English
If the initial segment up to n is linearly independent, then the nth Gram-Schmidt vector is nonzero.
Русский
Если в начальном отрезке до n ЛИНЕЙНО НИСХОДИТСЯ независимым, то вектор грам-Schmidt на позиции n ненулевой.
LaTeX
$$$$\text{If } \{ f(i) : i \le n \}\text{ is linearly independent, then } \mathrm{gramSchmidt}_{\mathbb{K}} f(n) \neq 0.$$$$
Lean4
theorem gramSchmidt_ne_zero_coe {f : ι → E} (n : ι) (h₀ : LinearIndependent 𝕜 (f ∘ ((↑) : Set.Iic n → ι))) :
gramSchmidt 𝕜 f n ≠ 0 := by
by_contra h
have h₁ : f n ∈ span 𝕜 (f '' Set.Iio n) :=
by
rw [← span_gramSchmidt_Iio 𝕜 f n, gramSchmidt_def' 𝕜 f, h, zero_add]
apply Submodule.sum_mem _ _
intro a ha
simp only [starProjection_singleton]
apply Submodule.smul_mem _ _ _
rw [Finset.mem_Iio] at ha
exact subset_span ⟨a, ha, by rfl⟩
have h₂ : (f ∘ ((↑) : Set.Iic n → ι)) ⟨n, le_refl n⟩ ∈ span 𝕜 (f ∘ ((↑) : Set.Iic n → ι) '' Set.Iio ⟨n, le_refl n⟩) :=
by
rw [image_comp]
simpa using h₁
apply LinearIndependent.notMem_span_image h₀ _ h₂
simp only [Set.mem_Iio, lt_self_iff_false, not_false_iff]
