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Reduced QR over the reals solves full-column-rank least squares without squaring the condition number

Statement

Let $A \in M_{m \times n} (R)$ with $m \geq n \geq 1$ and full column rank. Let

$A = \hat{Q} \hat{R}$

be a reduced QR factorisation, with ${\hat{Q}}^{T} \hat{Q} = I_{n}$ and $\hat{R}$ upper triangular. Then:

The least-squares minimiser of $∥ A x - b ∥_{2}$ is the unique solution of $\hat{R} x = {\hat{Q}}^{T} b .$
The spectral condition number of $\hat{R}$ equals that of $A$ , whereas the normal-equation matrix satisfies $κ_{2} (A^{T} A) = κ_{2} (A)^{2} .$ Thus the QR route solves with a factor of condition $κ_{2} (A)$ rather than the squared factor $κ_{2} (A)^{2}$ .

Facts & Assumptions

Given: A full-column-rank matrix $A \in M_{m \times n} (R)$ , a reduced QR factorisation $A = \hat{Q} \hat{R}$ , and a vector $b \in R^{m}$ .

[L1]

Reduced QR factorisation means ${\hat{Q}}^{T} \hat{Q} = I_{n}$ and $\hat{R}$ upper triangular (Full, reduced, and column-pivoted computational QR factorisations).

[L2]

Successive Householder or Givens transformations produce reduced QR (Successive Householder or Givens transformations produce full and reduced QR factorisations with the standard dense operation counts).

[L3]

Least-squares minimisers are exactly the solutions of the normal equation $A^{T} A x = A^{T} b$ (For a linear map $T : V \to W$ between finite-dimensional inner-product spaces, $x$ minimises $∥ T x - b ∥$ if and only if $T^{*} (T x - b) = 0$ , equivalently $T^{*} T x = T^{*} b$ ; minimisers exist and any two differ by an element of $\ker T$ ).

[L4]

For full-column-rank $A$ , the normal equations square the spectral condition number (For full-column-rank A, the normal equations square the spectral condition number).

[L5]

Spectral norm is unitarily invariant and is determined by singular values (Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison).

Proof

technique · orthogonal decomposition of the residual

1.1

With the reduced QR factorisation supplied by [L2] and shaped as in [L1], extend the columns of $\hat{Q}$ to an orthogonal basis $Q = [\hat{Q} {\hat{Q}}_{⊥}]$ . Then $∥ A x - b ∥_{2} = ∥ Q^{T} (A x - b) ∥_{2} = {∥ (\binom{\hat{R} x - {\hat{Q}}^{T} b}{- {\hat{Q}}_{⊥}^{T} b}) ∥}_{2},$ so $∥ A x - b ∥_{2}^{2} = ∥ \hat{R} x - {\hat{Q}}^{T} b ∥_{2}^{2} + ∥ {\hat{Q}}_{⊥}^{T} b ∥_{2}^{2} .$ The second term is independent of $x$ , so minimising $∥ A x - b ∥_{2}$ is equivalent to solving $\hat{R} x = {\hat{Q}}^{T} b$ .

L1L2L5algebra

2.1

Because $A$ has full column rank, the upper-triangular factor $\hat{R}$ has nonzero diagonal, so the triangular system of step 1.1 has a unique solution. Thus the least-squares minimiser is unique and equals that solution. This agrees with the normal equation of [L3].

step 1.1L1L3algebra

3.1

Since $Q$ is orthogonal and $A = Q (\binom{\hat{R}}{0})$ , [L5] implies that $A$ and $\hat{R}$ have the same positive singular values, hence the same spectral condition number. By [L4], the normal-equation matrix has condition number $κ_{2} (A)^{2}$ . Therefore the QR solve uses a factor with condition $κ_{2} (A)$ , not the squared condition of the normal equations.

step 2.1L4L5algebra

4.1

Steps 1.1-2.1 prove claim 1, and step 3.1 proves claim 2.

step 1.1step 2.1step 3.1∎

Depends on

Used by

Reduced QR avoids the condition-number squaring seen in the normal equations Example

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24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Section 3.3.3 (standard reference, not scraped)
David Bindel, CS 4220: Numerical Analysis, Least squares and QR (standard reference, not scraped)