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✓ 9 results · all verified · 6 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Square root covariance update error and stability

1 · Prerequisites

2 · Summary

Objective

This page gives a finite-horizon covariance-factor implementation of the canonical Kalman prediction and measurement maps. Rectangular factors, zero columns, singular state and noise covariances, and positive-definite innovations are allowed. The exact QR identities precede a local rounding bound for the represented joint covariance, including an explicit rational dimension-dependent envelope at sufficiently small unit roundoff; conditioned Schur-complement and triangular-solve estimates then propagate explicit errors over any fixed finite horizon.

The published Kalman items retain the Gaussian posterior and Joseph identity. This page uses those results as suppliers and compares the subtractive covariance expression only to explain cancellation. It does not claim that an unstructured QR perturbation preserves the original independent-noise model, that a rounded materialization of a factor product is positive semidefinite, or that the finite-horizon constants remain bounded uniformly in time.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

An explicit rational envelope for the scaled Householder QR error

Statement

Let q≥n≥1, u>0, and let κq be the local scaled-Householder coefficient of The scaled computed Householder step satisfies the local backward error premise at active row dimension q. Set βq,n=(1+κq)n−1,Cq,n=200qn. If Cq,nu<1, then βq,n≤ηq,n:=Cq,nu1−Cq,nu. Thus the QR backward-error factor can be bounded by the stated rational dimension-dependent expression in this precision range.

Facts & Assumptions

Given: Integers q≥n≥1, u>0, and 200qnu<1.

[F1]

For active row dimension r, the local constants are er=(1+u)r/2+3−1,dr=er+u(2+er),ar=1+u(1−u)r−1,gr=ru1−ru, cr=2((1+ar)(1+u)2(1+gr)−1),κr=max⁡{er+8dr,(1+u)cr+3u}. The scaled computed Householder step satisfies the local backward error premise

Proof

Given: The hypotheses in the Statement.

1.1F1givenalgebra

Put x=qu. The precision hypothesis gives u≤1/(200q) and 0<x<1/200. Since q/2+3≤7q/2, set s=(q/2+3)u≤7x/2<7/400. The power series gives 1+u≤eu and, for 0≤z<1, ez−1=∑j≥1zj/j!≤z/(1−z). Therefore eq≤es−1≤s1−s≤1400393x<185x. Also ueq≤(9/500)u, so dq=eq+2u+ueq≤185x+1009500u≤2809500x<5710x. Hence the first branch in [F1] satisfies eq+8dq<(18/5+8⋅57/10)x<50x.

1.2F1givenalgebra

Bernoulli's inequality gives (1−u)q≥1−qu=1−x; for integer q, this follows by induction from (1−u)(1−qu)=1−(q+1)u+qu2≥1−(q+1)u. Since u≤1/200 and x≤1/200, (1+u)3≤1+4u by expansion, and (1−x)−2≤1+3x because (1+3x)(1−x)2−1=x(1−5x+3x2)≥0. Thus (1+aq)(1+u)2(1+gq)≤(1+u)3(1−x)2≤(1+4u)(1+3x). As u≤x and u≤1/200, the last expression minus 1 is at most 4x+3x+12ux≤(7+3/50)x<8x. Consequently cq<16x and (1+u)cq+3u<(16+16/200+3)x<20x. Both branches in [F1] are therefore bounded by 50x, so κq≤50qu.

2.1F1step 1.1step 1.2givenalgebra∎

Put t=Cq,nu<1 and y=nκq. Step 1.2 gives 0≤y≤50qnu=t/4<1/4. Since (1+κq)n≤ey and ey−1≤y/(1−y) by the same power-series estimate as in step 1.1, βq,n≤y1−y≤t/41−t/4=t4−t≤t1−t=ηq,n. This proves the stated rational envelope.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A factor perturbation gives an explicit Gram-matrix error

Statement

Let G,E∈Rm×n and set G^=G+E. Then ∥G^G^T−GGT∥F≤2∥G∥F∥E∥F+∥E∥F2. The estimate is valid for all ranks, including G=0 and rank-deficient factors.

Facts & Assumptions

Given: Real matrices G,E of the same finite dimensions.

[F1]

The Frobenius norm is the Euclidean norm of the matrix entries, so it is invariant under transpose. The Frobenius norm ∥A∥F=(∑i,j∣aij∣2)1/2 on real or complex matrices

[F2]

The operator norm is the maximum stretch of a unit vector, and the Frobenius norm is the square root of the sum of squared singular values; therefore ∥A∥2≤∥A∥F for every real matrix A. The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison

Proof

Given: The hypotheses in the Statement.

1.1givenalgebra

Expanding G^G^T=(G+E)(G+E)T gives G^G^T−GGT=GET+EGT+EET.

