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
[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