return A.is_hermitian() and all( v >= 0 for v in A.eigenvalues() )
-def ldlt_naive(A):
- r"""
- Perform a pivoted `LDL^{T}` factorization of the Hermitian
- positive-semidefinite matrix `A`.
-
- This is a naive, recursive implementation that is inefficient due
- to Python's lack of tail-call optimization. The pivot strategy is
- to choose the largest diagonal entry of the matrix at each step,
- and to permute it into the top-left position. Ultimately this
- results in a factorization `A = PLDL^{T}P^{T}`, where `P` is a
- permutation matrix, `L` is unit-lower-triangular, and `D` is
- diagonal decreasing from top-left to bottom-right.
-
- ALGORITHM:
-
- The algorithm is based on the discussion in Golub and Van Loan, but with
- some "typos" fixed.
-
- OUTPUT:
-
- A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where,
-
- * `P` is a permutaiton matrix
- * `L` is unit lower-triangular
- * `D` is a diagonal matrix whose entries are decreasing from top-left
- to bottom-right
-
- SETUP::
-
- sage: from mjo.ldlt import ldlt_naive, is_positive_semidefinite_naive
-
- EXAMPLES:
-
- All three factors should be the identity when the original matrix is::
-
- sage: I = matrix.identity(QQ,4)
- sage: P,L,D = ldlt_naive(I)
- sage: P == I and L == I and D == I
- True
-
- TESTS:
-
- Ensure that a "random" positive-semidefinite matrix is factored correctly::
-
- sage: set_random_seed()
- sage: n = ZZ.random_element(5)
- sage: A = matrix.random(QQ, n)
- sage: A = A*A.transpose()
- sage: is_positive_semidefinite_naive(A)
- True
- sage: P,L,D = ldlt_naive(A)
- sage: A == P*L*D*L.transpose()*P.transpose()
- True
-
- """
- n = A.nrows()
-
- # Use the fraction field of the given matrix so that division will work
- # when (for example) our matrix consists of integer entries.
- ring = A.base_ring().fraction_field()
-
- if n == 0 or n == 1:
- # We can get n == 0 if someone feeds us a trivial matrix.
- P = matrix.identity(ring, n)
- L = matrix.identity(ring, n)
- D = A
- return (P,L,D)
-
- A1 = A.change_ring(ring)
- diags = A1.diagonal()
- s = diags.index(max(diags))
- P1 = copy(A1.matrix_space().identity_matrix())
- P1.swap_rows(0,s)
- A1 = P1.T * A1 * P1
- alpha1 = A1[0,0]
-
- # Golub and Van Loan mention in passing what to do here. This is
- # only sensible if the matrix is positive-semidefinite, because we
- # are assuming that we can set everything else to zero as soon as
- # we hit the first on-diagonal zero.
- if alpha1 == 0:
- P = A1.matrix_space().identity_matrix()
- L = P
- D = A1.matrix_space().zero()
- return (P,L,D)
-
- v1 = A1[1:n,0]
- A2 = A1[1:,1:]
-
- P2, L2, D2 = ldlt_naive(A2 - (v1*v1.transpose())/alpha1)
-
- P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [0*v1, P2]])
- L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [P2.transpose()*v1/alpha1, L2]])
- D1 = block_matrix(2,2, [[alpha1, ZZ(0)],
- [0*v1, D2]])
-
- return (P1,L1,D1)
-
-
-
-def ldlt_fast(A):
- r"""
- Perform a fast, pivoted `LDL^{T}` factorization of the Hermitian
- positive-semidefinite matrix `A`.
-
- This function is much faster than ``ldlt_naive`` because the
- tail-recursion has been unrolled into a loop.
- """
- ring = A.base_ring().fraction_field()
- A = A.change_ring(ring)
-
- # Keep track of the permutations in a vector rather than in a
- # matrix, for efficiency.
- n = A.nrows()
- p = list(range(n))
-
- for k in range(n):
- # We need to loop once for every diagonal entry in the
- # matrix. So, as many times as it has rows/columns. At each
- # step, we obtain the permutation needed to put things in the
- # right place, then the "next" entry (alpha) of D, and finally
- # another column of L.
- diags = A.diagonal()[k:n]
- alpha = max(diags)
-
- # We're working *within* the matrix ``A``, so every index is
- # offset by k. For example: after the second step, we should
- # only be looking at the lower 3-by-3 block of a 5-by-5 matrix.