2.1F1F2step 1.1algebra∎

For any compatible matrices A,B, applying the definition of the operator norm to each column gives ∥AB∥F2=∑j∥ABej∥22≤∥A∥22∑j∥Bej∥22=∥A∥22∥B∥F2≤∥A∥F2∥B∥F2. Thus ∥GET∥F,∥EGT∥F≤∥G∥F∥E∥F and ∥EET∥F≤∥E∥F2. The triangle inequality applied to step 1.1 proves the stated estimate.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

The block QR measurement update gives the canonical Kalman update

Statement

Let P−=L−(L−)T and R=LRLRT, with real factors of arbitrary rank. Let H∈Rp×n and suppose S=HP−HT+R≻0. Pad LR with zero columns to at least p columns and L− with zero columns to at least n columns. Denote the resulting factors by LRpad and Lpad−, and form Z=[LRpadHLpad−0Lpad−]. Then ZT has at least p+n rows. Take an exact reduced QR factorization ZT=UR with UTU=I and set L=RT. For each i=1,…,p, multiply column i of L by the sign of its diagonal entry. This makes the first p diagonal entries positive; the signs are well defined because S≻0. Partition the resulting L as L=[L110L21L22] at the observation and state dimensions. Then L11L11T=S,L21L11T=P−HT, and L22L22T=P−−P−HTS−1HP−. For innovation v=y−Hm−, the exact solve L11z=v gives m−+L21z=m−+P−HTS−1v. Thus the factor update has exactly the canonical one-step mean and covariance; L22 may be singular.

Facts & Assumptions

Given: AC, the real covariance factors and matrices, positive-definite innovation S, and exact QR of the padded prearray.

[A1]

AC is the choice assumption of the Gaussian conditional-law supplier; the block-matrix identities below add no choice. The Axiom of Choice

[F1]

For a real matrix with at least as many rows as columns, exact successive Householder QR gives ZT=UR with UTU=I; rank deficiency is permitted. Successive Householder or Givens transformations produce full and reduced QR factorisations with the standard dense operation counts

[F2]

Under AC, let U∼Nd(m,P), q∼Nn(0,Q), and r∼Np(0,R) be mutually independent Gaussian vectors, with positive dimensions and positive-semidefinite covariances. Set X=FU+q, Y=HX+r, m−=Fm, and P−=FPFT+Q. If S=HP−HT+R is positive definite, the everywhere-defined conditional law of X given Y=y has mean m−+P−HTS−1(y−Hm−) and covariance P−−P−HTS−1HP−, which also equals the Joseph expression. One-step Gaussian prediction and conditioning

Proof

Given: The hypotheses in the Statement.

1.1givenalgebra

Write rˉR and rˉ− for the padded column counts. They satisfy rˉR≥p and rˉ−≥n, so ZT has rˉR+rˉ−≥p+n rows, as required for the reduced QR. Padding either factor with zero columns leaves its Gram matrix unchanged. Multiplication of the prearray by its transpose therefore gives ZZT=[LRpad(LRpad)T+HLpad−(Lpad−)THTHLpad−(Lpad−)TLpad−(Lpad−)THTLpad−(Lpad−)T]=[SHP−P−HTP−].

2.1F1step 1.1algebra

Initially set L0=RT and partition it at dimensions p,n. Since ZT=UR and UTU=I, L0(L0)T=RTR=ZZT. Block multiplication gives L110(L110)T=S, L210(L110)T=P−HT, and L210(L210)T+L220(L220)T=P−. The positive-definite matrix S=L110(L110)T makes L110 invertible, so its diagonal entries are nonzero. Let D be the diagonal matrix whose entries are the signs of those entries, and set L=L0diag⁡(D,In). This flips the first p columns of L0, making the diagonal of L11=L110D positive. Because diag⁡(D,In) is orthogonal, LLT=L0(L0)T; moreover L21=L210D and L22=L220. Hence all three displayed block identities remain true.

3.1givenstep 2.1algebra

Because S≻0 and S=L11L11T, the square matrix L11 is invertible. From the cross-block identity, L21=P−HTL11−T; hence L21L21T=P−HTL11−TL11−1HP−=P−HTS−1HP−. Subtracting this equality from the lower-right identity in step 2.1 gives the stated posterior covariance. Since it is also L22L22T, it is positive semidefinite even when L22 is singular.

4.1A1F2step 2.1step 3.1givenalgebra∎

The solve L11z=v gives L21z=L21L11−1v=P−HTL11−TL11−1v=P−HTS−1v. When n,p≥1 and these blocks arise in a model satisfying [F2], these are its one-step Gaussian conditional update formulas; for zero-dimensional blocks this lemma asserts only the matrix identities. The matrix calculations themselves are deterministic and use no choice beyond [A1].

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

QR prediction preserves the canonical covariance factor

Statement

Let P=LLT with L∈Rd×r and let Q=LQLQT with LQ∈Rn×s, where zero columns and rank deficiency are allowed. For any real F∈Rn×d, form G=[FL  LQ] and append zero columns if needed so that the transpose has at least n rows. In exact arithmetic, if GT=UR is a reduced QR factorization with UTU=I, then L−:=RT satisfies

L−(L−)T=FPFT+Q.

