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Successful certified Bartels-Stewart substitution returns the exact Sylvester solution

Statement

Suppose Finite-budget Bartels-Stewart solve for a Sylvester equation returns success for A,B,C. Then its returned X satisfies AX+XB=C. If σ(A)∩(−σ(B))=∅, this X is the unique solution.

Facts & Assumptions

Given: A successful output X=UYV∗ from the algorithm, with its verified exact Schur identities and exact local block solves; the delegated rounded Schur calls were accepted only after those checks.

[F1]

Finite-budget Bartels-Stewart solve for a Sylvester equation: Success requires exact Schur certificates, and the block traversal solves TiiYij+YijSjj=Dij−∑k>iTikYkj−∑ℓ<jYiℓSℓj for every block pair.

[F2]
[F3]

A Sylvester equation is uniquely solvable exactly when the spectra are disjoint: The spectral-disjointness condition is equivalent to unique solvability for every right-hand side.

Proof

1.1F1algebra

By [F1], A=UTU∗, B=VSV∗, and D=U∗CV hold exactly. Multiplying AX+XB on the left by U∗ and on the right by V gives TY+YS.

2.1F1step 1.1algebra

In the (i,j) block, TY+YS equals TiiYij+YijSjj+∑k>iTikYkj+∑ℓ<jYiℓSℓj. The successful local solve from [F1] makes this block equal Dij, so every block of TY+YS equals the corresponding block of D.

3.1F2step 1.1step 2.1algebra

Therefore TY+YS=D=U∗CV. Multiplying by U on the left and V∗ on the right yields A(UYV∗)+(UYV∗)B=C. By [F2], this is SA,B(X)=C for the returned X=UYV∗.

4.1F3step 3.1∎

Under the stated spectral-disjointness condition, [F3] gives uniqueness for this right-hand side; hence the returned exact solution is the unique one.

Depends on

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Sources