- s = k + diags.index(alpha)
-
- # Move the largest diagonal element up into the top-left corner
- # of the block we're working on (the one starting from index k,k).
- # Presumably this is faster than hitting the thing with a
- # permutation matrix.
- #
- # Since "L" is stored in the lower-left "half" of "A", it's a
- # good thing that we need to permute "L," too. This is due to
- # how P2.T appears in the recursive algorithm applied to the
- # "current" column of L There, P2.T is computed recusively, as
- # 1 x P3.T, and P3.T = 1 x P4.T, etc, from the bottom up. All
- # are eventually applied to "v" in order. Here we're working
- # from the top down, and rather than keep track of what
- # permutations we need to perform, we just perform them as we
- # go along. No recursion needed.
- A.swap_columns(k,s)
- A.swap_rows(k,s)
-
- # Update the permutation "matrix" with the swap we just did.
- p_k = p[k]
- p[k] = p[s]
- p[s] = p_k
-
- # Now the largest diagonal is in the top-left corner of the
- # block below and to the right of index k,k. When alpha is
- # zero, we can just leave the rest of the D/L entries
- # zero... which is exactly how they start out.
- if alpha != 0:
- # Update the "next" block of A that we'll work on during
- # the following iteration. I think it's faster to get the
- # entries of a row than a column here?
- for i in range(n-k-1):
- for j in range(i+1):
- A[k+1+j,k+1+i] = A[k+1+j,k+1+i] - A[k,k+1+j]*A[k,k+1+i]/alpha
- A[k+1+i,k+1+j] = A[k+1+j,k+1+i] # keep it symmetric!
-
- for i in range(n-k-1):
- # Store the "new" (kth) column of L, being sure to set
- # the lower-left "half" from the upper-right "half"
- A[k+i+1,k] = A[k,k+1+i]/alpha
-
- MS = A.matrix_space()
- P = MS.matrix(lambda i,j: p[j] == i)
- D = MS.diagonal_matrix(A.diagonal())
-
- for i in range(n):
- A[i,i] = 1
- for j in range(i+1,n):
- A[i,j] = 0
-
- return P,A,D
-
-
-def block_ldlt_naive(A, check_hermitian=False):
+def block_ldlt(A):
r"""
Perform a block-`LDL^{T}` factorization of the Hermitian
matrix `A`.
- This is a naive, recursive implementation akin to
- ``ldlt_naive()``, where the pivots (and resulting diagonals) are
- either `1 \times 1` or `2 \times 2` blocks. The pivots are chosen
- using the Bunch-Kaufmann scheme that is both fast and numerically
- stable.
+ The standard `LDL^{T}` factorization of a positive-definite matrix
+ `A` factors it as `A = LDL^{T}` where `L` is unit-lower-triangular
+ and `D` is diagonal. If one allows row/column swaps via a
+ permutation matrix `P`, then this factorization can be extended to
+ some positive-semidefinite matrices `A` via the factorization
+ `P^{T}AP = LDL^{T}` that places the zeros at the bottom of `D` to
+ avoid division by zero. These factorizations extend easily to
+ complex Hermitian matrices when one replaces the transpose by the
+ conjugate-transpose.
+
+ However, we can go one step further. If, in addition, we allow `D`
+ to potentially contain `2 \times 2` blocks on its diagonal, then
+ every real or complex Hermitian matrix `A` can be factored as `A =
+ PLDL^{*}P^{T}`. When the row/column swaps are made intelligently,
+ this process is numerically stable over inexact rings like ``RDF``.
+ Bunch and Kaufman describe such a "pivot" scheme that is suitable
+ for the solution of Hermitian systems, and that is how we choose
+ our row and column swaps.
OUTPUT:
- A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where,
+ If the input matrix is Hermitian, we return a triple `(P,L,D)`
+ such that `A = PLDL^{*}P^{T}` and
- * `P` is a permutation matrix
- * `L` is unit lower-triangular
+ * `P` is a permutation matrix,
+ * `L` is unit lower-triangular,
* `D` is a block-diagonal matrix whose blocks are of size
one or two.
- """
- n = A.nrows()
-
- # Use the fraction field of the given matrix so that division will work
- # when (for example) our matrix consists of integer entries.
- ring = A.base_ring().fraction_field()
-
- if n == 0 or n == 1:
- # We can get n == 0 if someone feeds us a trivial matrix.