Facts & Assumptions

Given: The factor inputs, matrix F, zero-padding convention, and an exact reduced QR factorization of the transposed padded array.

[F1]

For a matrix with at least as many rows as columns, successive Householder transformations produce a reduced QR factorization whose orthogonal factor has orthonormal columns. Successive Householder or Givens transformations produce full and reduced QR factorisations with the standard dense operation counts

Proof

Given: The hypotheses in the Statement.

1.1givenalgebra

The padded zero columns do not change the Gram matrix, so GGT=[FL  LQ][FL  LQ]T=FLLTFT+LQLQT=FPFT+Q. The cross terms are absent because the two factors occupy disjoint column blocks.

2.1F1step 1.1algebra∎

By [F1], GT=UR and UTU=I. Thus L−(L−)T=RTR=RTUTUR=GGT=FPFT+Q. No inverse or full-rank assumption was used, so the identity includes zero and rank-deficient factors.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Innovation, gain, and posterior errors follow from joint covariance perturbations

Statement

Let S=ST∈Rp×p satisfy xTSx≥μ∥x∥22 for all x∈Rp, with μ>0. Let B∈Rn×p, D=DT∈Rn×n, and let symmetric ES and matrices EB,ED satisfy ∥ES∥F≤eS<μ,∥EB∥F≤eB,∥ED∥F≤eD. Put S^=S+ES, B^=B+EB, D^=D+ED, Φ=D−BS−1BT, and Φ^=D^−B^S^−1B^T. If b≥∥B∥2, define β=(μ−eS)−1. Then S^⪰(μ−eS)I≻0 and ∥S^−1∥2≤β,∥S^−1−S−1∥2≤μ−1βeS, ∥B^S^−1−BS−1∥2≤eBβ+bμ−1βeS, and ∥Φ^−Φ∥2≤eD+2beBβ+b2μ−1βeS+eB2β.

Facts & Assumptions

Given: The symmetric matrices, the lower bound μ, the perturbation bounds, and the finite-dimensional real operator and Frobenius norms.

[F1]

The operator norm is the maximum Euclidean stretch of a unit vector. The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||

[F2]

Compatible operator norms are submultiplicative. The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2

[F3]

The operator norm is the largest singular value and is at most the Frobenius norm. Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison

[F4]

For real vectors, Cauchy–Schwarz follows by expanding 0≤∥x−ty∥22 at t=(xTy)/∥y∥22 when y≠0; the y=0 case is immediate. Expanding ∥x+y∥22 then gives the Euclidean triangle inequality. By [F1], evaluating each unit vector gives the triangle inequality for the operator norm.

Proof

Given: The hypotheses in the Statement.

1.1F1F3F4givenalgebra

For every x, [F3] and [F4] give ∣xTESx∣≤∥x∥2∥ESx∥2≤∥ES∥2∥x∥22≤eS∥x∥22, hence xTS^x=xTSx+xTESx≥(μ−eS)∥x∥22. Thus S^ is positive definite. Also ∥S−1∥2≤μ−1 and ∥S^−1∥2≤(μ−eS)−1=β: if y=Sx (or y=S^x), then [F4] gives ∥x∥2∥y∥2≥xTy and the corresponding quadratic lower bound gives the stated inverse estimate.

2.1F2step 1.1algebra

The resolvent identity is S^−1−S−1=−S−1ESS^−1. Using [F2] and step 1.1 gives ∥S^−1−S−1∥2≤μ−1eSβ.

3.1F2F3step 1.1step 2.1algebra

Splitting the gain difference yields B^S^−1−BS−1=EBS^−1+B(S^−1−S−1). The operator norm is bounded by the Frobenius norm for EB, so [F2] and steps 1.1–1.2 imply the displayed gain bound.

4.1F2F3F4step 1.1step 2.1algebra∎

Expanding the Schur difference gives Φ^−Φ=ED−EBS^−1BT−BS^−1EBT−B(S^−1−S−1)BT−EBS^−1EBT. Apply the operator-norm triangle inequality from [F4], [F2], the Frobenius-to-operator bound [F3], and the inverse estimates of steps 1.1–1.2. The five terms are bounded by eD, eBβb, bβeB, b2μ−1βeS, and eB2β, respectively, which proves the stated posterior bound.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

An innovation solve residual gives a singular-value forward bound

Statement

Let p≥1 and T∈Rp×p be nonsingular, define

σmin⁡(T):=min⁡∥x∥2=1∥Tx∥2,

and let z∗=T−1y. For any computed vector z^ with residual r=Tz^−y,

∥z^−z∗∥2≤∥r∥2σmin⁡(T).

If S=TTT also satisfies xTSx≥ν∥x∥22 for all x, with ν>0, then σmin⁡(T)≥ν and therefore

∥z^−z∗∥2≤∥r∥2ν.

Facts & Assumptions

Given: The real nonsingular matrix T, vector y, computed vector z^, and, for the second bound, the stated quadratic lower bound ν>0.

