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.