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A truncated SVD separates discarded-direction bias from retained data noise

Statement

Let $A \in F^{m \times n}$ , where $F \in {R, C}$ , with singular value decomposition $A = U Σ V^{*}$ , where the columns $(v_{1}, \dots, v_{n})$ of $V$ and $(u_{1}, \dots, u_{m})$ of $U$ are orthonormal bases and the nonzero singular values satisfy $σ_{1} \geq \dots \geq σ_{r} > 0$ . Thus $A v_{i} = σ_{i} u_{i}$ for $i \leq r$ and $A v_{i} = 0$ for $i > r$ . The remaining $v_{r + 1}, \dots, v_{n}$ span $\ker A : = {v \in F^{n} : A v = 0}$ . Fix a threshold $τ \geq 0$ and put $I_{τ} = {i \leq r : σ_{i} > τ}$ and $J_{τ} = {1, \dots, n} ∖ I_{τ}$ . For $x^{†} \in F^{n}$ , $e \in F^{m}$ , and $b = A x^{†} + e$ , the truncated pseudoinverse from The truncated singular-value pseudoinverse obtained by discarding singular values below a declared numerical-rank threshold obeys

$x^{†} - A_{τ}^{+} b = \sum_{i \in J_{τ}} ⟨ x^{†}, v_{i} ⟩ v_{i} - \sum_{i \in I_{τ}} \frac{⟨ e, u_{i} ⟩}{σ_{i}} v_{i} .$

Thus discarded singular directions, including $\ker A$ , form the truncation bias, while only retained directions amplify the data perturbation. If $I_{τ} \neq \emptyset$ , then

$∥ A_{τ}^{+} e ∥ \leq (\max_{i \in I_{τ}} \frac{1}{σ_{i}}) ∥ e ∥;$

if $I_{τ} = \emptyset$ , then $A_{τ}^{+} = 0$ and the noise term is zero.

Facts & Assumptions

Given: A finite matrix $A \in F^{m \times n}$ , its singular value decomposition, $τ \geq 0$ , a target $x^{†} \in F^{n}$ , and data $b = A x^{†} + e$ with $e \in F^{m}$ .

[F1]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: A finite-dimensional map has orthonormal bases $(e_{i})$ and $(f_{i})$ with $T v = \sum_{i = 1}^{r} s_{i} ⟨ v, e_{i} ⟩ f_{i}$ and $s_{1} \geq \dots \geq s_{r} > 0$ .

[F2]

The truncated singular-value pseudoinverse obtained by discarding singular values below a declared numerical-rank threshold: The truncated filter uses coefficient $σ_{i}^{- 1}$ when $σ_{i} > τ$ and coefficient $0$ when $σ_{i} \leq τ$ .

[F3]

The induced length is a norm: The norm induced by the inner product satisfies $∥ v ∥^{2} = ⟨ v, v ⟩$ .

[F4]

Real and complex inner product spaces, with the inner product linear in the first argument: The inner product is linear in its first argument and conjugate symmetric.

Proof

1.1givenF2algebra

By linearity and $b = A x^{†} + e$ , $x^{†} - A_{τ}^{+} b = (I - A_{τ}^{+} A) x^{†} - A_{τ}^{+} e$ .

2.1F1F2step 1.1algebra

The definition in [F2] and the SVD in [F1] give $A_{τ}^{+} A x^{†} = \sum_{i \in I_{τ}} ⟨ x^{†}, v_{i} ⟩ v_{i}$ ; expanding $x^{†}$ in the full orthonormal basis $(v_{i})$ therefore makes $(I - A_{τ}^{+} A) x^{†} = \sum_{i \in J_{τ}} ⟨ x^{†}, v_{i} ⟩ v_{i}$ . Likewise, $A_{τ}^{+} e = \sum_{i \in I_{τ}} σ_{i}^{- 1} ⟨ e, u_{i} ⟩ v_{i}$ .

3.1F1F2F3F4step 2.1algebra∎

If $I_{τ}$ is empty, [F2] makes $A_{τ}^{+} = 0$ . Otherwise orthonormality of the retained $v_{i}$ and [F1] give $∥ A_{τ}^{+} e ∥^{2} = \sum_{i \in I_{τ}} ∣ ⟨ e, u_{i} ⟩ ∣^{2} / σ_{i}^{2} \leq (\max_{i \in I_{τ}} σ_{i}^{- 1})^{2} \sum_{i \in I_{τ}} ∣ ⟨ e, u_{i} ⟩ ∣^{2}$ . Since $(u_{1}, \dots, u_{m})$ is an orthonormal basis of $F^{m}$ , write $e = \sum_{i = 1}^{m} ⟨ e, u_{i} ⟩ u_{i}$ ; conjugate symmetry and orthonormality give $∥ e ∥^{2} = \sum_{i = 1}^{m} ∣ ⟨ e, u_{i} ⟩ ∣^{2}$ . Consequently $\sum_{i \in I_{τ}} ∣ ⟨ e, u_{i} ⟩ ∣^{2} \leq ∥ e ∥^{2}$ . Taking nonnegative square roots proves the noise bound and the stated bias/noise split.

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Sources

Per Christian Hansen, Introduction to Inverse Problems, Chapter 4: Regularization Methods (standard reference, not scraped)
Per Christian Hansen, Introduction to Inverse Problems, Chapter 5: Regularization Parameter Choice (standard reference, not scraped)
Guido Clason, Regularization of Inverse Problems, full lecture notes (standard reference, not scraped)