Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2
Statement
For compatible linear maps between finite-dimensional real or complex inner product spaces,
For every such linear map ,
Facts & Assumptions
Given: Compatible finite-dimensional linear maps and between real or complex inner product spaces.
The operator norm is defined by the maximum over unit vectors (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).
The operator norm equals the 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).
If is a singular value decomposition, then (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).
Proof
For every unit vector in the domain of , [L1] gives . Taking the maximum over all unit yields .
If are the singular values of , then [L3] shows that has eigenvalues and is already non-negative. Hence its singular values are , so [L2] gives .
Depends on
- The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||
- The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector
- Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)