally an $\frac{n(n+1)}2$-dimensional vector space, and the positive semi-definite matrices $K$ form a convex cone in this space: If $A\in K$, $cA\in K$ for $c\ge 0$, and if $A,B\in K$, then $A+B\in K$.

Using the standard Euclidean metric on $K\subset\Bbb R^{n(n+1)/2}$, the shortest path joining any two matrices in $K$ will be the straight line segment joining them.

  • $\begingroup$ Hi @TedShifrin , you mean that A(t) = A(0) + t* (A(1)-A(0)), the simple linear relationship? $\endgroup$ – Excalibur Jun 12 '14 at 17:58
  • $\begingroup$ Yes, that's what I'm suggesting for the "obvious" geometric structure on this set of matrices. $\endgroup$ – Ted Shifrin Jun 12 '14 at 17:59
  • $\begingroup$ Hi @TedShifrin , thanks very much! I updated my question, can you check it? $\endgroup$ – Excalibur Jun 12 '14 at 18:09
  • $\begingroup$ You're now looking at matrices of rank $\le r$, and it's a very complex criterion that you end up with something positive semidefinite, symmetric in terms if $B$ and $C$. I don't know what metric you're imposing. And then you refer to the surface $z=xy$ in $\Bbb R^3$. First, the geodesics on this surface are quite complicated; second, I don't see the relevance. Even if you could reduce to this low dimension correctly, piecewise-smooth geodesics in a Riemannian manifold are never geodesics: Rounding corners makes a shorter math. In summary, with your edit, I suspect the problem is very hard. $\endgroup$ – Ted Shifrin Jun 12 '14 at 21:42

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