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A trace eigenspace objective and a Lanczos start-angle bound

Example

Let $A = diag (5, 2, 0)$ and minimize $f (x) = - x^{T} A x$ on $St (3, 1) = S^{2}$ . The top eigenspace is $span (e_{1})$ , so the optimum is $- 5$ . Start Lanczos at $b = (1, 1, 1)^{T} / \sqrt{3}$ and use the two-dimensional Krylov space. The normalized vector generated by $p (A) b$ , with $p (λ) = (λ - 1) / 4$ , has Rayleigh quotient $41 / 9$ for $A$ , objective value $- 41 / 9$ , and objective gap $4 / 9$ . The Chebyshev start-angle theorem bounds the largest Ritz-value gap by $5 / 8$ . Direct arithmetic also gives $4 / 9 \leq 5 / 8$ for this explicit candidate; the candidate need not maximize the Rayleigh quotient on the Krylov space.

Facts & Assumptions

Given: The diagonal matrix $A$ , unit starting vector $b$ , and the Stiefel trace objective above.

[F1]

The Stiefel gradient is the tangent projection of the ambient Euclidean gradient (Projected Euclidean gradients are the Stiefel and Grassmann gradients).

[F2]

For a simple top eigenvalue, a nonzero top-eigenspace component of the start vector gives the stated Chebyshev Ritz-value error bound (Lanczos largest Ritz values obey a Chebyshev start-angle bound, including degenerate branches).

Proof

technique · explicit polynomial filtering and exact Rayleigh-quotient arithmetic

1.1F1F2givenalgebra

The ambient gradient of $f$ is $- 2 A x$ . At $x = e_{2}$ , the Stiefel projection in [F1] is zero, so the eigendirection is stationary; at the global minimizer $e_{1}$ , $f (e_{1}) = - 5$ . For the displayed start, $p (A) b = (1, 1 / 4, - 1 / 4)^{T} / \sqrt{3}$ is nonzero and belongs to $K_{2} (A, b) = span {b, A b}$ .

2.1givenstep 1.1algebra

The squared norm of this vector is $(1 + 1 / 16 + 1 / 16) / 3 = 3 / 8$ , and its $A$ -quadratic form is $(5 + 2 / 16) / 3 = 41 / 24$ . Dividing gives Rayleigh quotient $41 / 9$ , hence objective $- 41 / 9$ and gap $5 - 41 / 9 = 4 / 9$ .

3.1F2givenstep 1.1step 2.1algebra∎

Here $λ_{1} = 5$ , $λ_{2} = 2$ , $λ_{3} = 0$ , the top multiplicity is one, and $τ = ∥ (I - P) b ∥ / ∥ P b ∥ = \sqrt{2}$ . Since step 1.1 places the candidate in $K_{2}$ , its quotient shows $θ_{2} \geq 41 / 9$ . For Krylov dimension $m = 2$ , the theorem [F2] gives $λ_{1} - θ_{2} \leq \frac{(5 - 0) (\sqrt{2})^{2}}{T_{1} (1 + 2 (5 - 2) / (2 - 0))^{2}} = \frac{10}{4^{2}} = \frac{5}{8} .$ Since $θ_{2} \geq 41 / 9$ , its Ritz gap obeys $0 \leq λ_{1} - θ_{2} \leq 4 / 9 \leq 5 / 8$ , so the candidate comparison and source bound agree. Also $b$ and $A b$ are independent because the third coordinate of $A b$ is zero while that of $b$ is nonzero, so the two-dimensional Krylov space is attained without breakdown.

Depends on

Used by

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Sources

Yousef Saad, Iterative Methods for Sparse Linear Systems, second edition (standard reference, not scraped)
David Bindel, Matrix Computations (CS 6210), Cornell University, Lectures 38–39 (2012) (standard reference, not scraped)