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CG's positive-denominator hypothesis fails on an indefinite system while MINRES can start

Example

For $A_{-} = diag (1, - 1)$ , $b = (1, 1)^{T}$ , and $x_{0} = 0$ , evaluate the formal first CG denominator expression $b^{T} A_{-} b$ outside the CG definition's SPD domain, and compute the first MINRES least-squares problem and the exact solution's membership in $K_{2} (A_{-}, b)$ . Contrast this with the first CG denominator for $A_{+} = diag (1, 2)$ and the same $b, x_{0}$ , and verify that one exact CG step decreases the $A_{+}$ -energy error norm from $\sqrt{3 / 2}$ to $1 / \sqrt{6}$ .

Facts & Assumptions

Given: The two real diagonal systems and starting data in the Example.

[F1]

The conjugate-gradient recurrence: for real SPD $A$ , $x_{0} = 0$ , and $b \neq 0$ , one has $p_{0} = r_{0} = b$ and the first CG step uses $α_{0} = (b^{T} b) / (p_{0}^{T} A p_{0})$ . Substitution of an indefinite matrix into this formula is only a formal attempt to extend it beyond its defined domain.

[F2]

Finite-precision CG has a recursive-residual gap and an attainable-accuracy bound: its hypotheses require a symmetric positive-definite matrix; it makes no claim for the indefinite system $A_{-}$ .

[F3]

The MINRES iterate from the Lanczos tridiagonal least-squares problem: MINRES step $1$ minimizes $∥ β e_{1} - {\overset{ˉ}{T}}_{1} y ∥_{2}$ , where $β = ∥ r_{0} ∥_{2}$ , when Lanczos completes its first column without breakdown.

[F4]

The Krylov subspace $K_{m} (A, b) = span {b, A b, \dots, A^{m - 1} b}$ : $K_{2} (A, b) = span {b, A b}$ .

[F5]

The energy inner product $⟨ u, v ⟩_{A} = ⟨ A u, v ⟩$ and energy norm $∥ u ∥_{A}$ for a Hermitian positive-definite matrix: for real SPD $A$ , $∥ e ∥_{A}^{2} = e^{T} A e$ .

Verification

1.1givenF1F2algebra

Formally setting $p_{0} = b$ for $A_{-}$ gives $A_{-} p_{0} = (1, - 1)^{T}$ , so $p_{0}^{T} A_{-} p_{0} = 1 - 1 = 0$ while $b^{T} b = 2$ . Thus attempting the first CG update formula would divide by zero. Since $A_{-}$ is indefinite, [F1] does not define a CG step for it, and the SPD theorem [F2] does not apply.

1.2givenF3F4algebra

The normalized first Lanczos vector is $q_{1} = b / \sqrt{2}$ . Then $q_{1}^{T} A_{-} q_{1} = 0$ and $A_{-} q_{1} = (1, - 1)^{T} / \sqrt{2}$ has norm $1$ , so the first subdiagonal coefficient is $β_{1} = 1 > 0$ . Hence [F3] applies at step $1$ , with $β = ∥ r_{0} ∥_{2} = \sqrt{2}$ and ${\overset{ˉ}{T}}_{1} = (0, 1)^{T}$ . The squared objective is $∥ (\sqrt{2}, 0)^{T} - (0, y)^{T} ∥_{2}^{2} = 2 + ∣ y ∣^{2}$ , uniquely minimized at $y = 0$ even over $C$ . Also $b$ and $A_{-} b = (1, - 1)^{T}$ are linearly independent, so [F4] gives $K_{2} (A_{-}, b) = R^{2}$ . The exact solution $A_{-}^{- 1} b = (1, - 1)^{T}$ therefore lies in $K_{2}$ ; no step- $2$ MINRES claim is needed after the invariant-subspace breakdown.

2.1givenF1F5algebra∎

For $A_{+}$ , the first CG denominator is $b^{T} A_{+} b = 1 + 2 = 3 > 0$ and $α_{0} = 2 / 3$ . The exact solution is $x_{*} = A_{+}^{- 1} b = (1, 1 / 2)^{T}$ , so $e_{0} = x_{*} - x_{0} = (1, 1 / 2)^{T}$ and [F5] gives $∥ e_{0} ∥_{A_{+}}^{2} = 1 + 2 (1 / 2)^{2} = 3 / 2$ . The first iterate is $x_{1} = x_{0} + α_{0} b = (2 / 3, 2 / 3)^{T}$ , hence $e_{1} = x_{*} - x_{1} = (1 / 3, - 1 / 6)^{T}$ and $∥ e_{1} ∥_{A_{+}}^{2} = (1 / 3)^{2} + 2 (1 / 6)^{2} = 1 / 6$ . Thus the energy-error norm decreases from $\sqrt{3 / 2}$ to $1 / \sqrt{6}$ , a factor of $1 / 3$ . The recursive residual is $r_{1} = b - α_{0} A_{+} b = (1 / 3, - 1 / 3)^{T} \neq 0$ . This verifies the positive first denominator and energy decrease in this SPD instance, contrasting with step 1.1.

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Sources

Youcef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §§6.6–6.8, printed pp. 194–204 (standard reference, not scraped)
David Bindel, Matrix Computations (CS 6210), Cornell University, Fall 2012, Lecture 35: Some minimization problems (standard reference, not scraped)