- # For block-LDLT, n=2 is a base case.
- P = matrix.identity(ring, n)
- L = matrix.identity(ring, n)
- D = A
- return (P,L,D)
-
- alpha = (1 + ZZ(17).sqrt()) * ~ZZ(8)
- A1 = A.change_ring(ring)
-
- # Bunch-Kaufmann step 1, Higham step "zero." We use Higham's
- # "omega" notation instead of Bunch-Kaufman's "lamda" because
- # lambda means other things in the same context.
- column_1_subdiag = [ a_i1.abs() for a_i1 in A1[1:,0].list() ]
- omega_1 = max([ a_i1 for a_i1 in column_1_subdiag ])
-
- if omega_1 == 0:
- # "There's nothing to do at this step of the algorithm,"
- # which means that our matrix looks like,
- #
- # [ 1 0 ]
- # [ 0 B ]
- #
- # We could still do a pivot_one_by_one() here, but it would
- # pointlessly subract a bunch of zeros and multiply by one.
- B = A1[1:,1:]
- one = matrix(ring, 1, 1, [1])
- P2, L2, D2 = block_ldlt_naive(B)
- P1 = block_diagonal_matrix(one, P2)
- L1 = block_diagonal_matrix(one, L2)
- D1 = block_diagonal_matrix(one, D2)
- return (P1,L1,D1)
-
- def pivot_one_by_one(M, c=None):
- # Perform a one-by-one pivot on "M," swapping row/columns "c".
- # If "c" is None, no swap is performed.
- if c is not None:
- P1 = copy(M.matrix_space().identity_matrix())
- P1.swap_rows(0,c)
- M = P1.T * M * P1
-
- # The top-left entry is now our 1x1 pivot.
- C = M[1:n,0]
- B = M[1:,1:]
-
- P2, L2, D2 = block_ldlt_naive(B - (C*C.transpose())/M[0,0])
-
- if c is None:
- P1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [0*C, P2]])
- else:
- P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [0*C, P2]])
+ If the input matrix is not Hermitian, the output from this function
+ is undefined.
- L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [P2.transpose()*C/M[0,0], L2]])
- D1 = block_matrix(2,2, [[M[0,0], ZZ(0)],
- [0*C, D2]])
+ SETUP::
- return (P1,L1,D1)
+ sage: from mjo.ldlt import block_ldlt
+ EXAMPLES:
- if A1[0,0].abs() > alpha*omega_1:
- return pivot_one_by_one(A1)
+ This three-by-three real symmetric matrix has one positive, one
+ negative, and one zero eigenvalue -- so it is not any flavor of
+ (semi)definite, yet we can still factor it::
+
+ sage: A = matrix(QQ, [[0, 1, 0],
+ ....: [1, 1, 2],
+ ....: [0, 2, 0]])
+ sage: P,L,D = block_ldlt(A)
+ sage: P
+ [0 0 1]
+ [1 0 0]
+ [0 1 0]
+ sage: L
+ [ 1 0 0]
+ [ 2 1 0]
+ [ 1 1/2 1]
+ sage: D
+ [ 1| 0| 0]
+ [--+--+--]
+ [ 0|-4| 0]
+ [--+--+--]
+ [ 0| 0| 0]
+ sage: P.transpose()*A*P == L*D*L.transpose()
+ True
- r = 1 + column_1_subdiag.index(omega_1)
+ This two-by-two matrix has no standard factorization, but it
+ constitutes its own block-factorization::
- # If the matrix is Hermitian, we need only look at the above-
- # diagonal entries to find the off-diagonal of maximal magnitude.
- omega_r = max( a_rj.abs() for a_rj in A1[:r,r].list() )
+ sage: A = matrix(QQ, [ [0,1],
+ ....: [1,0] ])
+ sage: block_ldlt(A)
+ (
+ [1 0] [1 0] [0 1]
+ [0 1], [0 1], [1 0]
+ )
- if A1[0,0].abs()*omega_r >= alpha*(omega_1**2):
- return pivot_one_by_one(A1)
+ The same is true of the following complex Hermitian matrix::
- if A1[r,r].abs() > alpha*omega_r:
- # Higham step (3)
- # Another 1x1 pivot, but this time swapping indices 0,r.
