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14 results · all verified · 0 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 14 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Matrix Norms, Condition Numbers and Numerical Stability

1 · Prerequisites

2 · Summary

The published p-norms on Rn and their norm axioms, the operator norm and the singular value decomposition with the rank and Eckart–Young consequences, matrix multiplication and invertibility, the geometric series, and the least-squares normal equation supply the machinery this page uses; the published spectral page owns singular values and the operator 2-norm, and this page cites them rather than re-minting them.

The page separates the three questions that numerical analysis keeps apart. First it fixes the vocabulary of problem conditioning: absolute and relative forward error, and the absolute and relative local condition numbers of a problem map. Second it builds the matrix-norm tools: the norm induced by a published vector p-norm with its compatibility, submultiplicativity and normalisation; the induced 1-norm as the maximum column sum and the induced -norm as the maximum row sum; and the Frobenius norm with the spectral and Frobenius singular-value formulas, unitary invariance and the sharp rank comparison. Third it runs the linear-system spine: the condition number κp(A), the right-hand-side and matrix perturbation bounds, the reciprocal spectral condition number as the relative distance to singularity, residual-to-forward-error bounds, and normwise and componentwise backward error with their explicit residual formulas.

The page then makes the arithmetic model explicit: the standard relative floating-point model and unit roundoff with its stated exclusions, forward and backward stability for a problem family, the product-of-errors lemma giving γn=nu/(1nu), the componentwise and normwise dot-product bounds, and the first-order principle that conditioning times backward error controls forward error. It closes by separating exact rank from numerical rank: exact rank is locally constant at maximal-rank matrices and discontinuous at nonmaximal-rank ones, numerical rank is declared against a norm, scale and tolerance, and in the spectral norm it counts the singular values above the declared threshold.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Absolute and relative forward error for an approximate value of a problem map

Definition

Let X and Y be normed spaces over R (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), let f:XY be a map, and let x be an element of the domain of f. Let y^Y be an approximate value of f at x: an element of the target space meant to stand for f(x).

The absolute forward error of y^ at x is the norm

y^f(x),

and, provided f(x)0, the relative forward error of y^ at x is

y^f(x)f(x).

The exact value f(x) is called the exact answer, and y^ the computed (or approximate) answer. Forward error is the error in the answer itself; it is measured in the target space Y and knows nothing about how y^ was produced.

Remarks

  • Why the relative version needs f(x)0. Relative error divides by f(x); when f(x)=0 the division is not defined, and the only sensible quantity is the absolute error. Every statement on this page that writes a relative forward error carries the hypothesis f(x)0.

  • Approximate values are plain elements of Y. No machine arithmetic is assumed here: y^ can come from hand calculation, a rounded formula, or an algorithm. The floating-point model is introduced separately in The standard relative floating-point model and unit roundoff, and nothing in this definition depends on it.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Absolute and relative local condition numbers of a problem map

Definition

Let X and Y be normed spaces over R (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) and let f:XY be a map. For xX and δ>0 write

Pf,x(δ)  :=  {hX:0<h<δ, x+hdomf}.

Define the worst amplification over perturbations of size at most δ by

Sf,x(δ)  :=  {0,Pf,x(δ)=,sup{f(x+h)f(x)h:hPf,x(δ)},Pf,x(δ).

so Sf,x(δ)[0,+] in every case.

For each fixed x the map δSf,x(δ) is nondecreasing as δ increases, so the infimum over all δ>0 exists in [0,+]. The absolute (local) condition number of f at x is

κabs(f,x)  :=  infδ>0Sf,x(δ)  =  limδ0+Sf,x(δ)    [0,+].

If x0 and f(x)0, the relative (local) condition number of f at x is

κrel(f,x)  :=  κabs(f,x)xf(x)    [0,+].

For the computed value y^=f(x+h) of Absolute and relative forward error for an approximate value of a problem map, the quantity h is the absolute backward error. When x0, the quotient h/x is the relative backward error: the amount by which the data must be moved, relative to its size, for the computed answer to be exactly correct. At x=0 only the absolute backward error is defined. The condition number converts backward error into a bound on forward error; that conversion is Local conditioning times backward error controls forward error to first order.

Remarks

  • The infimum form is the limit of the supremum form. Because Sf,x is monotone in δ, the infimum over δ>0 equals the one-sided limit limδ0+Sf,x(δ), which is the familiar definition of a condition number as the worst relative amplification of infinitesimal perturbations. The infimum formulation is used below because it is an honest value of the library's ordered arithmetic with no separate limit apparatus.

  • The value is allowed to be +. An ill-posed (discontinuous) map has infinite condition number at a point of discontinuity; κabs=+ is a legitimate value and no arithmetic is performed with it. The first-order conversion theorem only yields a finite bound when κabs<+.

  • Conditioning belongs to the problem, not to the algorithm. The condition number depends only on f, the point x, and the two norms; no algorithm enters its definition. The separate notion of stability belongs to an algorithm under an arithmetic model and is defined in Forward and backward stability for a problem family under an arithmetic model.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The matrix norm induced by a published vector p-norm

Definition

Let m,nN, let pQ with p1, and let A=(aij)Mm×n(R) be a real matrix (Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes), regarded as the linear map xAx from Rn to Rm with the published p-norms (The p-norms xp for rational p1, and x). The matrix p-norm induced by p is

Ap  :=  sup{Axp  :  xRn, xp1}.

When n1, the following three displayed quantities are the same number:

Ap=supxp1Axp=supxp=1Axp=supx0Axpxp,

the equalities following from positive homogeneity of p (Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page, axiom (N2)) and the equality 0p=0.

The value is a finite nonnegative real. For n1 the norms p and 2 are equivalent by For n1 all norms on Rn are equivalent, so there is C>0 with xCxp for every x. For every i<m, (Ax)ij<naijxj(j<naij)xC(j<naij)xp, so Axp(i<m(j<naij)p)1/pCxp; hence the supremum over the unit ball is at most the finite constant (i<m(j<naij)p)1/pC.

The induced infinity norm is a separate named case. When m,n1, define

A  :=  sup{Ax  :  xRn, x1}.

As in the rational-p case, positive homogeneity gives

A=supx1Ax=supx=1Ax=supx0Axx.

The supremum is finite because for every i<m one has (Ax)ij<naijxj(j<naij)x, so

Axmaxi<mj<naijx.

At n=0 the convention is explicit. The only element of R0 is the zero vector, the only matrix AMm×0(R) is the empty matrix, and the unit-ball definition gives Ap=0. The unit sphere and the set of nonzero vectors are empty, so the two ratio formulas above are not asserted in this case.

The induced quantity is a norm on the matrix space Mm×n(R): for each rational p1, and likewise for the separately named -case when m,n1, separation, absolute homogeneity and the triangle inequality follow from the same three axioms of the underlying vector norm (Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page) through the defining supremum. The compatibility with matrix-vector multiplication, the submultiplicativity and the value on identity matrices are Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1; the explicit value at p=1 and the separate -formula are The induced 1-norm is the maximum column sum and the induced infinity-norm is the maximum row sum.

Remarks

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1

Statement

Let pQ with p1 and let m,n,qN, with the induced p-norms of The matrix norm induced by a published vector p-norm.

  1. Compatibility. For every AMm×n(R) and every xRn, Axp    Apxp.
  2. Submultiplicativity. For every AMm×n(R) and BMn×q(R), ABp    ApBp.
  3. Normalisation. For every n1, Inp=1, where In is the identity matrix of Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes; at n=0 the unique empty matrix I0M0(R) has I0p=0, the convention of The matrix norm induced by a published vector p-norm.

