English
Over a field K, a matrix M is singular iff there exists a nonzero vector v with M v = 0.
Русский
Пусть K — поле; матрица M сингулярна тогда и только тогда, когда существует ненулевой вектор v с M v = 0.
LaTeX
$$\(\\exists v \\neq 0 \\; (M \\cdot v = 0)) \\iff \\det(M) = 0\\)$$
Lean4
/-- This holds for all integral domains (see `Matrix.exists_mulVec_eq_zero_iff`),
not just fields, but it's easier to prove it for the field of fractions first. -/
theorem exists_mulVec_eq_zero_iff_aux {K : Type*} [DecidableEq n] [Field K] {M : Matrix n n K} :
(∃ v ≠ 0, M *ᵥ v = 0) ↔ M.det = 0 := by
constructor
· rintro ⟨v, hv, mul_eq⟩
contrapose! hv
exact eq_zero_of_mulVec_eq_zero hv mul_eq
· contrapose!
intro h
have : Function.Injective (Matrix.toLin' M) := by
simpa only [← LinearMap.ker_eq_bot, ker_toLin'_eq_bot_iff, not_imp_not] using h
have :
M * LinearMap.toMatrix' ((LinearEquiv.ofInjectiveEndo (Matrix.toLin' M) this).symm : (n → K) →ₗ[K] n → K) = 1 :=
by
refine Matrix.toLin'.injective (LinearMap.ext fun v => ?_)
rw [Matrix.toLin'_mul, Matrix.toLin'_one, Matrix.toLin'_toMatrix', LinearMap.comp_apply]
exact (LinearEquiv.ofInjectiveEndo (Matrix.toLin' M) this).apply_symm_apply v
exact Matrix.det_ne_zero_of_right_inverse this