Required Number of LU Factorization is $1$ If the System is Separable

12 Views Asked by At

I am reading the paper about the deformation transfer, which is to implant the vertex transformation at one mesh into another one.

At page 4 of the paper, it has stated as:

"... Furthermore, the system is separable in the spatial dimension of the vertices. Thus, for each source/target pair, we compute and store the LU factorization of $A^TA$ only once."

The system mentioned in the above quote is $A^TA\tilde{x} = A^Tc$ where $\tilde{x}$ is a vector of the possible(unknown) target mesh's vertex locations, and $c$ is a vector containing entries from the source transformations.

First, I'd like to know to which concept (or definition) of the above quoted separability refers.

Second, in case the first question's separability is well defined, why does the separability guarantees the LU decomposition to be performed only once?