[F1]

The operator norm is the maximum Euclidean stretch on unit vectors. The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||

[F2]

For every linear map A, ∥ATA∥2=∥A∥22. The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2

[F3]

Euclidean inner products satisfy ∣xTy∣≤∥x∥2∥y∥2. Cauchy–Schwarz: ∣⟨x,y⟩∣≤∥x∥ ∥y∥, with equality exactly for dependent pairs

Proof

Given: The hypotheses in the Statement.

1.1F1givenalgebra

Set M=∥T−1∥2. For any vector w, ∥w∥2=∥T−1Tw∥2≤M∥Tw∥2 by the operator norm definition. Thus ∥Tw∥2≥∥w∥2/M.

1.2F3givenalgebra

Suppose S=TTT has the stated lower bound. If b=Sx, then ν∥x∥22≤xTSx=xTb≤∥x∥2∥b∥2 by [F3]. For x≠0 this yields ∥S−1b∥2≤∥b∥2/ν; it is also true for x=0. Thus ∥S−1∥2≤1/ν.

2.1F1step 1.1givenalgebra

Since p≥1 and T−1 is invertible, M>0. By [F1], the maximum defining M is attained at a unit vector u. Set v=T−1u/M. Then ∥v∥2=1 and ∥Tv∥2=1/M. Combining this with step 1.1 shows that the minimum in the definition of σmin⁡(T) exists and equals 1/M.

3.1step 1.1step 2.1givenalgebra

Since Tz∗=y, r=T(z^−z∗). Step 1.1 gives ∥z^−z∗∥2≤M∥r∥2, and step 2.1 gives M=1/σmin⁡(T). This proves the first residual bound.

4.1F2step 2.1step 3.1step 1.2algebra∎

Since (T−1)TT−1=(TTT)−1=S−1, [F2] and step 1.2 give ∥T−1∥22=∥S−1∥2≤1/ν. Step 2.1 then yields σmin⁡(T)=1/∥T−1∥2≥ν. Substituting this lower bound into step 3.1 proves the final residual estimate.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite-horizon square-root Kalman filter by rectangular factors and QR

Definition

Fix a finite horizon T≥0, positive state dimensions dk for 0≤k≤T, and positive observation dimensions pk for 1≤k≤T. Inputs are real finite matrices Fk∈Rdk×dk−1, Hk∈Rpk×dk, observation vectors yk∈Rpk, and an initial mean m0∈Rd0. For each covariance supply a rectangular real factor L0∈Rd0×r0,LQ,k∈Rdk×rQ,k,LR,k∈Rpk×rR,k, where each column count is any finite nonnegative integer. The represented covariances are P0=L0L0T, Qk=LQ,kLQ,kT, and Rk=LR,kLR,kT. Thus zero columns and rank-deficient factors are allowed. The admissible statistical inputs are those of Linear-Gaussian state-space models with independent noises for which the canonical innovation Sk=HkPk−HkT+Rk is positive definite at each step. The The Axiom of Choice assumption is inherited only for the interpretation as a posterior; the finite matrix procedure itself uses no probability-space choice.

Choose a unit roundoff u in the no-underflow operationwise model of Operationwise real and complex floating point arithmetic with rounded square roots and fused multiply-add. During an update, whenever an entry a of a supplied matrix, factor, or vector is read as an arithmetic operand or copied into a QR prearray, first convert it to fl⁡(a) in that format. Return range-failure at that read if conversion fails under the same no-underflow rule. Read operands from left to right; for products and block copies, visit output entries column by column, with increasing row index within each column. Repeat the conversion on each such read; the supplied arrays themselves remain unchanged. The formulas below use these converted entries during execution, while P0,Qk,Rk,Fk,Hk,yk,m0 denote the original exact target data in comparisons with the canonical recursion. In particular, conversion errors must be included in any certified local error bounds. For T=0, return the input pair immediately without reading or converting its entries. Every matrix or vector product below is evaluated as left-to-right dot products. A nonzero nonnormal intermediate, nonfinite result, zero innovation diagonal, or failed factorization returns a status and the first failing step; it is not coerced into a covariance estimate. Each arithmetic and factorization failure has a declared return; there is no iterative convergence loop or unbounded refinement search.

The QR operation used below is the following unpivoted reduced procedure on A∈Rm×n with floating-point entries and m≥n≥1. Initialize a working array R(0)=A and an empty ordered reflector list. For j=1,…,n, take x=Rj:m,j(j−1) and B=Rj:m,j+1:n(j−1), call Scaled Householder reflector formation and application on (x,B), and, on success, replace the active block by its returned C+. If the call fails, return its first status and step. Store the exact orthogonal reflector represented by the returned record, or the identity for its identity branch. After n successful steps, return the computed upper-triangular block R=R1:n,1:n(n) and the ordered reflector list representing the exact product Qfull whose first n columns are the reduced Q factor. Do not materialize a rounded dense Q. The conventions are those of Full, reduced, and column-pivoted computational QR factorisations.

