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

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Under separated moduli and leading-minor hypotheses, unshifted QR drives the strict lower triangle to zero and orders the eigenvalues on the diagonal

Statement

Let $F \in {R, C}$ and let $A = X Λ X^{- 1} \in M_{n} (F)$ be diagonalisable, with $∣ λ_{1} ∣ > ∣ λ_{2} ∣ > \dots > ∣ λ_{n} ∣ > 0$ , and suppose every leading principal minor of $X^{- 1}$ is nonzero. At every step choose the QR factorisation $A_{k} = Q_{k} R_{k}$ with each diagonal entry of $R_{k}$ positive real. Then $(A_{k})_{i j} ⟶ 0 (i > j), (A_{k})_{j j} ⟶ λ_{j} (1 \leq j \leq n) .$ Thus the iterates converge to triangular form with the eigenvalues ordered on the diagonal. The upper-triangular entries need not themselves converge.

Facts & Assumptions

Given: A diagonalisable invertible matrix $A = X Λ X^{- 1}$ with distinct eigenvalue moduli, nonzero leading principal minors of $X^{- 1}$ , and the positive-real-diagonal QR convention from the statement.

[L1]

Unshifted QR is orthonormalised simultaneous iteration, and $A_{k} = {\hat{Q}}_{k}^{*} A {\hat{Q}}_{k}$ (Unshifted QR is orthonormalised simultaneous iteration, and every QR iterate is unitarily similar to the original matrix).

[L2]

Subspace iteration converges to the dominant invariant subspace under a spectral gap and nondegenerate initial projection (Subspace iteration converges to the dominant invariant subspace when a spectral gap separates the wanted and unwanted eigenvalues).

[L3]

A diagonalisable matrix admits an eigenbasis (A diagonalisable endomorphism is one admitting a basis of eigenvectors, equivalently a diagonal matrix representation).

Proof

technique · direct

1.1

For each $j = 1, \dots, n - 1$ , apply [L2] to the first $j$ columns of the simultaneous-iteration frame from [L1]. In eigenvector coordinates, the initial frame is $X^{- 1} [e_{1} \dots e_{j}]$ ; its leading $j \times j$ coefficient block is the leading principal block of $X^{- 1}$ and is invertible by hypothesis.

L1L2L3algebra

2.1

Because $∣ λ_{j} ∣ > ∣ λ_{j + 1} ∣$ for every $j$ , each dominant $j$ -dimensional invariant subspace is unique. Step 1.1 therefore shows that, for every $j$ , the span $S_{j, k}$ of the first $j$ columns of ${\hat{Q}}_{k}$ converges to $E_{j} : = span (v_{1}, \dots, v_{j})$ . Thus the orthonormal frames converge flag-by-flag to the ordered eigenvector flag, even though individual frame vectors may retain varying signs or phases.

step 1.1algebra

3.1

By [L1], $A_{k} = {\hat{Q}}_{k}^{*} A {\hat{Q}}_{k}$ . Since $E_{j}$ is $A$ -invariant and $S_{j, k} \to E_{j}$ , the component of $A (S_{j, k})$ orthogonal to $S_{j, k}$ tends to zero. In the ${\hat{Q}}_{k}$ coordinates this component is the block of $A_{k}$ below the first $j$ columns, so $(A_{k})_{i j} \to 0$ whenever $i > j$ .

L1step 2.1algebra

3.2

The trace of the leading $j \times j$ block of $A_{k}$ is the trace of the compression of $A$ to $S_{j, k}$ . By $S_{j, k} \to E_{j}$ , it tends to the trace of $A ∣_{E_{j}}$ , namely $λ_{1} + \dots + λ_{j}$ . Subtracting the corresponding limit for $j - 1$ gives $(A_{k})_{j j} \to λ_{j}$ .

step 2.1algebra

4.1

Steps 3.1 and 3.2 prove that the strict lower triangle tends to zero and the diagonal tends to $(λ_{1}, \dots, λ_{n})$ . No convergence of the upper-triangular entries is asserted.

step 3.1step 3.2∎

Depends on

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Sources

Per-Olof Persson, The QR Algorithm I (standard reference, not scraped)
Andrew Stuart and Jochen Voss, Matrix Analysis and Algorithms (standard reference, not scraped)