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A successful rounded Cholesky factorization has componentwise and normwise backward error bounds

Statement

Let n≥1 and let A=A∗ be positive definite in Cn×n. Suppose the rounded Cholesky algorithm succeeds with L^. Put m=20n+20 and g=γm=mu1−mu; assume mu<1/5. For E=L^L^∗−A, which is Hermitian, and with ∣M∣ denoting entrywise absolute value for matrices, ∣E∣≤100g ∣L^∣ ∣L^∣∗entrywise,∥E∥F≤100g∥L^∥F2,∥E∥2≤100g∥L^∥F2. Thus cn(u)u=100mu1−mu. If also 100n g≤1/2, then ∥E∥F≤200n g∥A∥F,∥E∥2≤200n g∥A∥F. These bounds apply on success only; positive definiteness does not guarantee rounded success.

Facts & Assumptions

Given: successful rounded recurrences, the normal-range model, and the exact Cholesky convention.

[F1]

Failure-aware rounded Cholesky factorization: success supplies positive rounded pivots and rounded updates using six-real-operation complex products.

[F2]

A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique: Hermitian positive-definite input has an exact lower triangular factor with positive diagonal.

[F3]

If |delta_i| <= u and n u < 1, then prod (1+delta_i) = 1 + theta_n with |theta_n| <= gamma_n: a chain of k successful operations has the gamma_k relative bound when ku<1.

Proof

1.1givenF1F3algebra

For one complex product zw, write its real exact products as ac,bd,ad,bc. Each rounded product has error at most u times its operand product; each final rounded real sum or difference contributes at most u times the sum of the two operand magnitudes. Hence the real-component error is at most (2u+u2)(∣ac∣+∣bd∣) and the imaginary-component error at most (2u+u2)(∣ad∣+∣bc∣). Since u<1/200, their Euclidean combination is at most 6u∣z∣∣w∣. A sequential sum of at most n complex products contributes at most γn(1+6u) times the absolute product sum; adding product errors gives 6u+γn(1+6u)=(n+6)u1−nu≤γn+6≤g. Each diagonal term ∣l^jk∣2 uses two rounded squares and one rounded addition; sequentially summing at most n terms uses at most 4n rounded real operations, so the same expansion bounds the diagonal sum error by γ4nSj≤gSj.

2.1step 1.1F1algebra

For i>j, let Sij=∑k<jl^ikl^jk‾, Tij=∑k≤j∣l^ik∣∣l^jk∣, and S^ij be the computed dot product. Step 1.1 gives ∣S^ij−Sij∣≤gTij and ∣S^ij∣≤(1+g)Tij. The subtraction error is at most u(∣aij∣+∣S^ij∣), and the two component divisions imply ∣l^ijl^jj−r^ij∣≤u∣r^ij∣; also ∣r^ij∣≤(1+u)(∣aij∣+∣S^ij∣). Therefore ∣Eij∣≤gTij+u(2+u)(∣aij∣+(1+g)Tij). Since ∣aij∣≤Tij+∣Eij∣, moving u(2+u)∣Eij∣ to the left gives ∣Eij∣≤g+u(2+u)(2+g)1−u(2+u)Tij. Here u≤g/m, g<1/4, m≥40, and u<1/200, so the numerator coefficient is less than 1.12g and the denominator exceeds 0.989. The quotient is less than 1.14g<100g.

2.2step 1.1F1F3algebra

On the diagonal put Sj=∑k<j∣l^jk∣2. Step 1.1 and the pivot subtraction give pj=ajj−Sj+ϵj with ∣ϵj∣≤u∣ajj∣+(g+u(1+g))Sj≤u∣ajj∣+1.26gSj. The rounded square root has l^jj=pj(1+δj), ∣δj∣≤u, so ∣l^jj2−pj∣≤(2u+u2)pj≤3upj. From ∣ajj∣≤Sj+pj+∣ϵj∣ we get (1−u)∣ajj∣≤(1+1.26g)Sj+pj. Since u<1/200 and g<1/4, this implies ∣ajj∣≤1.33Sj+1.01pj≤2110(Sj+pj). Also pj=l^jj2/(1+δj)2≤1.011l^jj2, so 2110(Sj+pj)≤115(Sj+l^jj2). Thus ∣Ejj∣≤u∣ajj∣+1.26gSj+3upj≤1.32gSj+0.14gl^jj2≤100g(Sj+l^jj2), using u≤g/40. Hermitian symmetry gives the corresponding conjugate entries, proving the componentwise estimate.

3.1step 2.1step 2.2F2F4F5algebra∎

By [F5], Cauchy-Schwarz applied to each row-column inner product gives ∥∣L^∣∣L^∣∗∥F2≤∑i,j(∑k∣l^ik∣2)(∑k∣l^jk∣2)=∥L^∥F4. The entrywise bound therefore yields ∥E∥F≤100g∥L^∥F2, and [F4] gives the spectral bound. For the relative consequence, take traces in L^L^∗=A+E: ∥L^∥F2≤n∥A∥F+n∥E∥F≤n∥A∥F+100n g∥L^∥F2. The additional condition gives ∥L^∥F2≤2n∥A∥F, yielding both stated relative norm bounds.

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