Initialize m0∣0=m0 and L0∣0=L0. At step k, perform the following operations in order.

  1. Compute the rounded mean prediction m^k∣k−1=fl⁡(Fkm^k−1∣k−1). Form the rounded factor product X^k=fl⁡(FkLk−1∣k−1). Concatenate G^k=[X^k  LQ,k] and append zero columns until its column count is qk=max⁡(dk,rk−1∣k−1+rQ,k). Apply the above reduced Householder QR procedure to G^kT∈Rqk×dk. On success let Rkp be its reduced upper factor and set Lk∣k−1=(Rkp)T. No diagonal positivity or full-rank condition is imposed on this prediction factor.

  2. Compute v^k=fl⁡(yk−fl⁡(Hkm^k∣k−1)). Pad LR,k with zero columns to rˉk=max⁡(pk,rR,k) columns. Compute Y^k=fl⁡(HkLk∣k−1) and form the block prearray Z^k=[LR,kpadY^k0Lk∣k−1]. Apply the above reduced Householder QR procedure to Z^kT∈R(rˉk+dk)×(pk+dk). Its row count is at least its column count: the prediction step gives Lk∣k−1∈Rdk×dk, and rˉk≥pk. On success transpose its reduced upper factor and partition the resulting lower factor as L^k=[L^11,k0L^21,kL^22,k], with block sizes pk and dk. If any diagonal entry of L^11,k is zero, return innovation-factorization-failure. Otherwise, for each i=1,…,pk, multiply column i of the whole factor L^k by the sign of its diagonal entry. These exact sign changes make L^11,k have positive diagonal and preserve L^kL^kT and its lower block structure. Use the resulting factor blocks below. No invertibility or positive diagonal is required of L^22,k.

  3. Solve L^11,kz^k=v^k by forward substitution in the order of Forward and backward substitution for triangular linear systems. Set m^k∣k=fl⁡(m^k∣k−1+L^21,kz^k),Lk∣k=L^22,k. Do not form a covariance downdate or an explicit matrix inverse. The factor-defined covariance Lk∣kLk∣kT is positive semidefinite as an exact product of the returned entries; no claim is made that a separately rounded materialization of that product remains positive semidefinite.

  4. If any arithmetic operation or QR call returns a failure status, return the first such failure and its step; if the leading innovation block has a zero diagonal, return innovation-factorization-failure. Otherwise the finite loop returns the predicted and updated mean/factor arrays with success. The success status records arithmetic and factorization completion only.

For T=0 the loop is empty and the output is (m0,L0) with no update status. In exact arithmetic, the result implements the covariance and mean map of Exact linear-Gaussian Kalman filter. The prediction factor identity is supplied by QR prediction preserves the canonical covariance factor, and the measurement block identity by The block QR measurement update gives the canonical Kalman update. Under the inherited AC assumption its statistical interpretation is the one in The finite-horizon linear-Gaussian Kalman filter is exact. The nearby covariance supplied by an unstructured QR backward error may introduce cross-correlations between model blocks. Therefore no backward-stability claim within the original independent-noise state-space model, and no uniform-in-time accuracy or stability claim, is part of this procedure.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Product and Householder QR errors bound the represented joint covariance

Statement

Let G∈Rn×q, with q≥n≥1, be an exact prediction or measurement prearray. Work in the operationwise no-underflow model with 0<u<1 and qu<1. Let G^ be its computed version and assume ∥G^−G∥F≤eform+ein, where eform,ein≥0, eform bounds rounded products used to form the array, and ein bounds separately specified input-factor uncertainty. If the successful scaled-Householder QR of G^T∈Rq×n has reduced upper factor R and L=RT, set κq=κq(u),βq,n=(1+κq)n−1,ϵ=eform+ein+βq,n∥G^∥F, where κq is the explicit local coefficient in The scaled computed Householder step satisfies the local backward error premise. Then ∥LLT−GGT∥F≤2∥G∥Fϵ+ϵ2. If 200qnu<1, the rational envelope lemma An explicit rational envelope for the scaled Householder QR error gives βq,n≤ηq,n:=200qnu/(1−200qnu). In that regime the same conclusion holds with ϵ=eform+ein+ηq,n∥G^∥F. For one left-to-right rounded product UV^ with positive inner dimension r≥1 and ru<1, a valid choice is eform=γr∥U∥F∥V∥F. A product with zero inner dimension is exactly zero and contributes zero to eform; for several disjoint product blocks, sum the nonempty-product bounds. This conclusion allows rank loss and does not imply that the nearby joint covariance has the original independent-noise block structure.

Facts & Assumptions

Given: The exact and computed arrays, operationwise round-to-nearest arithmetic with all successful nonzero intermediates normal, ru<1 for each nonempty product, and successful scaled-Householder QR with qu<1.

