if A.nrows() == 0:
return True # vacuously
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 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!
-
- 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
projects / sage.d.git / blobdiff