Facts & Assumptions

Given: A rational p1, natural numbers m,n,q, matrices AMm×n(R), BMn×q(R), and vectors xRn.

[L1]

p is a norm on Rn and on Rm: in particular 0p=0 and λyp=λyp (Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page).

[L2]

The induced norm is Ap=sup{Ayp:yp1} (The matrix norm induced by a published vector p-norm).

Proof

technique · direct
1.1

For x=0 the compatibility claim reads 0Ap0, which is true by [L1] and [L2].

L1L2algebra
1.2

For x0 the vector xp>0 by [L1], so Axp/xp=A(x/xp)p by homogeneity of [L1].

L1algebra
1.3

For x=0 both sides of claim 2 are 0, using [L1] and [L3].

L1L3algebra
1.4

For n1 every x0 attains ratio one: by [L3], Inxp=xp, so Inxp/xp=1, and the supremum in [L2] is therefore 1, which is claim 3 for n1.

L2L3algebra
1.5

At n=0 the space R0 has only the zero vector, so the definition of The matrix norm induced by a published vector p-norm assigns I0p=0, which is the stated exceptional value.

L2algebra
2.1

The scaled vector x/xp has p-norm 1, so A(x/xp)pAp by [L2]; combining step 1.2 with this bound gives AxpApxp, which with step 1.1 proves claim 1.

step 1.2L2algebra
3.1

For x0 associativity of [L3] gives (AB)xp=A(Bx)p, and applying claim 1 first to A at the vector Bx and then to B at the vector x gives A(Bx)pApBxpApBpxp.

step 2.1L3algebra
4.1

Dividing step 3.1 by the positive number xp and taking the supremum over all nonzero x in the definition of [L2] gives ABpApBp; with step 1.3 this is claim 2.

step 3.1L1L2algebra
5.1

Claims 1, 2 and 3 are steps 2.1, 4.1, 1.4 and 1.5 respectively.

step 2.1step 4.1step 1.4step 1.5
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The induced 1-norm is the maximum column sum and the induced infinity-norm is the maximum row sum

Statement

Let m,nN with m1 and n1, and let A=(aij)Mm×n(R), with the induced 1-norm and the separately defined induced -norm of The matrix norm induced by a published vector p-norm.

  1. Maximum column sum. A1=maxj<ni<maij.
  2. Maximum row sum. A=maxi<mj<naij.

Facts & Assumptions

Given: Natural numbers m,n1, a matrix A=(aij)Mm×n(R), and vectors x=(xj)Rn.

[L1]

The 1-norm and the -norm are x1=j<nxj and x=max{xj:j<n} (The p-norms xp for rational p1, and x); each is a norm, in particular the triangle inequality and xjx hold (Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page).

[L2]

The induced norm is the supremum of the ratio over nonzero vectors, and compatibility gives AxpApxp (The matrix norm induced by a published vector p-norm, Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1).

[L3]

The standard basis vector ejRn satisfies ej(j)=1 and ej(k)=0 for kj, and finite sums are evaluated pointwise; in particular (Aej)i=aij (The standard list e:nFn with ei(i)=1F and ei(j)=0F for ji is an ordered basis of Fn; hence dimFFn=n, and F0 is the zero space with basis and dimension 0).

Proof

technique · direct
1.1

Write C:=maxj<ni<maij; the maximum exists because it is a maximum of the nonempty finite set of column sums of [L1].

givenL1
1.2

Expanding the matrix-vector product entrywise and using the triangle inequality and absolute multiplicativity of [L1] gives Ax1=i<mj<naijxji<mj<naijxj.

L1algebra
1.3

Write R:=maxi<mj<naij, a maximum of a nonempty finite set of row sums by [L1].

L1
2.1

Exchanging the finite double sum in step 1.2 and bounding every column sum by C from step 1.1 gives Ax1j<n(i<maij)xjCj<nxj=Cx1.

step 1.2step 1.1L1algebra
2.2

Choose j<n attaining the maximum column sum, so i<maij=C, which exists by step 1.1.

step 1.1choose
2.3

For every i<m, (Ax)i=j<naijxjj<naijxj(j<naij)xRx, using the triangle inequality, xjx and the definition of R.

L1step 1.3algebra
2.4

Choose i<m attaining the maximum row sum: j<naij=R, which exists by step 1.3.

step 1.3choose
3.1

By [L2] the induced norm is the supremum of the ratio over nonzero x, so step 2.1 gives A1C.

step 2.1L2
3.2

For the standard basis vector ej of [L3], ej1=1 and Aej1=i<maij=C, so the ratio Aej1/ej1=C is attained and A1C.

L1L3step 2.2
3.3

Taking the maximum over i<m in step 2.3 gives AxRx, and by [L2] the induced -norm is at most R.

step 2.3L1L2
3.4

Define xRn by xj:=1 when aij0 and xj:=1 when aij<0; then x=1 and (Ax)i=j<naijxj=j<naij=R, by [L1] and [L3].

L1L3step 2.4algebra
4.1

Steps 3.1 and 3.2 give A1=C, which is claim 1.

step 3.1step 3.2
4.2

The vector x of step 3.4 attains the ratio R, so AR; with step 3.3 this gives A=R, which is claim 2.

step 3.4step 3.3L2
5.1

Claims 1 and 2 are steps 4.1 and 4.2.

step 4.1step 4.2
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The Frobenius norm AF=(i,jaij2)1/2 on real or complex matrices

Definition

Let F be R or C, let m,nN and let A=(aij)Mm×n(F). The Frobenius norm of A is

AF  :=  (i<mj<naij2)1/2,

where is the absolute value on R (Absolute value in an ordered field) for F=R and the modulus on C (Real and imaginary parts, complex conjugation, and modulus) for F=C, and the outer power is the nonnegative real square root.

The displayed quantity is a norm. Regard A as the element of the coordinate space Fmn whose j+in-th entry is aij. The standard coordinate inner product of The standard formulas x,y=k<nxkyk on Rn and k<nxkyk on Cn are inner products satisfies A,A=i,jaijaij, which equals i,jaij2 because zz=z2 (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive). Thus AF is exactly the inner-product norm of The norm v=v,v induced by a real or complex inner product evaluated at A, and the three norm axioms hold for F because they hold for every inner-product norm. For real matrices the same sum is i,jaij2, since x2=x2 for real x by the two cases of Absolute value in an ordered field.

The Frobenius norm is the Euclidean norm of the full entry list, not the largest factor by which the matrix stretches a single vector. It can coincide with an induced 2-norm in one-row or one-column cases, but in general it is a different matrix norm from the induced operator norms used on this page. Its relation to the spectral norm, its unitary invariance and its singular-value formula are Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison.

Remarks

  • The name records the entry-sum definition. Different sources write AF, AF, or A2,HS; this library uses F exclusively and does not fold it into the induced p-norm notation.

  • Zero-sized shapes are included. At m=0 or n=0 the double sum is the empty sum 0, so the unique empty matrix has Frobenius norm 0; this is the value of the inner-product norm of the zero space.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison

Statement

Let F be R or C, let AMm×n(F), and let σ1σ2σr>0 be the positive singular values of A, where r=rankA; write σ1=0 when r=0.