[F1]

A left-to-right computed dot product of length r has componentwise relative error at most γr=ru/(1−ru), and its absolute error is at most γr∑j∣xjyj∣. The computed floating-point dot product obeys componentwise and normwise error bounds

[F2]

Successful scaled-Householder QR of an m×n real matrix, m≥n, gives A+ΔA=Q^R^ with ∥ΔA∥F≤((1+κm)n−1)∥A∥F; the coefficient comes from the local computed-reflector bound. The scaled Householder QR implementation is backward stable

[F3]

For any factor perturbation E, the Gram-matrix error is at most 2∥G∥F∥E∥F+∥E∥F2. A factor perturbation gives an explicit Gram-matrix error

[F4]

If 200qnu<1, then βq,n≤200qnu/(1−200qnu). An explicit rational envelope for the scaled Householder QR error

[F5]

For real vectors a,b, if b≠0 then expanding 0≤∥a−tb∥22 at t=(aTb)/∥b∥22 gives ∣aTb∣≤∥a∥2∥b∥2; if b=0 the inequality is immediate.

Proof

Given: The hypotheses in the Statement.

1.1F1F5givenalgebra

Exact-zero operations satisfy the relative model with error factor zero, so the rounded-factor expansion underlying [F1] also covers zero products and partial sums in the present model. An empty inner product is exact zero. For each entry of a nonempty product UV, [F1] therefore gives an error at most γr∑j∣UijVjk∣. Apply [F5] to the vectors (∣Uij∣)j and (∣Vjk∣)j, then square and sum over entries: ∑i,k(∑j∣Uij∣∣Vjk∣)2≤∑i,k(∑j∣Uij∣2)(∑j∣Vjk∣2)=∥U∥F2∥V∥F2. Therefore ∥UV^−UV∥F≤γr∥U∥F∥V∥F. Disjoint product blocks add by the Frobenius norm definition, and separately bounded input-factor errors add by the triangle inequality.

1.2F2givenalgebra

Apply [F2] to G^T, which has q≥n rows and n columns. Transposing its factorization gives G^+ΔT=RTQ^T=LQ^T,∥ΔT∥F≤βq,n∥G^∥F. Set E=(G^−G)+ΔT. Then ∥E∥F≤ϵ. Since Q^TQ^=I, the transposed factorization also gives LLT=LQ^TQ^LT=(G+E)(G+E)T.

2.1F3F4step 1.2algebra∎

Apply [F3] to the factor identity and perturbation E from step 1.2. It gives ∥LLT−GGT∥F≤2∥G∥Fϵ+ϵ2 for the stated βq,n choice. Under 200qnu<1, [F4] makes this epsilon no larger than the version using ηq,n. Since t↦2∥G∥Ft+t2 is increasing for t≥0, the same covariance bound follows with the rational-envelope epsilon. The argument permits zero singular values and places no structural restriction on the perturbation of GGT.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

The square-root Kalman procedure terminates with a finite status and operation-count bound

Statement

For every finite admissible input to Finite-horizon square-root Kalman filter by rectangular factors and QR with finite state and observation dimensions, the procedure returns after finitely many arithmetic operations. For T>0 its outcome is one of success, range-failure, or innovation-factorization-failure. Let r0,rQ,k,rR,k be the input factor column counts, put ρ0=r0 and ρk−1=dk−1 for 2≤k≤T, and set qk=max⁡(dk,ρk−1+rQ,k),rˉk=max⁡(pk,rR,k). In a unit-cost model for scalar arithmetic, comparisons, status checks, and entry copies, the operation count for T>0 is O ⁣(∑k=1T[dkdk−1(ρk−1+1)+qkdk2+(rˉk+dk)(pk+dk)2]). For T=0 the procedure returns its input pair in O(1) operations. These are termination, status, and operation-count results only; they assert neither that the numerical factorization succeeds nor that an error bound holds.

Facts & Assumptions

Given: The finite horizon, finite dimensions, finite input matrices and vectors, and the declared finite-status behavior of the component procedures.

[F1]

The algorithm has a loop over the finite horizon and each prediction, measurement, factorization, and solve has a declared finite status. Finite-horizon square-root Kalman filter by rectangular factors and QR

[F2]

Each scaled Householder step has finite loops and returns either its declared identity/reflector result or its first range-failure status. The scaled real-column Householder procedure terminates with defined statuses

[F3]

The forward-substitution definition specifies one recursive row update for each component when the diagonal is nonzero, so its solve schedule has finitely many updates. Forward and backward substitution for triangular linear systems

[F4]

Each scalar arithmetic operation, comparison, status check, and entry copy has unit cost in the operation-count convention of the Statement.

Proof

Given: The hypotheses in the Statement.

1.1F1givenalgebra

Every matrix or vector product in [F1] is evaluated by finitely many left-to-right dot products because all dimensions are finite. Each scalar operation, including conversion of a supplied entry at its read, either returns a finite value in the declared arithmetic model or the first range-failure status. Concatenation, zero padding, and block partitioning use finitely many entries.

