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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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Over an infinite field, a finite linear system has no solution, exactly one solution, or infinitely many solutions according to its pivots

Statement

Let F be an infinite field. A finite system over F has no solutions when its augmented column contains a pivot, exactly one solution when it is consistent and every variable column contains a pivot, and infinitely many solutions when it is consistent and has a nonpivot variable. Consistency is not implied by the pivot condition on the variable columns: over any field the system with matrix (10) and right-hand side (01) has a pivot in its single variable column and also a pivot in its augmented column, and has no solution.

Facts & Assumptions

Given: A finite system over an infinite field F.

[L1]

RREF detects inconsistency and parametrises solutions by arbitrary values of the nonpivot variables (Reduced row echelon form detects consistency and parametrises every solution by the nonpivot variables).

[L2]

A set is finite when it is equinumerous with a natural number (The cardinality ∣A∣ of a finite set).

[L3]

A field supplies addition, multiplication and distinct 0,1, and each nonzero scalar is invertible (Field).

[L4]

An infinite set is one that is not finite; countability is a separate property and is not assumed here (Finite, countably infinite, countable, uncountable).

Proof

technique · direct
1.1

If the augmented column is a pivot column, [L1] gives no solution.

L1L2L3
2.1

If there is no augmented pivot and no free variable, [L1] determines every variable uniquely, so there is exactly one solution.

step 1.1L1
3.1

If a free variable exists, fix all other free variables and let that one range through F. The parametrisation of [L1] assigns distinct solutions to distinct scalars, injecting the infinite set F into the solution set; hence the solution set is not finite and is infinite.

step 2.1L1L4∎

Depends on

Used by

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Sources