LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

How 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.

Column vectorization is an isometry and represents the Sylvester map by a Kronecker sum

Statement

Let $F$ be $R$ or $C$ , let $m, n \geq 1$ , let $A \in F^{m \times m}$ , $B \in F^{n \times n}$ , and let $X = [x_{1} \dots x_{n}] \in F^{m \times n}$ . Define $vec (X)$ by stacking the columns $x_{1}, \dots, x_{n}$ . For compatible matrices, define $P \otimes Q$ as the block matrix whose $(i, j)$ block is $p_{i j} Q$ . Then

$∥ vec (X) ∥_{2} = ∥ X ∥_{F}, vec (A X + X B) = (I_{n} \otimes A + B^{T} \otimes I_{m}) vec (X) .$

Facts & Assumptions

Given: The dimensions and matrices in the statement.

[F1]

The Frobenius norm $∥ A ∥_{F} = (\sum_{i, j} ∣ a_{i j} ∣^{2})^{1 / 2}$ on real or complex matrices: $∥ X ∥_{F}^{2} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} ∣ x_{i j} ∣^{2}$ .

Proof

1.1givenF1algebra

By the definition of column stacking, $∥ vec (X) ∥_{2}^{2} = \sum_{j = 1}^{n} \sum_{i = 1}^{m} ∣ x_{i j} ∣^{2} = ∥ X ∥_{F}^{2}$ by [F1]. Both sides are nonnegative, so taking square roots proves the isometry claim.

1.2givenalgebra

The $j$ th block of $(I_{n} \otimes A) vec (X)$ is $A x_{j}$ , which is the $j$ th column of $A X$ . Hence $vec (A X) = (I_{n} \otimes A) vec (X)$ .

1.3givenalgebra

The $j$ th column of $X B$ is $\sum_{ℓ = 1}^{n} b_{ℓ j} x_{ℓ}$ . The $j$ th block of $(B^{T} \otimes I_{m}) vec (X)$ is the same sum because its $(j, ℓ)$ block is $b_{ℓ j} I_{m}$ . Thus $vec (X B) = (B^{T} \otimes I_{m}) vec (X)$ .

2.1step 1.2step 1.3algebra∎

Adding the identities in steps 1.2 and 1.3 gives the displayed Sylvester-map representation, completing both claims.

Depends on

The Frobenius norm $\lVert A\rVert_F=(\sum_{i,j}|a_{ij}|^2)^{1/2}$ on real or complex matrices

Used by

Dependency tree · two levels

6 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

Valeria Simoncini, Computational Methods for Linear Matrix Equations, SIAM Review 58(3), 2016 (standard reference, not scraped)
Elias Jarlebring, Numerical Methods for Lyapunov Equations, lecture notes in numerical linear algebra, 2017 (standard reference, not scraped)