1.2F1F2F3givenalgebra

Each QR call processes a finite number of columns and, within a column, a finite number of active entries. By [F2], each reflector step terminates with an identity, reflector, or range-failure result; hence a QR call either returns its reduced factor or a finite failure status. The lower-triangular innovation solve has finitely many forward-substitution row updates by [F3]; its diagonal is nonzero because [F1] returns failure before the solve otherwise.

1.3F1F2F4givenalgebra

A reduced Householder QR call on an M×N array (M≥N≥1) has N passes. Pass j forms a reflector using O(M−j+1) scalar operations and applies it to at most N−j trailing columns using O((M−j+1)(N−j)) operations; scans and status checks have no larger order. Summing over j gives O(MN2). Hence the prediction and measurement QR calls at step k cost O(qkdk2) and O((rˉk+dk)(pk+dk)2), respectively. A failure can only shorten these loops.

2.1F1step 1.1step 1.2inductionalgebra

The outer loop has at most T steps, and for T=0 the algorithm returns its initial pair immediately. By steps 1.1 and 1.2, every entered step terminates; therefore the whole procedure returns after finitely many operations with a declared status. No success or accuracy guarantee is asserted by this termination result.

3.1F1F3F4step 1.3givenalgebra∎

The products Fkm^k−1∣k−1 and FkLk−1∣k−1 cost O(dkdk−1) and O(dkdk−1ρk−1). The product HkLk∣k−1 costs O(pkdk2); the innovation product, triangular solve, factor sign changes, and all block copying/padding cost at most O((rˉk+dk)(pk+dk)2+qkdk2) because pk,dk≥1, qk≥rQ,k, and rˉk≥rR,k. Input conversions add at most one scalar rounding per supplied operand read or copied entry, so they do not change these orders. Together with step 1.3, summing over k gives the stated count. For T=0 there is only the constant-time return, with no entry conversions.

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The square-root filter has an explicit conditioned finite-horizon error bound

Statement

Fix a finite horizon T and let (mk∣k,Pk∣k) and (mk−,Pk−) be the exact real-arithmetic prediction and update recursions of Exact linear-Gaussian Kalman filter. Let Sk=HkPk−HkT+Rk,Bk=Pk−HkT,vk=yk−Hkmk−. Suppose each innovation has a certified lower bound xTSkx≥μk∥x∥22(x∈Rpk),μk>0. Let the procedure Finite-horizon square-root Kalman filter by rectangular factors and QR output represented covariances P^k∣k=L^k∣kL^k∣kT and means m^k∣k. At step k, assume certified local bounds ∥P^k−−(FkP^k−1∣k−1FkT+Qk)∥2≤ekP,∥J^k−Jk(P^k−)∥2≤ekJ,∥m^k−−Fkm^k−1∣k−1∥2≤ekm,∥v^k−(yk−Hkm^k−)∥2≤ekv,∥L^11,kz^k−v^k∥2≤ρk,∥m^k∣k−(m^k−+L^21,kz^k)∥2≤eku, where Jk(P)=[HkPHkT+RkHkPPHkTP] and J^k=L^kL^kT is the joint covariance represented by the measurement QR blocks. Let fk≥∥Fk∥2, hk≥∥Hk∥2, bk≥∥Bk∥2, and rk≥∥vk∥2 be certified finite upper bounds. Choose Δ0≥∥m^0∣0−m0∥2+∥P^0∣0−P0∥2.

Define recursively ckH=1+hk2,εk=ckH(fk2Δk−1+ekP)+ekJ. If εk≤μk/2, put Kk#=1+2bk/μk,CkK=2μk(1+bkμk),CkP=2+4bkμk+2(bkμk)2, Ak=(1+hkKk#)fk+(CkKrk+CkP)ckHfk2, ηk=(1+hkKk#)ekm+(CkKrk+CkP)(ckHekP+ekJ)+Kk#(ekv+ρk)+eku,Δk=AkΔk−1+ηk. Then each S^k is positive definite, the factor update is defined, and, with zk∗=L^11,k−1v^k, ∥z^k−zk∗∥2≤ρkσmin⁡(L^11,k)≤2μk ρk, where σmin⁡(T)=min⁡∥x∥2=1∥Tx∥2. Also ∥m^T∣T−mT∣T∥2+∥P^T∣T−PT∣T∥2≤ΔT. Equivalently, ΔT=(∏k=1TAk)Δ0+∑j=1T(∏k=j+1TAk)ηj, where an empty product is 1. The bound is finite-horizon and explicitly conditioned on the innovation margins and data scales. It implies no uniform-in-time stability or accuracy claim.

Facts & Assumptions

Given: The exact and computed finite-horizon recursions, local defect bounds in the Statement, and εk≤μk/2 for each step.