- return pivot_one_by_one(A1,r)
+ sage: A = matrix(QQbar, [ [ 0,I],
+ ....: [-I,0] ])
+ sage: block_ldlt(A)
+ (
+ [1 0] [1 0] [ 0 I]
+ [0 1], [0 1], [-I 0]
+ )
- # Higham step (4)
- # If we made it here, we have to do a 2x2 pivot.
- P1 = copy(A1.matrix_space().identity_matrix())
- P1.swap_rows(1,r)
- A1 = P1.T * A1 * P1
+ TESTS:
- # The top-left 2x2 submatrix is now our pivot.
- E = A1[:2,:2]
- C = A1[2:n,0:2]
- B = A1[2:,2:]
+ All three factors should be the identity when the original matrix is::
- if B.nrows() == 0:
- # We have a two-by-two matrix that we can do nothing
- # useful with.
- P = matrix.identity(ring, n)
- L = matrix.identity(ring, n)
- D = A1
- return (P,L,D)
+ sage: set_random_seed()
+ sage: n = ZZ.random_element(6)
+ sage: I = matrix.identity(QQ,n)
+ sage: P,L,D = block_ldlt(I)
+ sage: P == I and L == I and D == I
+ True
- P2, L2, D2 = block_ldlt_naive(B - (C*E.inverse()*C.transpose()))
+ Ensure that a "random" real symmetric matrix is factored correctly::
- P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [0*C, P2]])
+ sage: set_random_seed()
+ sage: n = ZZ.random_element(6)
+ sage: A = matrix.random(QQ, n)
+ sage: A = A + A.transpose()
+ sage: P,L,D = block_ldlt(A)
+ sage: A == P*L*D*L.transpose()*P.transpose()
+ True
- L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [P2.transpose()*C*E.inverse(), L2]])
- D1 = block_diagonal_matrix(E,D2)
+ Ensure that a "random" complex Hermitian matrix is factored correctly::
- return (P1,L1,D1)
+ sage: set_random_seed()
+ sage: n = ZZ.random_element(6)
+ sage: F = NumberField(x^2 +1, 'I')
+ sage: A = matrix.random(F, n)
+ sage: A = A + A.conjugate_transpose()
+ sage: P,L,D = block_ldlt(A)
+ sage: A == P*L*D*L.conjugate_transpose()*P.conjugate_transpose()
+ True
+ Ensure that a "random" complex positive-semidefinite matrix is
+ factored correctly and that the resulting block-diagonal matrix is
+ in fact diagonal::
-def block_ldlt(A):
- r"""
- Perform a block-`LDL^{T}` factorization of the Hermitian
- matrix `A`.
+ sage: set_random_seed()
+ sage: n = ZZ.random_element(6)
+ sage: F = NumberField(x^2 +1, 'I')
+ sage: A = matrix.random(F, n)
+ sage: A = A*A.conjugate_transpose()
+ sage: P,L,D = block_ldlt(A)
+ sage: A == P*L*D*L.conjugate_transpose()*P.conjugate_transpose()
+ True
+ sage: diagonal_matrix(D.diagonal()) == D
+ True
- OUTPUT:
+ The factorization should be a no-op on diagonal matrices::
- A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where,
+ sage: set_random_seed()
+ sage: n = ZZ.random_element(6)
+ sage: A = matrix.diagonal(random_vector(QQ, n))
+ sage: I = matrix.identity(QQ,n)
+ sage: P,L,D = block_ldlt(A)
+ sage: P == I and L == I and A == D
+ True
- * `P` is a permutation matrix
- * `L` is unit lower-triangular
- * `D` is a block-diagonal matrix whose blocks are of size
- one or two.
"""
# We have to make at least one copy of the input matrix so that we
# right-hand corner of "A".
for i in range(n-k-1):
for j in range(i+1):
- A[k+1+j,k+1+i] = ( A[k+1+j,k+1+i] -
- A[k,k+1+j]*A[k,k+1+i]/A[k,k] )
- A[k+1+i,k+1+j] = A[k+1+j,k+1+i] # keep it symmetric!
+ A[k+1+i,k+1+j] = ( A[k+1+i,k+1+j] -
+ 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.
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
# 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:
# [ 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
# 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.
# 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+j,k+2+i] = A[k+2+j,k+2+i] - schur_complement[j,i]
- A[k+2+i,k+2+j] = A[k+2+j,k+2+i] # keep it symmetric!
+ 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
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]
projects / sage.d.git / blobdiff