- """
- 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):
- 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.
-
- OUTPUT:
-
- A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where,
-
- * `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]])
-
- 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]])
-
- return (P1,L1,D1)
-
-
- if A1[0,0].abs() > alpha*omega_1:
- return pivot_one_by_one(A1)
-
- r = 1 + column_1_subdiag.index(omega_1)
-
- # 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() )
-
- if A1[0,0].abs()*omega_r >= alpha*(omega_1**2):
- return pivot_one_by_one(A1)
-
- 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)
-
- # 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
-
- # The top-left 2x2 submatrix is now our pivot.
- E = A1[:2,:2]
- C = A1[2:n,0:2]
- B = A1[2:,2:]
-
- 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)
-
- P2, L2, D2 = block_ldlt_naive(B - (C*E.inverse()*C.transpose()))
-
- P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [0*C, P2]])
-
- L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
- [P2.transpose()*C*E.inverse(), L2]])
- D1 = block_diagonal_matrix(E,D2)
-
- return (P1,L1,D1)
-
-
-def block_ldlt(A):
- r"""
- Perform a block-`LDL^{T}` factorization of the Hermitian
- matrix `A`.
-
- 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:
-
- 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,
- * `D` is a block-diagonal matrix whose blocks are of size
- one or two.
-
- If the input matrix is not Hermitian, the output from this function
- is undefined.
-
- EXAMPLES:
-
- 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.T*A*P == L*D*L.T
- True
-
- This two-by-two matrix has no standard factorization, but it
- constitutes its own block-factorization::
-
- sage: A = matrix(QQ, [ [0,1],
- ....: [1,0] ])
- sage: block_ldlt(A)
- (
- [1 0] [1 0] [0 1]
- [0 1], [0 1], [1 0]
- )
-
- The same is true of the following complex Hermitian matrix::
-
- sage: A = matrix(QQbar, [ [ 0,I],
- ....: [-I,0] ])
- sage: block_ldlt(A)
- (
- [1 0] [1 0] [ 0 I]
- [0 1], [0 1], [-I 0]
- )
-
- TESTS:
-
- All three factors should be the identity when the original matrix is::
-
- 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
-
- Ensure that a "random" real symmetric matrix is factored correctly::
-
- 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.transpose()
- sage: P,L,D = block_ldlt(A)
- sage: A == P*L*D*L.transpose()*P.transpose()
- True
-
- Ensure that a "random" complex Hermitian matrix is factored correctly::
-
- 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.transpose()*P.transpose()
- True
-
- Ensure that a "random" complex positive-semidefinite matrix is
- factored correctly and that the resulting block-diagonal matrix is
- in fact diagonal::
-
- 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.transpose()*P.transpose()
- True
- sage: diagonal_matrix(D.diagonal()) == D
- True
-
- The factorization should be a no-op on diagonal matrices::