Diagonalisation: Schmidt vs eigenvalue - when to use which? In physics we encounter diagonalisation of matrices or operators in a variety of areas. But there are different kinds, the main two being Schmidt decomposition and eigenvalue diagonalisation.
The two are in general different: Schmidt decomposition also works for non-square matrices while eigenvalue diagonalisation is restricted to square matrices. Even for square matrices they are not generally the same. On the other hand they coincide for some cases (Symmetric square matrices I believe. Or is it hermitian?).
My question is when do we use which one and why?
For example when dealing with Hamiltonians on finite dimensional Hilbert (not sure if infinite works too) spaces we seem to always be using the eigenvalue diagonalisation. My suspicion is that this is because we need the left and right states/vectors of the diagonal matrix to be the same since otherwise the eigenvalues could not be interpreted as the energies of the system and the states would not be the stationary ones.
For density matrices of multipartite systems we seem to be using Schmidt decomposition most of the time. My suspicion in this case that this is because the Schmidt values have the interpretation of probabilities (since they are always positive).
 A: Schmidt decomposition is in general a singular value decomposition (SVD) and it is applied on wave vectors and not on density matrices.
While dealing with bi-partite wave vectors we use SVD because there is no restriction that the size of the two systems in question are the same. So the matrix of the wave vector coefficients can be rectangular and SVD can be computed for rectangular matricies.
Since Hamiltonians are Hermitian matrices their left and right singular vectors are the same. One reason to prefer eigenvalues over singular values is that singular values are always constrained to be positive while there is no such constraint for the energy of a system.
 For  Hermitian matrices the singular values are the absolute values of the eigenvalues.
Note:
The left and right singular vectors of a matrix contain the eigenvectors of $AA^\dagger$ and $A^\dagger A$.  So if $[A, A^\dagger] = 0$ then SVD and eigenvalue diagonalization match. The matrices which satisfy this are called normal matrices. 
Edit: In addition to being normal, $A$ must also be positive semi-definite (All eigenvalues are $\geq 0$)  for SVD and eigen-decomposition to match.  
