From bf58ed8a630a51f4a521e5b4952bd10303c13696 Mon Sep 17 00:00:00 2001 From: Michael Orlitzky Date: Sun, 4 Oct 2020 09:40:50 -0400 Subject: [PATCH] mjo/ldlt.py: make block_ldlt() work on Hermitian matrices, too. --- mjo/ldlt.py | 45 +++++++++++++++++++++------------------------ 1 file changed, 21 insertions(+), 24 deletions(-) diff --git a/mjo/ldlt.py b/mjo/ldlt.py index 09b3bb0..729e011 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -435,15 +435,13 @@ def block_ldlt(A): for i in range(n-k-1): for j in range(i+1): A[k+1+i,k+1+j] = ( A[k+1+i,k+1+j] - - A[k,k+1+i]*A[k,k+1+j]/A[k,k] ) - A[k+1+j,k+1+i] = A[k+1+i,k+1+j] # keep it symmetric! + A[k+1+i,k]*A[k,k+1+j]/A[k,k] ) + A[k+1+j,k+1+i] = A[k+1+i,k+1+j].conjugate() # stay hermitian! for i in range(n-k-1): # Store the new (kth) column of "L" within the lower- - # left-hand corner of "A", being sure to set the lower- - # left entries from the upper-right ones to avoid - # collisions. - A[k+i+1,k] = A[k,k+1+i]/A[k,k] + # left-hand corner of "A". + A[k+i+1,k] /= A[k,k] # No return value, only the desired side effects of updating # p, d, and A. @@ -460,7 +458,9 @@ def block_ldlt(A): if k == (n-1): # Handle this trivial case manually, since otherwise the # algorithm's references to the e.g. "subdiagonal" are - # meaningless. + # meaningless. The corresponding entry of "L" will be + # fixed later (since it's an on-diagonal element, it gets + # set to one eventually). d.append( matrix(ring, 1, [[A[k,k]]]) ) k += 1 continue @@ -469,10 +469,8 @@ def block_ldlt(A): # kth column. This occurs prior to Step (1) in Higham, # but is part of Step (1) in Bunch and Kaufman. We adopt # Higham's "omega" notation instead of B&K's "lambda" - # because "lambda" can lead to some confusion. Beware: - # the subdiagonals of our matrix are being overwritten! - # So we actually use the corresponding row entries instead. - column_1_subdiag = [ a_ki.abs() for a_ki in A[k,k+1:].list() ] + # because "lambda" can lead to some confusion. + column_1_subdiag = [ a_ki.abs() for a_ki in A[k+1:,k].list() ] omega_1 = max([ a_ki for a_ki in column_1_subdiag ]) if omega_1 == 0: @@ -482,7 +480,9 @@ def block_ldlt(A): # [ 0 B ] # # and we can simply skip to the next step after recording - # the 1x1 pivot "1" in the top-left position. + # the 1x1 pivot "a" in the top-left position. The entry "a" + # will be adjusted to "1" later on to ensure that "L" is + # (block) unit-lower-triangular. d.append( matrix(ring, 1, [[A[k,k]]]) ) k += 1 continue @@ -508,9 +508,8 @@ def block_ldlt(A): # B&K's Step (3) where we find the largest off-diagonal entry # (in magniture) in column "r". Since the matrix is Hermitian, # we need only look at the above-diagonal entries to find the - # off-diagonal of maximal magnitude. (Beware: the subdiagonal - # entries are being overwritten.) - omega_r = max( a_rj.abs() for a_rj in A[:r,r].list() ) + # off-diagonal of maximal magnitude. + omega_r = max( a_rj.abs() for a_rj in A[r,k:r].list() ) if A[k,k].abs()*omega_r >= alpha*(omega_1**2): # Step (2) in Higham or Step (4) in B&K. @@ -540,22 +539,22 @@ def block_ldlt(A): # We don't actually need the inverse of E, what we really need # is C*E.inverse(), and that can be found by setting # - # C*E.inverse() == X <====> XE == C. + # X = C*E.inverse() <====> XE = C. # - # The latter can be found much more easily by solving a system. - # Note: I do not actually know that sage solves the system more + # Then "X" can be found easily by solving a system. Note: I + # do not actually know that sage solves the system more # intelligently, but this is still The Right Thing To Do. CE_inverse = E.solve_left(C) - schur_complement = B - (CE_inverse*C.transpose()) + schur_complement = B - (CE_inverse*C.conjugate_transpose()) # Compute the Schur complement that we'll work on during # the following iteration, and store it back in the lower- # right-hand corner of "A". for i in range(n-k-2): for j in range(i+1): - A[k+2+i,k+2+j] = A[k+2+i,k+2+j] - schur_complement[i,j] - A[k+2+j,k+2+i] = A[k+2+j,k+2+i] - schur_complement[j,i] + A[k+2+i,k+2+j] = schur_complement[i,j] + A[k+2+j,k+2+i] = schur_complement[j,i] # The on- and above-diagonal entries of "L" will be fixed # later, so we only need to worry about the lower-left entry @@ -565,9 +564,7 @@ def block_ldlt(A): for i in range(n-k-2): for j in range(2): # Store the new (k and (k+1)st) columns of "L" within - # the lower-left-hand corner of "A", being sure to set - # the lower-left entries from the upper-right ones to - # avoid collisions. + # the lower-left-hand corner of "A". A[k+i+2,k+j] = CE_inverse[i,j] -- 2.44.2