  1. Spectral formula. A2=σ1, where 2 is the operator norm of The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||.
  2. Frobenius formula. AF=(σ12++σr2)1/2.
  3. Unitary invariance. For every unitary UMm(F) and VMn(F) (over F=R: orthogonal matrices), UAV2=A2,UAVF=AF.
  4. Sharp rank comparison. A2    AF    rA2. The lower inequality is an equality exactly when r1; the upper inequality is an equality exactly when σ1==σr, that is, when all nonzero singular values coincide. In particular AFmin(m,n)A2.

Facts & Assumptions

Given: A matrix AMm×n(F) over F=R or C, with singular values σ1σr>0.

[L1]

There is a singular value decomposition A=UΣV with U,V unitary (orthogonal over R) and Σ the diagonal matrix of the singular values (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

The operator norm of a map between finite-dimensional real or complex inner product spaces equals 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 (The rank of a linear map is the number of its nonzero singular values).

[L4]

A unitary (orthogonal) operator preserves norms: Qv=v for every vector v (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

[L5]

Compositions and adjoints of unitary operators are unitary (Orthogonal and unitary operators form groups, and their determinants have modulus one).

Proof

technique · direct
1.1

Claim 1 is [L2] applied to the map xAx: A2=σ1.

L2
1.2

Left invariance of the Frobenius norm. For unitary QMm(F), each column of QA is the image under Q of the corresponding column of A, so by [L4] (QA) column j2=A column j2 for every j<n; summing over j gives QAF2=j<n(QA)j22=j<nAj22=AF2, using the entry formula of The Frobenius norm AF=(i,jaij2)1/2 on real or complex matrices.

L4algebra
1.3

Right invariance of the Frobenius norm. For unitary ZMn(F), the i-th row of AZ is rowi(A)Z, and by [L4] applied to Z, rowi(A)Z2=rowi(A)2 for every i<m; summing over i gives AZF2=i<mrowi(A)22=AF2.

L4algebra
1.4

Singular values of UAV. If A=U0ΣV0 is the decomposition of [L1], then UAV=(UU0)Σ(VV0)=(UU0)Σ(V0V), and by [L5] the factors UU0 and V0V are unitary, so this is a singular value decomposition of UAV with the same singular values as A.

L1L5algebra
2.1

From [L1], A=UΣV, so steps 1.2 and 1.3 give AF2=UΣVF2=ΣF2, and the diagonal entries of Σ are σ1,,σr followed by zeros, so ΣF2=σ12++σr2 by the entry formula of The Frobenius norm AF=(i,jaij2)1/2 on real or complex matrices. This is claim 2.

L1step 1.2step 1.3algebra
3.1

Claim 3 follows: by step 1.4 the singular values of UAV are those of A, so claim 1 gives UAV2=σ1=A2 and claim 2 gives UAVF=(jσj2)1/2=AF.

step 1.4step 1.1step 2.1
3.2

The lower inequality. The maximum σ1 is at most the Euclidean total (σ12++σr2)1/2, each term being nonnegative, so claims 1 and 2 give A2AF. Equality holds exactly when σ2==σr=0, that is exactly when r1 by [L3].

step 1.1step 2.1L3algebra
3.3

The upper inequality. Each σjσ1, so σ12++σr2rσ12, and claims 1 and 2 give AFrσ1=rA2. Equality holds exactly when jr(σ12σj2)=0, a sum of nonnegative terms, hence exactly when every σj=σ1.

step 1.1step 2.1algebra
4.1

Since rmin(m,n) by [L3], the upper bound also gives AFmin(m,n)A2.

step 3.3L3
5.1

Claims 1, 2, 3 and 4 are steps 1.1, 2.1, 3.1 and steps 3.2, 3.3 and 4.1.

step 1.1step 2.1step 3.1step 3.2step 3.3step 4.1
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system

Definition

Let nN with n1, let either pQ with p1 or p=, and let AGLn(R) be an invertible real matrix (Invertible matrices and the general linear group GLn(F)), so that the linear system Ax=b has the unique solution x=A1b for every right-hand side bRn. The condition number of the linear system (equivalently, of the matrix A) with respect to the induced p-norm of The matrix norm induced by a published vector p-norm, using its separately defined infinity case when p=, is

κp(A)  :=  ApA1p    [1,+).

The value is at least one and is never infinite. The inverse A1 is the unique two-sided inverse of A (Invertible matrices and the general linear group GLn(F)), so AA1=In and A1A=In. For rational finite p, submultiplicativity and normalisation are Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1. For p=, the same properties follow directly from the defining supremum: AxAx and applying this estimate twice gives submultiplicativity, while Inx=x gives In=1. Hence in either case

1=Inp=AA1pApA1p=κp(A).

Equivalently, κp(A) is the product of the largest factor by which A stretches a nonzero vector and the largest factor by which A1 stretches one. Applying the same estimate to A1 in place of A gives κp(A1)=κp(A).

Remarks

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Right-hand-side perturbations in Ax = b are amplified by the condition number

Statement

Let n1, let pQ with p1, let AGLn(R), let b,δbRn with b0, let x=A1b, and let δxRn satisfy A(x+δx)=b+δb. Then

δxpxp    κp(A)δbpbp.

In words: the relative perturbation of the solution is at most the condition number κp(A) of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system times the relative perturbation of the right-hand side.

Facts & Assumptions

Given: An invertible matrix AGLn(R) with n1, vectors b,δbRn with b0, x=A1b and δx=A1δb.

[L2]

Compatibility of the induced norm: AypApyp for every y (Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1).

[L3]

κp(A)=ApA1p (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

Proof

technique · direct
1.1

Subtracting the two equations gives Aδx=A(x+δx)Ax=(b+δb)b=δb, and multiplying on the left by A1 with [L1] gives δx=A1δb.

givenL1algebra
1.2

Since b=Ax, [L2] gives bp=AxpApxp, and b0 forces bp>0 and xp>0, so 1/xpAp/bp.

givenL2algebra
2.1

By [L2] applied to A1 at the vector δb, step 1.1 gives δxp=A1δbpA1pδbp.

step 1.1L2algebra
3.1

Multiplying the bound of step 2.1 by the bound of step 1.2 gives δxp/xpA1pApδbp/bp, and [L3] identifies the coefficient product as κp(A), which is the stated bound.

step 2.1step 1.2L3algebra
4.1

Step 3.1 proves the stated bound.

step 3.1

Remarks

  • The bound is sharp at p=2. The constant κ2(A) is optimal, but the two equalities in the proof are attained on different singular directions: choose x along a right singular vector for the largest singular value of A, so b=Ax has norm σmaxx2, and choose δb along a left singular vector for the smallest singular value, so A1δb2=σmin1δb2. Then the bound is attained with coefficient σmax/σmin=κ2(A).
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Matrix perturbations in Ax = b obey a first-order condition-number bound

Statement

Let n1, let pQ with p1, let AGLn(R), let bRn with b0, let δAMn(R), let x=A1b, and suppose δxRn satisfies (A+δA)(x+δx)=b. Put

c  :=  A1pδAp  =  κp(A)δApAp.

If c<1 then

δxpxp    c1c.

To first order in the matrix perturbation,

δxpxp    κp(A)δApAp,

because c/(1c)=c+c2/(1c)=c+O(c2) as c0: the denominator 1c contributes only a second-order correction when the relative matrix perturbation is small.

Facts & Assumptions

Given: An invertible matrix AGLn(R) with n1, vectors b0, x=A1b, and δx with (A+δA)(x+δx)=b for a matrix δA; write c=A1pδAp.

