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For full-column-rank A, the normal equations square the spectral condition number

Statement

Let $m, n \in N$ with $n \geq 1$ and $m \geq n$ , and let $A \in M_{m \times n} (R)$ have full column rank (rank $n$ ), with singular values $σ_{1} \geq \dots \geq σ_{n} > 0$ . Define the spectral condition number of the rectangular matrix $A$ by

$κ_{2} (A) : = \frac{σ_{1}}{σ_{n}};$

for a square invertible $A$ this agrees with $∥ A ∥_{2} ∥ A^{- 1} ∥_{2}$ of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system. Then:

The normal-equation matrix $A^{T} A$ is invertible, with $κ_{2} (A^{T} A) = κ_{2} (A)^{2}$ ;
equivalently, $κ_{2} (A^{T} A) = \frac{σ_{1}^{2}}{σ_{n}^{2}} .$

The least-squares normal equation $A^{T} A x = A^{T} b$ of 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 $\ker T$ therefore has condition number exactly the square of the condition number of the least-squares problem itself.

Facts & Assumptions

Given: A matrix $A \in M_{m \times n} (R)$ of full column rank with $n \geq 1$ , $m \geq n$ , and singular values $σ_{1} \geq \dots \geq σ_{n} > 0$ .

[L1]

There is a singular value decomposition $A = U Σ V^{T}$ with orthogonal $U \in M_{m} (R)$ , $V \in M_{n} (R)$ and $Σ$ diagonal with entries $σ_{1}, \dots, σ_{n}$ on the $n \times n$ block (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

The operator norm of a linear map is its largest singular value (The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector).

[L3]

The rank of a linear map is the number of its positive singular values; in particular the full-column-rank hypothesis is exactly $σ_{n} > 0$ (The rank of a linear map is the number of its nonzero singular values).

[L4]

An orthogonal operator satisfies $T^{*} T = T T^{*} = I$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

[L5]

Least-squares solutions 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 $\ker T$ ).

Proof

technique · direct

1.1

From [L1], $A^{T} A = V Σ^{T} Σ V^{T}$ , and $Σ^{T} Σ = diag (σ_{1}^{2}, \dots, σ_{n}^{2})$ is diagonal with decreasing nonnegative entries; by [L4], $V$ is orthogonal with $V V^{T} = I_{n}$ .

L1L4algebra

2.1

The displayed product $V diag (σ_{j}^{2}) V^{T}$ is a singular value decomposition of $A^{T} A$ with singular values $σ_{1}^{2} \geq \dots \geq σ_{n}^{2}$ , since the two orthogonal factors are $V$ and $V$ .

step 1.1L1algebra

3.1

By [L3], full column rank means $σ_{n} > 0$ , so every singular value of $A^{T} A$ in step 2.1 is positive, and $A^{T} A$ has the inverse $V diag (σ_{j}^{- 2}) V^{T}$ : the two products with $A^{T} A$ give $I_{n}$ by [L4].

step 2.1L3L4algebra

4.1

By [L2] applied to $A^{T} A$ and to its inverse, $∥ A^{T} A ∥_{2} = σ_{1}^{2}$ and $∥ (A^{T} A)^{- 1} ∥_{2} = σ_{n}^{- 2}$ , the largest singular value of $V diag (σ_{j}^{- 2}) V^{T}$ being $σ_{n}^{- 2}$ ; hence $κ_{2} (A^{T} A) = σ_{1}^{2} σ_{n}^{- 2}$ .

step 2.1step 3.1L2algebra

5.1

By the definition of $κ_{2} (A)$ in the Statement, $κ_{2} (A)^{2} = σ_{1}^{2} / σ_{n}^{2}$ , which equals the value in step 4.1; this is claim 2, and claim 1 is its equivalence with the condition number of the square invertible matrix $A^{T} A$ from The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system, the displayed agreement for square $A$ being $∥ A^{- 1} ∥_{2} = σ_{n}^{- 1}$ by [L2] applied to the SVD $A^{- 1} = V Σ^{- 1} U^{T}$ of [L1].

step 4.1L1L2algebra

6.1

The normal equation of [L5] has coefficient matrix $A^{T} A$ , so its condition number is the one computed in step 5.1.

step 5.1L5

7.1

Claims 1 and 2 are step 5.1, and the final sentence is step 6.1.

step 5.1step 6.1∎

Remarks

The square is unavoidable for the normal-equation route. Forming $A^{T} A$ squares the condition number even though the least-squares problem itself has condition number $κ_{2} (A)$ ; computing with $A$ directly through a QR route keeps the factor $κ_{2} (A)$ instead. For $A = (\begin{array}{cc} 1 & 0 \\ 0 & 10^{- 8} \\ 0 & 0 \end{array})$ , the Gram-Schmidt factor is $R = diag (1, 10^{- 8})$ while $A^{T} A = diag (1, 10^{- 16})$ .

Depends on

Used by

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

Dependency tree · two levels

21 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

L. N. Trefethen and D. Bau III, Numerical Linear Algebra, Lecture 18 and Lecture 31 (standard reference, not scraped)