提取ldltfact的相关分解矩阵可能有点乏味.下面的示例显示了一个与问题中的示例相似的示例，并进行了最终测试，即提取的矩阵将恢复原始的因式分解后的矩阵:

Extracting the relevant factorization matrices of ldltfact can be a little tedious. The following example shows an example similar to the one in the question with a final test that the extracted matrices recover the original factorized one:

srand(1)
pre = sprand(10,10,0.5)
H = pre + pre' + speye(10,10)

fac = ldltfact(H; shift=0.0)
P = sparse(1:size(H,1),fac[:p],ones(size(H,1)))
LD = sparse(fac[:LD]) # this matrix contains both D and L embedded in it

L = copy(LD)
for i=1:size(L,1)
  L[i,i] = 1.0
end

D = sparse(1:size(L,1),1:size(L,1),diag(LD))

PHP = P*H*P'
LDL = L*D*L'

using Base.Test
@test PHP ≈ LDL

预期的输出(在Julia v0.6.3上是实际的):

The expected output (and actual on Julia v0.6.3):

julia> @test PHP ≈ LDL
Test Passed

希望这会有所帮助.