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Normal equations square the conditioning compared with QR on a badly scaled least-squares problem

Example

For $ε = 10^{- 8}$ the badly scaled least-squares matrix

$A = (\begin{array}{cc} 1 & 0 \\ 0 & ε \\ 0 & 0 \end{array})$

has $κ_{2} (A) = 10^{8}$ , while its normal-equation matrix $A^{T} A = diag (1, ε^{2})$ has $κ_{2} (A^{T} A) = 10^{16} = κ_{2} (A)^{2}$ . A Gram–Schmidt (QR) route works with the upper triangular $R$ whose singular values equal those of $A$ , so it keeps the conditioning at $κ_{2} (A) = 10^{8}$ instead of squaring it.

Facts & Assumptions

Given: The matrix $A = (\begin{array}{cc} 1 & 0 \\ 0 & ε \\ 0 & 0 \end{array})$ with $ε = 10^{- 8} > 0$ , and the least-squares problem of minimising $∥ A x - b ∥_{2}$ .

[L1]

For full-column-rank $A$ with singular values $σ_{1} \geq \dots \geq σ_{n} > 0$ , the normal-equation matrix satisfies $κ_{2} (A^{T} A) = κ_{2} (A)^{2}$ where $κ_{2} (A) = σ_{1} / σ_{n}$ (For full-column-rank A, the normal equations square the spectral condition number).

[L2]

Gram–Schmidt orthonormalisation applied to the independent columns of $A$ produces orthonormal columns $q_{1}, q_{2}$ spanning the same column space (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).

Verification

technique · direct

1.1

The columns of $A$ are orthogonal and have norms $1$ and $ε$ . Writing $A = U Σ V^{T}$ with $U = [e_{1} e_{2}]$ , $V = I_{2}$ , and the $2 \times 2$ matrix $Σ = diag (1, ε)$ shows that the singular values are the diagonal entries $1$ and $ε$ of $Σ$ .

algebra

1.2

By [L2], Gram-Schmidt applied to the columns $a_{1} = (1, 0, 0)^{T}$ and $a_{2} = (0, ε, 0)^{T}$ returns the already orthogonal unit vectors $q_{1} = (1, 0, 0)^{T}$ and $q_{2} = (0, 1, 0)^{T}$ after scaling by their norms. Writing $Q = [q_{1} q_{2}]$ , one has $A = Q R$ with $R = Q^{T} A = diag (1, ε)$ .

L2algebraconstruct

2.1

Hence $κ_{2} (A) = 1 / ε = 10^{8}$ , and $A^{T} A$ is the $2 \times 2$ diagonal matrix $diag (1, ε^{2})$ , whose singular values are $1$ and $ε^{2}$ .

step 1.1algebra

2.2

By step 1.2, $R = diag (1, ε)$ , so its singular values are $1$ and $ε$ and therefore $κ_{2} (R) = 1 / ε = 10^{8}$ .

step 1.2algebra

3.1

By [L1], $κ_{2} (A^{T} A) = κ_{2} (A)^{2} = 10^{16}$ : forming the normal equations squares the conditioning of the least-squares problem.

L1step 2.1

4.1

Steps 3.1 and 2.2 compare the two routes on the same badly scaled problem: the normal-equation matrix carries condition number $10^{16}$ , the square of $κ_{2} (A)$ , while the Gram–Schmidt triangular factor $R$ carries $10^{8}$ , the original conditioning of the problem.

step 3.1step 2.2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

L. N. Trefethen and D. Bau III, Numerical Linear Algebra, Lecture 18 and Lecture 19 (standard reference, not scraped)