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Under unreduced hypotheses, the Francis double-shift chase matches explicit shifted QR

Statement

Let $n \geq 2$ , let $H \in R^{n \times n}$ be unreduced upper Hessenberg, and let $p (z) = z^{2} - s z + t$ be real with $p (H)$ nonsingular and $p (H) e_{1} \neq 0$ . Assume the exact double-shift bulge chase from Exact finite Francis single- or double-shift Hessenberg QR step and the two successive shifted QR steps for the roots of $p$ both finish unreduced. Let $Q_{b}$ be the bulge-chase factor, and let $p (H) = Q_{e} R_{e}$ be a QR factorization with positive diagonal in $R_{e}$ . Then $Q_{b} = Q_{e} D$ for a diagonal sign matrix $D$ . Its first sign records the possible difference between the bulge reflector's normalization and the positive-diagonal QR normalization, and

$Q_{b}^{T} H Q_{b} = D^{T} (Q_{e}^{T} H Q_{e}) D .$

Thus the real Francis step agrees with the real-normalized explicit two-shift QR similarity up to diagonal signs; arbitrary complex QR normalizations introduce diagonal unitary phases, which reduce to signs when both accumulated factors are real. No statement is made for a deflated or singular-shift branch.

Facts & Assumptions

Given: The matrix, polynomial, nonzero and unreduced hypotheses in the statement.

[F1]

The real Francis double-shift polynomial is real and annihilates its trailing block: a real trailing $2 \times 2$ block supplies the real quadratic $p (z)$ whose roots form the shift pair.

[F2]

Shifted QR iteration preserves upper Hessenberg form: an exact shifted QR step on an upper-Hessenberg matrix remains upper Hessenberg.

[F3]

An unreduced Hessenberg similarity is determined up to column phases by its first unitary column: two unitary similarities of the same unreduced Hessenberg matrix that yield unreduced Hessenberg matrices and have the same first column differ only by diagonal phases.

[F4]

Exact finite Francis single- or double-shift Hessenberg QR step: the algorithm returns completed only when the output is upper Hessenberg and $Q e_{1}$ is proportional to $p (H) e_{1}$ .

[F5]

Unshifted QR iteration, shifted QR iteration, deflation, and the Wilkinson shift: an exact shifted QR step factors $H - μ I = Q R$ and updates $H_{+} = R Q + μ I = Q^{*} H Q$ .

Proof

1.1givenF1F5algebra

By [F1], write the roots of $p$ as $μ_{1}, μ_{2}$ with $μ_{1} + μ_{2} = s$ and $μ_{1} μ_{2} = t$ . Factor $H - μ_{1} I = Q_{1} R_{1}$ and let $H_{1} = R_{1} Q_{1} + μ_{1} I = Q_{1}^{*} H Q_{1}$ by [F5]. Factoring $H_{1} - μ_{2} I = Q_{2} R_{2}$ gives $p (H) = (H - μ_{2} I) (H - μ_{1} I) = Q_{1} (H_{1} - μ_{2} I) R_{1} = Q_{1} Q_{2} R_{2} R_{1}$ . Put $Q_{c} = Q_{1} Q_{2}$ and $R_{c} = R_{2} R_{1}$ , so $p (H) = Q_{c} R_{c}$ with $Q_{c}$ unitary and $R_{c}$ upper triangular. Since $p (H)$ is nonsingular, both $R_{c}$ and $R_{e}$ are nonsingular. From $Q_{e}^{*} Q_{c} = R_{e} R_{c}^{- 1}$ , the matrix $Q_{e}^{*} Q_{c}$ is upper triangular and unitary, hence diagonal unitary; write $Q_{c} = Q_{e} Θ$ for a diagonal unitary $Θ$ . Also $p (H) e_{1} = (R_{c})_{11} Q_{c} e_{1} = (R_{e})_{11} Q_{e} e_{1}$ , with $(R_{e})_{11} > 0$ .

2.1givenF2step 1.1algebra

By [F2], the two explicit shifted steps produce the unreduced upper-Hessenberg similarity $Q_{c}^{*} H Q_{c}$ . Since $Q_{c} = Q_{e} Θ$ , this equals $Θ^{*} (Q_{e}^{T} H Q_{e}) Θ$ . Diagonal unitary similarity preserves the Hessenberg zero pattern and the nonzero first-subdiagonal entries, so $Q_{e}^{T} H Q_{e}$ is also unreduced upper Hessenberg.

3.1givenF3F4step 1.1step 2.1algebra

By [F4], $Q_{b} e_{1}$ is proportional to the nonzero real vector $p (H) e_{1}$ . Step 1.1 and the positive diagonal of $R_{e}$ give $Q_{e} e_{1} = p (H) e_{1} / (R_{e})_{11}$ . Since both are real unit vectors, there is $ε \in {1, - 1}$ with $Q_{b} e_{1} = ε Q_{e} e_{1}$ . Set $S = diag (ε, 1, \dots, 1)$ . Then $Q_{e} S$ and $Q_{b}$ have the same first column, and $S (Q_{e}^{T} H Q_{e}) S$ remains unreduced upper Hessenberg by step 2.1. Apply [F3] to $Q_{b}$ and $Q_{e} S$ : their relative factor is a diagonal sign matrix with first entry $1$ . Therefore $Q_{b} = Q_{e} D$ for a diagonal sign matrix $D$ .

4.1step 3.1algebra∎

Substituting $Q_{b} = Q_{e} D$ gives $Q_{b}^{T} H Q_{b} = D^{T} (Q_{e}^{T} H Q_{e}) D$ , as claimed.

Depends on

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Sources

James Demmel, Applied Numerical Linear Algebra, §4.4.8 (standard reference, not scraped)
David Bindel, CS 6210 Matrix Computations, Hessenberg QR iteration (standard reference, not scraped)