[L2]

Compatibility: MypMpyp (Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1).

[L3]

κp(A)=ApA1p (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

Proof

technique · direct
1.1

Expanding the perturbed equation and subtracting Ax=b gives Aδx+δAx+δAδx=0, so Aδx=δA(x+δx).

givenalgebra
2.1

Multiplying step 1.1 on the left by A1 and using [L1] gives δx=A1δA(x+δx).

step 1.1L1algebra
3.1

Taking norms in step 2.1 and applying [L2] twice gives δxp=A1δA(x+δx)pA1pδApx+δxp=cx+δxp.

step 2.1L2algebra
4.1

The triangle inequality of the vector p-norm gives x+δxpxp+δxp, so step 3.1 yields δxpcxp+cδxp.

step 3.1algebra
5.1

Under the hypothesis c<1 the factor 1c is positive, and rearranging step 4.1 gives (1c)δxpcxp, hence δxp/xpc/(1c), the claimed exact bound; x0 because b0 and A is invertible.

step 4.1givenalgebra
6.1

The first-order form. Writing c=κp(A)δAp/Ap by [L3], one has c/(1c)=c+c2/(1c), so the exact bound of step 5.1 reads δxp/xpκp(A)δAp/Ap+O(c2) with the error term tending to zero quadratically in c; to first order the amplification is κp(A) times the relative matrix perturbation.

step 5.1L3algebra
7.1

Step 5.1 proves the exact bound and step 6.1 the first-order reading of it.

step 5.1step 6.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The reciprocal spectral condition number is the relative distance to singularity

Statement

Let n1, let AGLn(R), let σ1σn>0 be its singular values, and let 2 be the spectral (operator) norm. Then:

  1. Smallest singular value as a minimum. σn=min{Ax2:x2=1}.
  2. Distance to singularity. min{E2  :  EMn(R), A+E is singular}=σn, and the minimum is attained.
  3. Relative distance. minA+E singularE2A2=σnσ1=1κ2(A), where κ2(A) is the condition number of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system. Equivalently, A12=1/σn.

In words: the nearest singular matrix to A lies at spectral distance σn, and that distance relative to the scale A2=σ1 is exactly 1/κ2(A).

Facts & Assumptions

Given: An invertible real matrix AGLn(R) with n1, its singular values σ1σn>0, and a matrix EMn(R).

[L1]

There is a singular value decomposition A=UΣVT with orthogonal U,V and diagonal Σ of the singular values (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

The operator norm equals the largest singular value, attained at a right-singular vector (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 matrix is the number of its positive singular values (The rank of a linear map is the number of its nonzero singular values).

[L5]

Unitary invariance and the singular-value formulas of Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison: for orthogonal Q,Z the matrices QAZ and A have the same singular values, and A2=σ1.

Proof

technique · direct
1.1

Write A=UΣVT as in [L1]. For every unit vector x, expanding in the right-singular basis gives Ax22=j<nσj2(VTx)j2σn2j<n(VTx)j2=σn2, the last equality because VT is orthogonal.

L1algebra
1.2

Attainment. Let un and vn be the last columns of U and V from [L1], and put E:=σnunvnT. Then A+E=U(ΣσnenenT)VT has a zero on its last diagonal position, so its smallest singular value is 0; by [L3] its rank is at most n1, and by [L4] it is singular.

L1L3L4algebra
2.1

At the last right-singular vector, AVen2=σnUen2=σn, so the lower bound of step 1.1 is attained and σn=minx2=1Ax2, which is claim 1.

step 1.1L1algebra
3.1

Lower bound on the distance. If E makes A+E singular then A+E has rank at most n1, and by [L3] its smallest singular value is 0. Every unit vector x satisfies (A+E)x2Ax2Ex2σnE2, using claim 1 and the operator-norm bound Ex2E2x2 of [L2].

step 2.1L2L3algebra
4.1

If E2<σn then step 3.1 gives (A+E)x2>0 for every unit x, so the minimum of claim 1 applied to A+E is positive and every singular value of A+E is positive; by [L3] its rank is n, and by [L4] it is invertible. Hence no E with E2<σn makes A+E singular: the distance is at least σn.

step 3.1step 2.1L3L4
5.1

The rank-one matrix unvnT has the singular values 1,0,,0, so [L2] gives E2=σn; combined with step 4.1, the minimum distance is exactly σn and is attained, which is claim 2.

step 1.2L2
6.1

By [L5], A2=σ1; and A1=VΣ1UT by [L1], a matrix with singular values 1/σn1/σ1, so [L2] gives A12=1/σn. Hence κ2(A)=A2A12=σ1/σn by The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system, and claim 3 follows with claim 2.

step 5.1L1L2L5algebra
7.1

Claims 1, 2 and 3 are steps 2.1, 5.1 and 6.1.

step 2.1step 5.1step 6.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Residual bounds on forward error for a nonsingular linear system

Statement

Let n1, let pQ with p1, let AGLn(R), let bRn with b0, let x=A1b, and let x^Rn be an approximate solution with residual r:=bAx^ and error e:=xx^. Then r=Ae and e=A1r, and the following bounds hold:

  1. Absolute bounds. ep    A1prp,rp    Apep.
  2. Relative bounds. 1κp(A)rpbp    epxp    κp(A)rpbp, where κp(A) is the condition number of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system.

In words: a small residual bounds the forward error below by rp/Ap, but bounds it above only up to the condition number κp(A).

Facts & Assumptions

Given: An invertible matrix AGLn(R) with n1, b0, x=A1b, x^Rn, r=bAx^ and e=xx^.

[L1]

A1A=In (Invertible matrices and the general linear group GLn(F)), and matrix multiplication distributes over addition, so A(xx^)=AxAx^ (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[L2]

Compatibility: MypMpyp (Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1).

[L3]

κp(A)=ApA1p (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

Proof

technique · direct
1.1

Since Ax=b, the residual is r=bAx^=AxAx^=A(xx^)=Ae, using distributivity of [L1].

givenL1algebra
2.1

Multiplying r=Ae on the left by A1 and using [L1] gives e=A1r.

step 1.1L1algebra
2.2

The lower absolute bound. Applying [L2] to A at the vector e in step 1.1 gives rp=AepApep.

step 1.1L2algebra
3.1

The upper absolute bound. Applying [L2] to A1 at the vector r in step 2.1 gives ep=A1rpA1prp.

step 2.1L2algebra
3.2

The lower relative bound. From x=A1b and [L2], xpA1pbp, so 1/bpA1p/xp; combining with step 2.2 gives rp/bpApA1pep/xp=κp(A)ep/xp by [L3], which rearranges to the lower bound on ep/xp.

step 2.2L2L3algebra
4.1

The upper relative bound. From b=Ax and [L2], bpApxp, and b0 forces xp>0, so 1/xpAp/bp; multiplying by step 3.1 gives ep/xpA1pAprp/bp=κp(A)rp/bp by [L3].

step 3.1L2L3algebra
5.1

Claim 1 is steps 3.1 and 2.2, and claim 2 is steps 4.1 and 3.2.

step 3.1step 2.2step 4.1step 3.2
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Normwise and componentwise backward error for an approximate linear-system solution

Definition

Let n1, let pQ with p1, let A=(aij)GLn(R), let b=(bi)Rn, and let x^Rn be an approximate solution of the system Ax=b with residual r=bAx^.

Normwise backward error. The normwise backward error of x^ is

ηp(x^)  :=  inf{ε0  :  there are ΔAMn(R), ΔbRn with (A+ΔA)x^=b+Δb, ΔApεAp, Δbpεbp},

where p on matrices is the induced norm of The matrix norm induced by a published vector p-norm and on vectors the p-norm.

Componentwise backward error. The componentwise backward error of x^ is

ω(x^)  :=  inf{ε0  :  there are ΔA=(Δaij), Δb=(Δbi) with (A+ΔA)x^=b+Δb, Δaijεaij for all i,j<n, Δbiεbi for all i<n},

with the absolute value (Absolute value in an ordered field).

Both infima are over nonempty sets, so they are well defined. The single choice ΔA:=A and Δb:=b always satisfies the defining relation, because (A+ΔA)x^=0=b+Δb. It is admissible with ε=1 for both notions: normwise, ΔAp=Ap and Δbp=bp; componentwise, Δaij=aij and Δbi=bi. Hence each defining set is a nonempty subset of [0,) and therefore has an infimum.

In words: ηp(x^) is the smallest factor by which the data (A,b) must be perturbed, measured in the p-norm with the weights Ap and bp, for the computed answer x^ to be exactly right; ω(x^) measures the same perturbation entrywise. The explicit residual formulas for both are Residual formulas for normwise and componentwise backward error.

Remarks

  • Backward error is about the data, not the answer. An approximate solution with small backward error is the exact solution of a nearby problem; the size of the forward error xx^ is governed separately by the condition number through Local conditioning times backward error controls forward error to first order.

  • The two notions are not interchangeable. Componentwise admissibility controls every entry and preserves the zero pattern of the data, whereas normwise admissibility controls only the declared aggregate norms. Neither definition gives the same numerical tolerance as the other without dimension- and data-dependent comparison factors. Sparse matrices make the distinction especially visible: ω preserves their zero pattern but ηp does not.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Residual formulas for normwise and componentwise backward error

Statement

Let n1, let AGLn(R), let bRn, let x^Rn, and put r=bAx^.

  1. Normwise formula (spectral norm). η2(x^)=r2A2x^2+b2, with the convention 0/0:=0.
  2. Componentwise formula. With (Ax^)i:=j<naijx^j, ω(x^)=maxi<nri(Ax^)i+bi, where a term with ri=0 and denominator 0 is interpreted as 0. Every such degenerate term has ri=0, so the maximum is a finite real number.

Here η2 and ω are the backward errors of Normwise and componentwise backward error for an approximate linear-system solution, and 2 is the spectral (operator) norm, which on vectors is the Euclidean norm of Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page.

Facts & Assumptions

Given: An invertible matrix AGLn(R) with n1, vectors b,x^Rn, the residual r=bAx^, and the backward errors η2(x^), ω(x^).

[L1]

An admissible perturbation satisfies ΔAx^Δb=r and the stated norm or entrywise bounds (Normwise and componentwise backward error for an approximate linear-system solution).

[L2]

Compatibility of the induced spectral norm: My2M2y2 (Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1), and the vector 2-norm is the Euclidean norm with the triangle inequality (Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page).

[L5]

Absolute value: u+vu+v and uv=uv (Absolute value in an ordered field).

Proof

technique · direct
1.1

Lower bound for the normwise error. For any admissible pair of [L1], r2=ΔAx^Δb2ΔA2x^2+Δb2ε(A2x^2+b2), using [L1], [L2] and the triangle inequality of [L2].

L1L2algebra
1.2

Attainment for the normwise error. Put γ:=A2x^2+b2 and suppose first γ>0. If x^0, define ΔA:=(A2/γ)rx^T/x^2 and Δb:=(b2/γ)r; then ΔAx^Δb=r and, by [L3] applied row-wise, rx^T2=r2x^2, so ΔA2=ηA2 and Δb2=ηb2 with η:=r2/γ.

L1L3algebraconstruct
1.3

If x^=0 then r=b and γ=b2>0; the perturbations ΔA=0 and Δb=r satisfy ΔAx^Δb=r with Δb2=r2=ηb2, so the same η is attained.

L1algebraconstruct
1.4

If γ=0 then A2x^2=0 and b=0, so r=bAx^=0; the convention 0/0:=0 and the zero perturbations of [L1] give η2(x^)=0, matching the formula.

L1algebra
1.5

Lower bound for the componentwise error. For any admissible pair of [L1], ri=jΔaijx^jΔbijΔaijx^j+Δbiε(jaijx^j+bi) by [L5] and the entrywise bounds of [L1].

L1L5algebra
1.6

A term with denominator 0 has bi=0 and aijx^j=0 for every j<n, so ri=bi(Ax^)i=0; thus every degenerate term carries numerator 0, and the maximum is finite as claimed.

givenalgebra
1.7

Attainment for the componentwise error. Put di:=(Ax^)i+bi and ω:=maxi<nri/di with the convention of the statement. For each i<n with di>0 define Δbi:=ribi/di and, for each j<n, Δaij:=aijsgn(x^j)ri/di when x^j0 and Δaij:=0 when x^j=0; for di=0 set all these entries to 0.

constructalgebra
2.1

Hence every admissible ε is at least r2/(A2x^2+b2), so the infimum satisfies η2(x^)r2/(A2x^2+b2) when the denominator is positive.

step 1.1algebra
2.2

For every i<n with (Ax^)i+bi>0, step 1.5 gives εri/((Ax^)i+bi), so every admissible ε is at least the maximum over such i; hence ω(x^)maxi<nri/((Ax^)i+bi) with the stated convention.

step 1.5algebra
2.3

For di>0 one has (ΔAx^)i=jaijx^jri/di=ri(Ax^)i/di and Δbi=ribi/di, so (ΔAx^)iΔbi=ri; and Δaij=aijri/diωaij together with Δbiωbi, since riωdi. For di=0 every displayed entry vanishes and step 1.6 gives ri=0.

step 1.7step 1.6L5algebra
3.1

Combining steps 2.1, 1.2, 1.3 and 1.4, the infimum is attained and equals r2/(A2x^2+b2) under the stated convention, which is claim 1.

step 2.1step 1.2step 1.3step 1.4
3.2

The perturbations of step 1.7 are admissible with ε=ω by step 2.3, so ω(x^)ω; with step 2.2 the formula of claim 2 holds.

step 2.3step 2.2L1
4.1

Claim 1 is step 3.1 and claim 2 is step 3.2.

step 3.1step 3.2
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

The standard relative floating-point model and unit roundoff

Definition

A floating-point system. A floating-point system consists of a nonempty finite set FR of representable numbers with 0F and F=F, a positive real NmaxF (the largest normal magnitude), a positive real NminF (the smallest normal magnitude), with NminNmax, and a rounding map fl defined on 0 and on the real numbers t with NmintNmax, taking values in F and satisfying fl(0)=0.

Unit roundoff. A real u>0 is a unit roundoff for the system when every real t with NmintNmax satisfies

fl(t)=t(1+δ)for some δ with δu,

with the absolute value (Absolute value in an ordered field). The model quantifies all subsequent error analysis through the single constant u and says nothing about the internal bit representation beyond the bound δu.

The standard relative model. For each operation {+,,×,÷} and all x,yF such that xy is zero or is a real number in the normal range (NminxyNmax; for ÷ this includes y0), the computed result is

fl(xy)=(xy)(1+δ)for some δ with δu.

When xy=0, this identity is the exact-zero rule fl(0)=0 and one may take δ=0.

The exclusions are part of the model, and are stated explicitly.

