I have a matrix K that I represent with an LU factorization, such that K = L @ U, and L and U are lower and upper triangular 2-D tensors respectively. I’m currently testing the functionality of torch.linalg.lu_factor with pivot=False, which enforces P (the permutation matrix) to be the identity.
In my sample code I have the following:

L = torch.randn(3,3).tril()
L.diagonal()[:] = torch.ones(3) # To enforces 1's in the diagonal
U = torch.randn(3,3).triu()
K = L @ U

If then I do: LU_factor, pivots = torch.linalg.lu_factor(K, pivot=False)
I would expect LU_factor to be the same matrix as K, but it’s not. Moreover, if I do: Pu, Lu, Uu, = torch.lu_unpack(LU_factor, pivots)
Then Lu and Uu are now exactly L and U (as expected).

My question is, if by setting pivot=False in lu_factor I enforce P to be the identity, what other operations do lu_factor() and lu_unpack() perform under the hood that make LU_factor be different from L @ U?

Even when you turn off pivoting, lu_factor() still performs the non-trivial
LU factorization and returns the “packed” factorization in LU_factor. So LU_factor is not the original, unfactored K. (Why would it be?)

Here is a script that illustrates this by constructing the “packed” factorization
from your original L and U matrices:

import torch
print (torch.__version__)
_ = torch.manual_seed (2023)
L = torch.randn (3, 3, device = 'cuda').tril()
L.diagonal()[:] = torch.ones (3, device = 'cuda') # To enforces 1's in the diagonal
U = torch.randn (3, 3, device = 'cuda').triu()
K = L @ U
LU_factor, pivots = torch.linalg.lu_factor (K, pivot=False)
tind = torch.triu_indices (3, 3) # "pack" factorization into a single square matrix
LU_pack = L.clone()
LU_pack[tind[0], tind[1]] = U[tind[0], tind[1]]
print ('LU_factor = ...')
print (LU_factor)
print ('LU_pack = ...')
print (LU_pack)
print ('allclose:', torch.allclose (LU_factor, LU_pack))

Oh my bad! I thought LU_factor would be L@U and not L and U packed together (which makes a lot of sense now that you mention it, for many reasons). Thanks for the very quick reply.