- 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 permuts "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!