CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$M_n(F)$ is noncommutative for every $n\ge2$

Statement

For every field $F$ and every natural $n\ge2$ , the ring $M_n(F)$ is not commutative.

Facts & Assumptions

Given: A field $F$ and a natural $n\ge2$ .

[L1]

Matrix units satisfy $E_{ij}E_{k\ell}=\delta_{jk}E_{i\ell}$ ( $E_{ij}E_{k\ell}=\delta_{jk}E_{i\ell}$ ).

Proof

technique · direct

1.1

By [L1], $E_{01}E_{10}=E_{00}$ while $E_{10}E_{01}=E_{11}$ ; these products differ at entry $(0,0)$ because $1\ne0$ in a field.

givenL1

2.1

Thus two elements of $M_n(F)$ fail to commute, so the matrix ring is noncommutative.

step 1.1L1∎

Depends on

Used by

Two explicit 2 by 2 matrices do not commute Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 13 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

S. Axler, Linear Algebra Done Right, 4th ed., §3C (standard reference, not scraped)

Mn(F)M_n(F)Mn​(F) is noncommutative for every n≥2n\ge2n≥2