[F1]

The exact Kalman recursion uses Pk−=FkPk−1∣k−1FkT+Qk and mk∣k=mk−+BkSk−1vk, Pk∣k=Pk−−BkSk−1BkT. Exact linear-Gaussian Kalman filter

[F2]

The measurement QR blocks satisfy P^k∣k=D^k−B^kS^k−1B^kT and L^21,kL^11,k−1=B^kS^k−1 when S^k≻0. The block QR measurement update gives the canonical Kalman update

[F3]

Under its successful no-underflow QR and precision hypotheses (qu<1, and ru<1 for nonempty product inner dimensions), the prearray bound includes input-factor, rounded-product, and Householder QR contributions to ∥J^k−Jk(P^k−)∥2; the prediction array has the analogous bound for ekP. Product and Householder QR errors bound the represented joint covariance

[F4]

The Schur perturbation lemma proves inverse, gain and covariance estimates under Frobenius block-error bounds. Its resolvent and Schur expansions are algebraic identities; the operator-norm estimates needed here are derived directly in step 3.1. Innovation, gain, and posterior errors follow from joint covariance perturbations

[F5]

The operator norm is the maximum Euclidean stretch of a unit vector, and compatible operator norms are submultiplicative. The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2

[F6]

If T is nonsingular and r=Tz^−v, then ∥z^−T−1v∥2≤∥r∥2/σmin⁡(T); if TTT⪰νI, then σmin⁡(T)≥ν. An innovation solve residual gives a singular-value forward bound

Proof

Given: The hypotheses in the Statement.

1.1F1F5givenalgebra

Put δk−1P=∥P^k−1∣k−1−Pk−1∣k−1∥2 and δk−1m=∥m^k−1∣k−1−mk−1∣k−1∥2. By [F1], [F5] and the prediction defect, ∥P^k−−Pk−∥2≤fk2δk−1P+ekP,∥m^k−−mk−∥2≤fkδk−1m+ekm.

2.1F3F5step 1.1algebra

Let CH=[HkI]. Direct block multiplication gives Jk(P^k−)−Jk(Pk−)=CH(P^k−−Pk−)CHT. For every x, ∥CHx∥22=∥Hkx∥22+∥x∥22≤(1+hk2)∥x∥22, so [F5] and the joint defect imply ∥J^k−Jk(Pk−)∥2≤(1+hk2)(fk2δk−1P+ekP)+ekJ. If δk−1P≤Δk−1, the right side is at most εk. Each of its observation, cross, and state blocks therefore has operator error at most εk: every block is obtained by left- and right-multiplying by coordinate projections of operator norm at most one, so [F5] and submultiplicativity give the bound.

3.1F2F4F5givenstep 2.1algebra

Assume δk−1P≤Δk−1 and suppress k. Write ES=S^−S, EB=B^−B, and ED=D^−P−. Step 2.1 bounds their operator norms by ε≤μ/2. Cauchy–Schwarz gives ∣xTESx∣≤ε∥x∥22, hence S^⪰(μ−ε)I≻0. The same inequality applied to y=Sx and y=S^x gives ∥S−1∥2≤μ−1 and ∥S^−1∥2≤β:=(μ−ε)−1≤2/μ. The resolvent identity S^−1−S−1=−S−1ESS^−1 then bounds the inverse difference by μ−1βε. Expanding the gain and Schur differences as in [F4], now applying operator-norm triangle and product inequalities throughout, yields ∥B^S^−1−BS−1∥2≤βε+bμ−1βε≤CKε, ∥P^k∣k−Pk∣k∥2≤ε+2bβε+b2μ−1βε+βε2≤CPε. Here [F2] identifies the represented posterior as the Schur complement, and βε2≤ε. Finally, ∥B^S^−1∥2≤(b+ε)β≤1+2b/μ=K#. These estimates use operator block bounds directly, without a Frobenius conversion.

4.1F1F2F6givenstep 1.1step 3.1algebra

The innovation residual and computed-innovation defect give ∥v^k−vk∥2≤hk∥m^k−−mk−∥2+ekv. Let zk∗=L^11,k−1v^k. By [F6] and the S^k⪰(μk/2)I bound from step 3.1, ∥z^k−zk∗∥2≤2/μk ρk, as included in the Statement. Because L^11,kz^k=v^k+ξk with ∥ξk∥2≤ρk, [F2] gives L^21,kz^k=B^kS^k−1(v^k+ξk). Subtract the exact mean equation in [F1], insert and subtract B^kS^k−1vk, and use step 3.1. The result is ∥m^k∣k−mk∣k∥2≤(1+hkKk#)(fkδk−1m+ekm)+CkKεkrk+Kk#(ekv+ρk)+eku.

5.1givenstep 1.1step 2.1step 3.1step 4.1inductionalgebra∎

Add the covariance bound in step 3.1 to the mean bound in step 4.1. Using δk−1P,δk−1m≤Δk−1 and the definition of εk yields ∥m^k∣k−mk∣k∥2+∥P^k∣k−Pk∣k∥2≤AkΔk−1+ηk=Δk. Induction from the stated Δ0 proves the bound for each k. Repeated substitution of the scalar recurrence gives the displayed product-and-sum formula at T. Since each assumption and step concerns only the finite indices 1,…,T, no uniform-in-time conclusion follows.

5 · Examples, counterexamples and false statements

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