Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2\|Lh\|_2\le K\|h\|_2 for some K0K\ge0

Statement

For every linear L:RmRnL:\mathbb R^m\to\mathbb R^n there is a unique matrix A=(aij)i<n,j<mA=(a_{ij})_{i<n,j<m} such that (Lh)i=j<maijhj(Lh)_i=\sum_{j<m}a_{ij}h_j. Moreover there is K0K\ge0 with Lh2Kh2\|Lh\|_2\le K\|h\|_2 for every hh.

Facts & Assumptions

Given: A Euclidean linear map L:RmRnL:\mathbb R^m\to\mathbb R^n.

[L2]

The Euclidean norm of xx is x2=i<nxi2\|x\|_2=\sqrt{\sum_{i<n}x_i^2}, and it is a norm (The Euclidean inner product x,y=k<nxkyk\langle x,y\rangle = \sum_{k<n} x_k y_k on Rn\mathbb{R}^n).

Proof

technique · direct
1.1

Put aij:=(Lej)ia_{ij}:=(L e_j)_i. By [L1] and linearity, Lh=L(j<mhjej)=j<mhjLejLh=L(\sum_{j<m}h_je_j)=\sum_{j<m}h_jLe_j, so (Lh)i=j<maijhj(Lh)_i=\sum_{j<m}a_{ij}h_j.

L1L2
2.1

The columns LejLe_j determine every value in step 1.1, and evaluating the displayed formula at eje_j shows that every representing matrix has exactly these entries; thus the matrix is unique.

step 1.1L2algebra
3.1

Let K=(i<n,j<maij2)1/2K=(\sum_{i<n,j<m}a_{ij}^2)^{1/2}. Cauchy--Schwarz [L3] in each row and summing gives Lh22K2h22\|Lh\|_2^2\le K^2\|h\|_2^2, hence Lh2Kh2\|Lh\|_2\le K\|h\|_2.

step 1.1step 2.1L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 113 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources