TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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A real Wilkinson-shifted symmetric tridiagonal QR step preserves symmetric tridiagonal form away from ties

Statement

Let $T$ be a real symmetric tridiagonal matrix whose trailing $2 \times 2$ principal block is not at a Wilkinson tie, and let $μ$ be the corresponding Wilkinson shift. If $T - μ I = Q R$ is the real orthogonal QR factorisation built from the standard adjacent-row Givens eliminations, then

T_{+} : = R Q + μ I

is again a real symmetric tridiagonal matrix.

Facts & Assumptions

Given: A real symmetric tridiagonal matrix $T$ , away from the Wilkinson tie case, and the real orthogonal QR factorisation $T - μ I = Q R$ built from standard adjacent-row Givens eliminations with the corresponding Wilkinson shift $μ$ .

[L1]

Away from a Wilkinson tie, the trailing $2 \times 2$ block determines a unique Wilkinson shift (Unshifted QR iteration, shifted QR iteration, deflation, and the Wilkinson shift).

[L2]

A shifted QR factorisation built from standard adjacent-row Givens eliminations preserves upper Hessenberg form (Shifted QR iteration preserves upper Hessenberg form).

Proof

technique · direct

1.1

By [L1], the trailing $2 \times 2$ block determines a unique Wilkinson shift $μ$ .

L1given

1.2

By [L2], the shifted step $T_{+} = R Q + μ I$ is upper Hessenberg.

L2given

1.3

Because $T - μ I = Q R$ , one has $T_{+} = R Q + μ I = Q^{T} (T - μ I) Q + μ I = Q^{T} T Q .$ Thus $T_{+}$ is orthogonally similar to the symmetric matrix $T$ , hence is itself symmetric.

givenalgebra

2.1

A symmetric upper Hessenberg matrix has zero entries below the first subdiagonal and, by symmetry, also above the first superdiagonal. Therefore step 1.2 and step 1.3 show that $T_{+}$ is real symmetric tridiagonal.

step 1.2step 1.3∎

Depends on

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Sources

Per-Olof Persson, The QR Algorithm II (standard reference, not scraped)