Documentation

Mathlib.LinearAlgebra.Matrix.InvariantBasisNumber

Invertible matrices over a ring with invariant basis number are square. #

theorem rankCondition_iff_le_of_comp_eq_one {R : Type u_1} [Semiring R] :
RankCondition R โ†” โˆ€ (n m : โ„•) (f : (Fin n โ†’ R) โ†’โ‚—[R] Fin m โ†’ R) (g : (Fin m โ†’ R) โ†’โ‚—[R] Fin n โ†’ R), f โˆ˜โ‚— g = 1 โ†’ m โ‰ค n
theorem rankCondition_iff_matrix {R : Type u_1} [Semiring R] :
RankCondition R โ†” โˆ€ (n m : โ„•) (f : Matrix (Fin n) (Fin m) R) (g : Matrix (Fin m) (Fin n) R), g * f = 1 โ†’ m โ‰ค n
theorem invariantBasisNumber_iff_matrix {R : Type u_1} [Semiring R] :
InvariantBasisNumber R โ†” โˆ€ (n m : โ„•) (f : Matrix (Fin n) (Fin m) R) (g : Matrix (Fin m) (Fin n) R), f * g = 1 โ†’ g * f = 1 โ†’ n = m

The rank condition is left-right symmetric. Note that the strong rank condition is not left-right symmetric, see Remark (1.32) in ยง1.1D of [lam_1999].

theorem Matrix.square_of_invertible {n : Type u_1} {m : Type u_2} [Fintype n] [DecidableEq n] [Fintype m] [DecidableEq m] {R : Type u_3} [Semiring R] [InvariantBasisNumber R] (M : Matrix n m R) (N : Matrix m n R) (h : M * N = 1) (h' : N * M = 1) :
@[instance 100]

Nontrivial commutative semirings R satisfy the rank condition.

If R is moreover a ring, then it satisfies the strong rank condition, see commRing_strongRankCondition. It is unclear whether this generalizes to semirings.