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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 pQ with p1 and let m,n,qN, with the induced p-norms of The matrix norm induced by a published vector p-norm.

  1. Compatibility. For every AMm×n(R) and every xRn, Axp    Apxp.
  2. Submultiplicativity. For every AMm×n(R) and BMn×q(R), ABp    ApBp.
  3. Normalisation. For every n1, Inp=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 I0M0(R) has I0p=0, the convention of The matrix norm induced by a published vector p-norm.

Facts & Assumptions

Given: A rational p1, natural numbers m,n,q, matrices AMm×n(R), BMn×q(R), and vectors xRn.

[L1]

p is a norm on Rn and on Rm: in particular 0p=0 and λyp=λyp (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 Ap=sup{Ayp:yp1} (The matrix norm induced by a published vector p-norm).

Proof

technique · direct
1.1

For x=0 the compatibility claim reads 0Ap0, which is true by [L1] and [L2].

L1L2algebra
1.2

For x0 the vector xp>0 by [L1], so Axp/xp=A(x/xp)p by homogeneity of [L1].

L1algebra
1.3

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

L1L3algebra
1.4

For n1 every x0 attains ratio one: by [L3], Inxp=xp, so Inxp/xp=1, and the supremum in [L2] is therefore 1, which is claim 3 for n1.

L2L3algebra
1.5

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 I0p=0, which is the stated exceptional value.

L2algebra
2.1

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

step 1.2L2algebra
3.1

For x0 associativity of [L3] gives (AB)xp=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)pApBxpApBpxp.

step 2.1L3algebra
4.1

Dividing step 3.1 by the positive number xp and taking the supremum over all nonzero x in the definition of [L2] gives ABpApBp; with step 1.3 this is claim 2.

step 3.1L1L2algebra
5.1

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

step 2.1step 4.1step 1.4step 1.5

Depends on

Used by

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