Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29
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Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1

Statement

Let p∈Q with p≥1 and let m,n,q∈N, with the induced p-norms of The matrix norm induced by a published vector p-norm.

  1. Compatibility. For every A∈Mm×n(R) and every x∈Rn, ∥Ax∥p  ≤  ∥A∥p ∥x∥p.
  2. Submultiplicativity. For every A∈Mm×n(R) and B∈Mn×q(R), ∥AB∥p  ≤  ∥A∥p ∥B∥p.
  3. Normalisation. For every n≥1, ∥In∥p=1, where In is the identity matrix of Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes; at n=0 the unique empty matrix I0∈M0(R) has ∥I0∥p=0, the convention of The matrix norm induced by a published vector p-norm.

Facts & Assumptions

Given: A rational p≥1, natural numbers m,n,q, matrices A∈Mm×n(R), B∈Mn×q(R), and vectors x∈Rn.

[L1]

∥⋅∥p is a norm on Rn and on Rm: in particular ∥0∥p=0 and ∥λy∥p=∣λ∣∥y∥p (Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page).

[L2]

The induced norm is ∥A∥p=sup⁡{∥Ay∥p:∥y∥p≤1} (The matrix norm induced by a published vector p-norm).

Proof

technique · direct
1.1L1L2algebra

For x=0 the compatibility claim reads 0≤∥A∥p⋅0, which is true by [L1] and [L2].

1.2L1algebra

For x≠0 the vector ∥x∥p>0 by [L1], so ∥Ax∥p/∥x∥p=∥A(x/∥x∥p)∥p by homogeneity of [L1].

1.3L1L3algebra

For x=0 both sides of claim 2 are 0, using [L1] and [L3].

1.4L2L3algebra

For n≥1 every x≠0 attains ratio one: by [L3], ∥Inx∥p=∥x∥p, so ∥Inx∥p/∥x∥p=1, and the supremum in [L2] is therefore 1, which is claim 3 for n≥1.

1.5L2algebra

At n=0 the space R0 has only the zero vector, so the definition of The matrix norm induced by a published vector p-norm assigns ∥I0∥p=0, which is the stated exceptional value.

2.1step 1.2L2algebra

The scaled vector x/∥x∥p has p-norm 1, so ∥A(x/∥x∥p)∥p≤∥A∥p by [L2]; combining step 1.2 with this bound gives ∥Ax∥p≤∥A∥p∥x∥p, which with step 1.1 proves claim 1.

3.1step 2.1L3algebra

For x≠0 associativity of [L3] gives ∥(AB)x∥p=∥A(Bx)∥p, and applying claim 1 first to A at the vector Bx and then to B at the vector x gives ∥A(Bx)∥p≤∥A∥p∥Bx∥p≤∥A∥p∥B∥p∥x∥p.

4.1step 3.1L1L2algebra

Dividing step 3.1 by the positive number ∥x∥p and taking the supremum over all nonzero x in the definition of [L2] gives ∥AB∥p≤∥A∥p∥B∥p; with step 1.3 this is claim 2.

5.1step 2.1step 4.1step 1.4step 1.5∎

Claims 1, 2 and 3 are steps 2.1, 4.1, 1.4 and 1.5 respectively.

Depends on

Used by

Dependency tree · two levels

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Sources