  • Overflow is excluded: when xy>Nmax the result is not a real number of the form (xy)(1+δ) with δu.
  • Underflow is excluded: when 0<xy<Nmin the result falls into the subnormal or lost range and need not satisfy the relative bound.
  • Invalid operations are excluded: when xy is not a real number (for example 0/0), there is no real δ for the displayed identity.

Every item on this page that uses the model therefore carries the hypotheses that the relevant operations lie in the normal range, and statements about overflow, underflow or invalid operations are made only where the exclusion is mentioned.

Remarks

  • The running example. In the IEEE-754 binary64 system the unit roundoff is u=253, the largest normal magnitude is Nmax=(2252)210231.79810308, and the smallest normal magnitude is Nmin=21022. The counterexample page uses these three numbers to exhibit the excluded events.

  • The exclusions are needed, not decorative. Each excluded event genuinely produces a value that is not of the form (xy)(1+δ) with δu: overflow at ((3/2)21023)2 gives +, underflow at 21074/2 rounds to 0, and 0/0 gives NaN.

  • The model is an assumption, not a theorem about hardware. Statements proved under this model are conditional on the displayed inequalities; they say "if every operation rounds this accurately, then the following error bound holds". That is exactly what makes the bounds of The computed floating-point dot product obeys componentwise and normwise error bounds transportable to any hardware conforming to the model.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

Forward and backward stability for a problem family under an arithmetic model

Definition

Fix normed spaces (X,X) and (Y,Y) over R (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), a map f:XY, a class Idomf of inputs, an arithmetic model with unit roundoff u (The standard relative floating-point model and unit roundoff), and an algorithm that, for each input xI, returns an approximate value y^Y of Absolute and relative forward error for an approximate value of a problem map.

Backward stability. The algorithm is backward stable on I when there is a constant C0 such that for every xI there is an input x~domf with

y^=f(x~)andx~xXCuxX.

The computed answer is exactly right for a slightly perturbed input, and the perturbation is at most a fixed multiple C of the unit roundoff.

Forward stability. The algorithm is forward stable on I when there is a constant C0 such that for every xI with f(x)0,

y^f(x)Yf(x)YCu.

The standard linear-system specialization. Fix n1 and a rational p1. For the problem "given AGLn(R) and b, return x with Ax=b", one usually measures the two data components separately rather than forcing the pair (A,b) into the single fixed input norm above. An algorithm is called normwise backward stable on a declared class of systems when one constant C makes every returned x^ satisfy ηp(x^)Cu in the sense of Normwise and componentwise backward error for an approximate linear-system solution; the componentwise specialization requires ω(x^)Cu. Thus there are perturbations ΔA,Δb with (A+ΔA)x^=b+Δb and separately measured relative sizes at most Cu. These are specialized product-wise definitions, not literal instances of one undeclared norm on the data pairs. Stability remains a property of the algorithm on the problem class, while the condition number is a property of the problem itself.

Remarks

  • Backward stability does not by itself bound the forward error. It bounds the data perturbation; converting that into an answer bound costs one factor of the condition number, by Local conditioning times backward error controls forward error to first order. A backward-stable algorithm on an ill-conditioned problem can still return a large forward error: for A=diag(1,ε), b=(1,ε), and x^=(1,0), the backward error is O(ε) while the relative forward error is 1/2.

  • The constant C is part of the honest statement. The definition records "at most C times the unit roundoff" for some fixed C, not "exactly the unit roundoff"; polynomial growth in the dimension is folded into C, and a stability claim without its class I and its constant is not a stability claim.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

If |delta_i| <= u and n u < 1, then prod (1+delta_i) = 1 + theta_n with |theta_n| <= gamma_n

Statement

Let nN with n1, let u0, let δ0,,δn1R satisfy δiu for every i<n, and suppose nu<1. Then

i<n(1+δi)=1+θnwithθnγn:=nu1nu.

Facts & Assumptions

Given: A natural number n1, a real u0 with nu<1, and reals δ0,,δn1 with δiu; write γn:=nu/(1nu).

[L1]

The hypothesis nu<1 gives u<1, hence 1u>0 and 1+u>0; the unit roundoff vocabulary comes from The standard relative floating-point model and unit roundoff.

[L2]

Binomial expansion over the reals: (1+t)n=kn(nk)tk (The binomial theorem in R: (x+y)n=k<n+1ι ⁣(nk)xkynk), and the binomial coefficient satisfies (nk)nk for every kn.

[L3]

Geometric series: for a real r with 0r<1, k=0rk=1/(1r) (For r<1, k0rk=1/(1r), and for r1 the series diverges).

Proof

technique · direct
1.1

Since δiu, each factor satisfies 1u1+δi1+u, and the factors 1±u are nonnegative by [L1]; multiplying the inequalities across i<n gives (1u)ni<n(1+δi)(1+u)n.

L1algebra
1.2

Upper deviation. By [L2], (1+u)n1=k=1n(nk)ukk=1nnkuk, since (nk)nk and u0.

L2algebra
2.1

The geometric series of [L3] at r=nu, where 0nu<1, gives k=1(nu)k=nu/(1nu)=γn, so step 1.2 gives (1+u)n1γn.

step 1.2L3algebra
3.1

Lower deviation. By [L2], 1(1u)n=k=1n(nk)(u)k=k=1n(nk)(1)k+1ukk=1n(nk)ukγn, the last estimate being step 1.2 and step 2.1; hence (1u)n1γn.

L2step 1.2step 2.1algebra
4.1

Define θn:=i<n(1+δi)1. Step 1.1 gives (1u)n1θn(1+u)n1, and steps 2.1 and 3.1 give (1u)n1γn and (1+u)n1γn, so γnθnγn, that is θnγn.

step 1.1step 2.1step 3.1algebra
5.1

Step 4.1 is exactly the claim.

step 4.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The computed floating-point dot product obeys componentwise and normwise error bounds

Statement

Let n1 and let x=(xj),y=(yj)Rn, with the dot product xy=j<nxjyj computed from left to right in an arithmetic model with unit roundoff u satisfying nu<1 (The standard relative floating-point model and unit roundoff), where every product xjyj, every rounded product pj:=fl(xjyj), and every intermediate sum lies in the normal range. Let s1:=p0 and, for 1k<n, let sk+1:=fl(sk+pk). Then:

  1. Componentwise form. There are reals δ0,,δn1 with δjγn:=nu/(1nu) such that sn=j<nxjyj(1+δj).
  2. Normwise form. snxy    γnj<nxjyj    γnx1y1, where 1 is the published 1-norm (The p-norms xp for rational p1, and x, Each p is a norm on Rn, and the induced metrics are exactly d1, d2 and d of the published metric-spaces page) and the absolute value (Absolute value in an ordered field).

Facts & Assumptions

Given: A natural number n1, vectors x,yRn, a unit roundoff u with nu<1, the rounded products pj=fl(xjyj) for j<n, and the left-to-right computed sum sn above.

[L1]

Each product and each sum obeys the relative model: fl(t)=t(1+ε) with εu for every real t in the normal range (The standard relative floating-point model and unit roundoff).

[L2]

A product of factors (1+εi)i<m with mn and εiu for every i<m equals 1+δ with δγmγn, where γn=nu/(1nu) (If |delta_i| <= u and n u < 1, then prod (1+delta_i) = 1 + theta_n with |theta_n| <= gamma_n).

[L3]

x1=j<nxj, and xjyj=xjyj (The p-norms xp for rational p1, and x, Absolute value in an ordered field).

Proof

technique · direct
1.1

By [L1], for each j<n there is πj with pj=xjyj(1+πj) and πju; and for each k with 1k<n there is αk with sk+1=(sk+pk)(1+αk) and αku.

givenL1
2.1

Unrolling the recursion of step 1.1 gives sn=x0y0(1+π0)k=1n1(1+αk)+j=1n1xjyj(1+πj)k=jn1(1+αk), where every empty product is 1; each coefficient is the product of the error factor from the rounded multiplication pj and the error factors from the later additions through which that term passes.

step 1.1algebra
3.1

The coefficient of x0y0 in step 2.1 is a product of exactly n factors of the form 1+ε with εu, and for each j with 1j<n the coefficient of xjyj is a product of nj+1n such factors. By [L2], each coefficient therefore equals 1+δj with δjγn, so step 2.1 becomes sn=j<nxjyj(1+δj), which is claim 1.

step 2.1L2algebra
4.1

Subtracting xy=jxjyj from claim 1 gives snxy=jxjyjδj, so by the triangle inequality and [L3] snxyjxjyjδjγnjxjyj.

step 3.1L3algebra
5.1

Since jxjyj=jxjyj(jxj)(kyk)=x1y1 by [L3], step 4.1 gives the normwise form of claim 2.

step 4.1L3algebra
6.1

Claim 1 is step 3.1 and claim 2 is steps 4.1 and 5.1.

step 3.1step 4.1step 5.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Local conditioning times backward error controls forward error to first order

Statement

Let f:XY be a map between normed spaces, let xdomf, and let κ:=κabs(f,x) be the absolute local condition number of Absolute and relative local condition numbers of a problem map.

  1. Quantified form. If κ<+, then for every c>κ there is a δ>0 such that every hX with 0<h<δ and x+hdomf satisfies f(x+h)f(x)    ch.
  2. First-order form. If κ<+, then along admissible h, f(x+h)f(x)    (κ+o(1))has h0, that is: for every ε>0 there is δ>0 such that 0<h<δ implies f(x+h)f(x)(κ+ε)h.
  3. Relative form. If additionally x0 and f(x)0, and κrel:=κx/f(x), then along admissible h, f(x+h)f(x)f(x)    (κrel+o(1))hxas h0.

In the vocabulary of Forward and backward stability for a problem family under an arithmetic model: a computed value y^=f(x+h) has backward error h, and its forward error is, to first order in that backward error, at most the condition number times the backward error; the linear-system instance uses the backward error of Normwise and componentwise backward error for an approximate linear-system solution.

Facts & Assumptions

Given: Normed spaces X,Y, a map f:XY, a point xdomf, and κ=infδ>0Sf,x(δ) where Sf,x(δ)=sup{f(x+h)f(x)/h:0<h<δ, x+hdomf}.

[L1]

The absolute condition number is the infimum over δ>0 of the nondecreasing map δSf,x(δ) (Absolute and relative local condition numbers of a problem map).

[L2]

An infimum characterisation: if κ is the infimum of a set S[0,+], then for every c>κ there is an element sS with s<c; in particular for every c>κ there is δ>0 with Sf,x(δ)<c.

Proof

technique · direct
1.1

By [L2] applied to the set S={Sf,x(δ):δ>0} whose infimum is κ by [L1], every c>κ admits some δ0>0 with Sf,x(δ0)<c.

L1L2choose
2.1

By the definition of Sf,x(δ0) as a supremum, every admissible h with 0<h<δ0 satisfies f(x+h)f(x)/hSf,x(δ0)<c, hence f(x+h)f(x)ch, which is claim 1 with δ:=δ0.

step 1.1givenalgebra
3.1

For every ε>0 the number c:=κ+ε is strictly larger than κ, so claim 1 supplies δ>0 with f(x+h)f(x)(κ+ε)h for all admissible h of norm below δ; this is exactly the stated bound, which is claim 2.

step 2.1algebra
4.1

For the relative form, divide the inequality of claim 2 by the fixed positive number f(x) and multiply by the fixed positive number x: f(x+h)f(x)/f(x)(κ+ε)h/f(x)=(κx/f(x)+εx/f(x))h/x, and the error term is o(1)h/x because the positive constants x,f(x) are fixed, which is claim 3.

step 3.1givenalgebra
5.1

Claims 1, 2 and 3 are steps 2.1, 3.1 and 4.1.

step 2.1step 3.1step 4.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

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

Statement

Let m,nN with n1 and mn, and let AMm×n(R) have full column rank (rank n), with singular values σ1σn>0. Define the spectral condition number of the rectangular matrix A by

κ2(A):=σ1σn;

for a square invertible A this agrees with A2A12 of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system. Then:

  1. The normal-equation matrix ATA is invertible, with κ2(ATA)=κ2(A)2;
  2. equivalently, κ2(ATA)=σ12σn2.

The least-squares normal equation ATAx=ATb of For a linear map T:VW between finite-dimensional inner-product spaces, x minimises Txb if and only if T(Txb)=0, equivalently TTx=Tb; minimisers exist and any two differ by an element of kerT therefore has condition number exactly the square of the condition number of the least-squares problem itself.

Facts & Assumptions

Given: A matrix AMm×n(R) of full column rank with n1, mn, and singular values σ1σn>0.

[L1]

There is a singular value decomposition A=UΣVT with orthogonal UMm(R), VMn(R) and Σ diagonal with entries σ1,,σn on the n×n block (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L3]

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

[L4]

Proof

technique · direct
1.1

From [L1], ATA=VΣTΣVT, and ΣTΣ=diag(σ12,,σn2) is diagonal with decreasing nonnegative entries; by [L4], V is orthogonal with VVT=In.

L1L4algebra
2.1

The displayed product Vdiag(σj2)VT is a singular value decomposition of ATA with singular values σ12σn2, since the two orthogonal factors are V and V.

step 1.1L1algebra
3.1

By [L3], full column rank means σn>0, so every singular value of ATA in step 2.1 is positive, and ATA has the inverse Vdiag(σj2)VT: the two products with ATA give In by [L4].

step 2.1L3L4algebra
4.1

By [L2] applied to ATA and to its inverse, ATA2=σ12 and (ATA)12=σn2, the largest singular value of Vdiag(σj2)VT being σn2; hence κ2(ATA)=σ12σn2.

step 2.1step 3.1L2algebra
5.1

By the definition of κ2(A) in the Statement, κ2(A)2=σ12/σn2, 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 ATA from The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system, the displayed agreement for square A being A12=σn1 by [L2] applied to the SVD A1=VΣ1UT of [L1].

step 4.1L1L2algebra
6.1

The normal equation of [L5] has coefficient matrix ATA, 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 ATA squares the condition number even though the least-squares problem itself has condition number κ2(A); computing with A directly through a QR route keeps the factor κ2(A) instead. For A=(10010800), the Gram-Schmidt factor is R=diag(1,108) while ATA=diag(1,1016).
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

Exact rank is locally constant at full-rank matrices and discontinuous at nonmaximal-rank matrices

Statement

Let m,nN with m1 and n1, let AMm×n(R), let k:=min(m,n), and let 2 be the spectral norm.

  1. Locally constant at full rank. If rankA=k and σk is the smallest singular value of A (which is positive), then every BMm×n(R) with BA2<σk satisfies rankB=k.
  2. Discontinuous at nonmaximal rank. If rankA=r<k, then for every ε>0 there is BMm×n(R) with BA2<ε and rankB=r+1.

