ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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The rank-one truncation of an SVD realises the Eckart-Young minimiser

Example

For the matrix

$A = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \end{array}),$

the rank-one truncation

$A_{1} = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 0 & 0 \end{array})$

is a best rank-at-most-one approximation in operator norm, and the error is $1$ .

Facts & Assumptions

Given: The matrix $A$ above and its rank-one truncation $A_{1}$ .

[L1]

The matrix $A$ has singular values $2, 1, 0$ (The singular value decomposition of an explicit 2x3 matrix).

[L2]

The rank- $1$ truncation of an SVD is a best rank-at-most- $1$ approximation, with error equal to the next singular value (The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition).

Verification

technique · direct

1.1

By [L1], deleting the second singular direction replaces $A$ by $A_{1} = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 0 & 0 \end{array})$ , which has rank $1$ , and $A - A_{1} = (\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \end{array})$ has operator norm $1$ .

L1algebra

2.1

The next singular value after the retained one is $1$ , so [L2] says exactly that no rank-at-most-one matrix lies closer to $A$ in operator norm than $A_{1}$ . Therefore $A_{1}$ is the Eckart-Young minimiser.

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)