LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$

Statement

For every linear $L : R^{m} \to R^{n}$ there is a unique matrix $A = (a_{i j})_{i < n, j < m}$ such that $(L h)_{i} = \sum_{j < m} a_{i j} h_{j}$ . Moreover there is $K \geq 0$ with $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for every $h$ .

Facts & Assumptions

Given: A Euclidean linear map $L : R^{m} \to R^{n}$ .

[L1]

The coordinate list of $x \in F^{n}$ with respect to the ordered basis $e$ is its ordinary coordinate list, and $x = \sum_{j < n} x_{j} e_{j}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

The Euclidean norm of $x$ is $∥ x ∥_{2} = \sqrt{\sum_{i < n} x_{i}^{2}}$ , and it is a norm (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L3]

Cauchy--Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

technique · direct

1.1L1L2

Put $a_{i j} : = (L e_{j})_{i}$ . By [L1] and linearity, $L h = L (\sum_{j < m} h_{j} e_{j}) = \sum_{j < m} h_{j} L e_{j}$ , so $(L h)_{i} = \sum_{j < m} a_{i j} h_{j}$ .

2.1step 1.1L2algebra

The columns $L e_{j}$ determine every value in step 1.1, and evaluating the displayed formula at $e_{j}$ shows that every representing matrix has exactly these entries; thus the matrix is unique.

3.1step 1.1step 2.1L3∎

Let $K = (\sum_{i < n, j < m} a_{i j}^{2})^{1 / 2}$ . Cauchy--Schwarz [L3] in each row and summing gives $∥ L h ∥_{2}^{2} \leq K^{2} ∥ h ∥_{2}^{2}$ , hence $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ .

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis I, §8.3