Facts & Assumptions

Given: Natural numbers m,n1, a matrix AMm×n(R) with rank r and singular values σ1σr>0, and k=min(m,n).

[L1]

There is a singular value decomposition A=UΣVT with orthogonal U,V and diagonal Σ; likewise every matrix BMm×n(R) admits one (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

The rank of a matrix is the number of its positive singular values (The rank of a linear map is the number of its nonzero singular values).

[L3]

The spectral norm of a matrix equals its largest singular value, and My2M2y2 for every vector y (Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison).

Proof

technique · direct
1.1

If rankA=k, then by [L2] there are exactly k positive singular values, so the smallest one, σk, is positive.

L2
1.2

Full-column-rank lower bound. Suppose nm and BMm×n(R) has singular values τ1τn. Writing a unit vector x in the right-singular basis of the decomposition from [L1] gives Bx22=j<nτj2(VTx)j2τn2, attained at the last right-singular vector. Hence τn=minx2=1Bx2.

L1algebra
1.3

Suppose rankA=r<k; then both r<m and r<n, so the decomposition of [L1] has a left singular vector ur+1Rm and a right singular vector vr+1Rn beyond the positive singular values.

L1givenalgebra
1.4

For ε>0 put Bε:=A+εur+1vr+1T. Then Bε=U(Σ+εer+1er+1T)VT is a singular value decomposition of Bε with positive singular values σ1,,σr,ε, so by [L2] rankBε=r+1.

L1L2algebraconstruct
2.1

Assume first that rankA=k and nm. Applying step 1.2 to A gives Ax2σk for every unit vector x, and [L3] gives Ex2E2 for E:=BA; so Bx2=Ax+Ex2Ax2Ex2σkE2. If E2<σk, then Bx2>0 for every unit vector x, so step 1.2 applied to B gives its smallest singular value τn>0. Therefore [L2] gives rankB=n=k.

step 1.2L2L3algebra
2.2

Fix ε>0 and set η:=ε/2. Step 1.4 gives rankBη=r+1. By [L3], the spectral norm of ηur+1vr+1T is η times the largest singular value of the rank-one matrix ur+1vr+1T, whose only positive singular value is 1; hence BηA2=η<ε. This proves the discontinuity claim.

step 1.4L3algebra
3.1

Assume next that rankA=k and m<n. Then AT has full column rank m and the same positive singular values as A by [L1], so its smallest singular value is again σk. The same transpose relation holds for BT, and [L3] gives BTAT2=BA2. Applying step 2.1 to AT and BT yields rankBT=m, hence rankB=m=k.

L1L3step 2.1algebra
4.1

Steps 2.1 and 3.1 prove the local-constancy claim.

step 2.1step 3.1
5.1

Claim 1 is step 4.1 and claim 2 is step 2.2.

step 4.1step 2.2
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

Numerical rank relative to a declared norm, scale, and tolerance

Definition

Let m,nN, let F be R or C, let be a declared norm on the matrix space Mm×n(F), let AMm×n(F), and let τ0 be a declared tolerance. The numerical rank of A at tolerance τ is

rankτ(A)  :=  min{rankB  :  BMm×n(F), ABτ}.

The minimum exists: it is a minimum of a nonempty subset of the finite set {0,1,,min(m,n)}, the value rankA being attained at B=A.

Scale. The tolerance may be declared absolutely, as a real number τ0, or relative to a scale, as τ=τ0A for a declared factor τ00. Both enter the definition through the single threshold τ; the scale only fixes which threshold is meant, and every statement about rankτ is a statement about that one threshold.

Why the exact-rank discontinuity forces the parameter. Exact rank is locally constant at the full-rank matrices but discontinuous at every nonmaximal-rank matrix (Exact rank is locally constant at full-rank matrices and discontinuous at nonmaximal-rank matrices); an exact rank read from noisy data is therefore unstable. Numerical rank replaces exact rank by the smallest exact rank attainable within the declared norm tolerance τ: it is, by construction, constant on every open ball of radius below the relevant gaps, and it counts as rank exactly those directions that perturbation of size at most τ cannot erase.

Remarks

  • The choice of norm is part of the data. Different norms give different distances to the low-rank sets, hence possibly different numerical ranks for the same matrix and tolerance. The spectral specialisation is In the spectral norm, numerical rank counts the singular values above the declared threshold; the dependence on the declared tolerance itself is genuine: for A=diag(1,1/2,0), the tolerances 3/4 and 1/4 give the numerical ranks 1 and 2 respectively.

  • Numerical rank is not a substitute for exact rank. It answers a different question: which rank is stable under perturbations of size at most τ, not what the exact rank is. In the spectral norm, if A0, the two ranks agree exactly when τ is strictly smaller than the smallest nonzero singular value of A; for A=0 they agree for every τ0. Once τ reaches a positive singular value, the numerical rank can drop.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

In the spectral norm, numerical rank counts the singular values above the declared threshold

Statement

Let m,nN with m1 and n1, let AMm×n(R) with singular values σ1σmin(m,n)0, let τ0, and let 2 be the spectral norm. Then

rankτ(A)=#{j{1,,min(m,n)}  :  σj>τ},

where rankτ(A) is the numerical rank of Numerical rank relative to a declared norm, scale, and tolerance. Consequently, if two tolerances τ<τ have a singular value σj with τ<σjτ, then rankτ(A)>rankτ(A): moving the threshold past a singular value changes the numerical rank.

Facts & Assumptions

Given: Natural numbers m,n1, a matrix AMm×n(R) with singular values σ1σmin(m,n)0, and a tolerance τ0.

[L1]

Numerical rank: rankτ(A)=min{rankB:AB2τ} (Numerical rank relative to a declared norm, scale, and tolerance).

[L2]

Eckart–Young: for 0k<dimRn=n, the least spectral-norm distance from A to the matrices of rank at most k is σk+1, with the convention σk+1=0 when k exceeds the number of positive singular values (The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition).

[L3]

The rank of a matrix is the number of its positive singular values (The rank of a linear map is the number of its nonzero singular values).

Proof

technique · direct
1.1

For kmin(m,n) the matrix A itself has rank at most min(m,n)k by [L3], so the distance to rank at most k is 0; with the convention of [L2] this agrees with σk+1=0. For 0k<min(m,n)n, [L2] gives the distance σk+1. Hence for every k0, min{AB2:rankBk}=σk+1.

L2L3algebra
1.2

Let q:=#{j{1,,min(m,n)}:σj>τ}. If q<min(m,n) then σq+1τ while σq>τ; if q=min(m,n) then all singular values exceed τ and σmin(m,n)+1:=0τ. In both cases min{k0:σk+1τ}=q, because every k<q has σk+1σq>τ.

algebra
2.1

By step 1.1, there exists a matrix B with rankBk and AB2τ exactly when σk+1τ. Therefore [L1] gives rankτ(A)=min{k0:σk+1τ}.

step 1.1L1algebra
3.1

Steps 2.1 and 1.2 give rankτ(A)=q, which is the displayed formula.

step 2.1step 1.2
4.1

If τ<σjτ then j is counted at tolerance τ but not at tolerance τ, and no larger index is counted at τ; hence rankτ(A)>rankτ(A), the stated consequence.

step 3.1algebra
5.1

Step 3.1 proves the formula and step 4.1 the consequence.

step 3.1step 4.1

5 · Examples, counterexamples and false statements

None yet.

Sources