TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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The reciprocal spectral condition number is the relative distance to singularity

Statement

Let $n \geq 1$ , let $A \in {GL}_{n} (R)$ , let $σ_{1} \geq \dots \geq σ_{n} > 0$ be its singular values, and let $∥ \cdot ∥_{2}$ be the spectral (operator) norm. Then:

Smallest singular value as a minimum. $σ_{n} = \min {∥ A x ∥_{2} : ∥ x ∥_{2} = 1}$ .
Distance to singularity. $\min {∥ E ∥_{2} : E \in M_{n} (R), A + E is singular} = σ_{n},$ and the minimum is attained.
Relative distance. $\frac{\min_{A + E singular} ∥ E ∥_{2}}{∥ A ∥_{2}} = \frac{σ_{n}}{σ_{1}} = \frac{1}{κ_{2} (A)},$ where $κ_{2} (A)$ is the condition number of The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system. Equivalently, $∥ A^{- 1} ∥_{2} = 1 / σ_{n}$ .

In words: the nearest singular matrix to $A$ lies at spectral distance $σ_{n}$ , and that distance relative to the scale $∥ A ∥_{2} = σ_{1}$ is exactly $1 / κ_{2} (A)$ .

Facts & Assumptions

Given: An invertible real matrix $A \in {GL}_{n} (R)$ with $n \geq 1$ , its singular values $σ_{1} \geq \dots \geq σ_{n} > 0$ , and a matrix $E \in M_{n} (R)$ .

[L1]

There is a singular value decomposition $A = U Σ V^{T}$ 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).

[L4]

A square matrix is invertible exactly when it has rank $n$ (Invertible matrix theorem: invertibility, full pivot rank, RREF $I$ , trivial nullspace and unique solvability are equivalent).

[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 $Q A Z$ and $A$ have the same singular values, and $∥ A ∥_{2} = σ_{1}$ .

Proof

technique · direct

1.1L1algebra

Write $A = U Σ V^{T}$ as in [L1]. For every unit vector $x$ , expanding in the right-singular basis gives $∥ A x ∥_{2}^{2} = \sum_{j < n} σ_{j}^{2} ∣ (V^{T} x)_{j} ∣^{2} \geq σ_{n}^{2} \sum_{j < n} ∣ (V^{T} x)_{j} ∣^{2} = σ_{n}^{2}$ , the last equality because $V^{T}$ is orthogonal.

1.2L1L3L4algebra

Attainment. Let $u_{n}$ and $v_{n}$ be the last columns of $U$ and $V$ from [L1], and put $E^{*} : = - σ_{n} u_{n} v_{n}^{T}$ . Then $A + E^{*} = U (Σ - σ_{n} e_{n} e_{n}^{T}) V^{T}$ has a zero on its last diagonal position, so its smallest singular value is $0$ ; by [L3] its rank is at most $n - 1$ , and by [L4] it is singular.

2.1step 1.1L1algebra

At the last right-singular vector, $∥ A V e_{n} ∥_{2} = ∥ σ_{n} U e_{n} ∥_{2} = σ_{n}$ , so the lower bound of step 1.1 is attained and $σ_{n} = \min_{∥ x ∥_{2} = 1} ∥ A x ∥_{2}$ , which is claim 1.

3.1step 2.1L2L3algebra

Lower bound on the distance. If $E$ makes $A + E$ singular then $A + E$ has rank at most $n - 1$ , and by [L3] its smallest singular value is $0$ . Every unit vector $x$ satisfies $∥ (A + E) x ∥_{2} \geq ∥ A x ∥_{2} - ∥ E x ∥_{2} \geq σ_{n} - ∥ E ∥_{2}$ , using claim 1 and the operator-norm bound $∥ E x ∥_{2} \leq ∥ E ∥_{2} ∥ x ∥_{2}$ of [L2].

4.1step 3.1step 2.1L3L4

If $∥ E ∥_{2} < σ_{n}$ then step 3.1 gives $∥ (A + E) x ∥_{2} > 0$ 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 $∥ E ∥_{2} < σ_{n}$ makes $A + E$ singular: the distance is at least $σ_{n}$ .

5.1step 1.2L2

The rank-one matrix $u_{n} v_{n}^{T}$ has the singular values $1, 0, \dots, 0$ , so [L2] gives $∥ E^{*} ∥_{2} = σ_{n}$ ; combined with step 4.1, the minimum distance is exactly $σ_{n}$ and is attained, which is claim 2.

6.1step 5.1L1L2L5algebra

By [L5], $∥ A ∥_{2} = σ_{1}$ ; and $A^{- 1} = V Σ^{- 1} U^{T}$ by [L1], a matrix with singular values $1 / σ_{n} \geq \dots \geq 1 / σ_{1}$ , so [L2] gives $∥ A^{- 1} ∥_{2} = 1 / σ_{n}$ . Hence $κ_{2} (A) = ∥ A ∥_{2} ∥ A^{- 1} ∥_{2} = σ_{1} / σ_{n}$ by The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system, and claim 3 follows with claim 2.

7.1step 2.1step 5.1step 6.1∎

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

Depends on

Used by

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

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Sources

James Demmel, Math 221 Lecture 3: Vector Norms, Matrix Norms, and Condition Numbers (standard reference, not scraped)
L. N. Trefethen and D. Bau III, Numerical Linear Algebra, Lecture 12 (standard reference, not scraped)