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Theorem List for Metamath Proof Explorer - 19801-19900   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremreslmhm2b 19801 Expansion of the codomain of a homomorphism. (Contributed by Stefan O'Rear, 3-Feb-2015.) (Revised by Mario Carneiro, 5-May-2015.)
𝑈 = (𝑇s 𝑋)    &   𝐿 = (LSubSp‘𝑇)       ((𝑇 ∈ LMod ∧ 𝑋𝐿 ∧ ran 𝐹𝑋) → (𝐹 ∈ (𝑆 LMHom 𝑇) ↔ 𝐹 ∈ (𝑆 LMHom 𝑈)))

Theoremlmhmeql 19802 The equalizer of two module homomorphisms is a subspace. (Contributed by Stefan O'Rear, 7-Mar-2015.)
𝑈 = (LSubSp‘𝑆)       ((𝐹 ∈ (𝑆 LMHom 𝑇) ∧ 𝐺 ∈ (𝑆 LMHom 𝑇)) → dom (𝐹𝐺) ∈ 𝑈)

Theoremlspextmo 19803* A linear function is completely determined (or overdetermined) by its values on a spanning subset. (Contributed by Stefan O'Rear, 7-Mar-2015.) (Revised by NM, 17-Jun-2017.)
𝐵 = (Base‘𝑆)    &   𝐾 = (LSpan‘𝑆)       ((𝑋𝐵 ∧ (𝐾𝑋) = 𝐵) → ∃*𝑔 ∈ (𝑆 LMHom 𝑇)(𝑔𝑋) = 𝐹)

Theorempwsdiaglmhm 19804* Diagonal homomorphism into a structure power. (Contributed by Stefan O'Rear, 24-Jan-2015.)
𝑌 = (𝑅s 𝐼)    &   𝐵 = (Base‘𝑅)    &   𝐹 = (𝑥𝐵 ↦ (𝐼 × {𝑥}))       ((𝑅 ∈ LMod ∧ 𝐼𝑊) → 𝐹 ∈ (𝑅 LMHom 𝑌))

Theorempwssplit0 19805* Splitting for structure powers, part 0: restriction is a function. (Contributed by Stefan O'Rear, 24-Jan-2015.)
𝑌 = (𝑊s 𝑈)    &   𝑍 = (𝑊s 𝑉)    &   𝐵 = (Base‘𝑌)    &   𝐶 = (Base‘𝑍)    &   𝐹 = (𝑥𝐵 ↦ (𝑥𝑉))       ((𝑊𝑇𝑈𝑋𝑉𝑈) → 𝐹:𝐵𝐶)

Theorempwssplit1 19806* Splitting for structure powers, part 1: restriction is an onto function. The only actual monoid law we need here is that the base set is nonempty. (Contributed by Stefan O'Rear, 24-Jan-2015.)
𝑌 = (𝑊s 𝑈)    &   𝑍 = (𝑊s 𝑉)    &   𝐵 = (Base‘𝑌)    &   𝐶 = (Base‘𝑍)    &   𝐹 = (𝑥𝐵 ↦ (𝑥𝑉))       ((𝑊 ∈ Mnd ∧ 𝑈𝑋𝑉𝑈) → 𝐹:𝐵onto𝐶)

Theorempwssplit2 19807* Splitting for structure powers, part 2: restriction is a group homomorphism. (Contributed by Stefan O'Rear, 24-Jan-2015.)
𝑌 = (𝑊s 𝑈)    &   𝑍 = (𝑊s 𝑉)    &   𝐵 = (Base‘𝑌)    &   𝐶 = (Base‘𝑍)    &   𝐹 = (𝑥𝐵 ↦ (𝑥𝑉))       ((𝑊 ∈ Grp ∧ 𝑈𝑋𝑉𝑈) → 𝐹 ∈ (𝑌 GrpHom 𝑍))

Theorempwssplit3 19808* Splitting for structure powers, part 3: restriction is a module homomorphism. (Contributed by Stefan O'Rear, 24-Jan-2015.)
𝑌 = (𝑊s 𝑈)    &   𝑍 = (𝑊s 𝑉)    &   𝐵 = (Base‘𝑌)    &   𝐶 = (Base‘𝑍)    &   𝐹 = (𝑥𝐵 ↦ (𝑥𝑉))       ((𝑊 ∈ LMod ∧ 𝑈𝑋𝑉𝑈) → 𝐹 ∈ (𝑌 LMHom 𝑍))

Theoremislmim 19809 An isomorphism of left modules is a bijective homomorphism. (Contributed by Stefan O'Rear, 21-Jan-2015.)
𝐵 = (Base‘𝑅)    &   𝐶 = (Base‘𝑆)       (𝐹 ∈ (𝑅 LMIso 𝑆) ↔ (𝐹 ∈ (𝑅 LMHom 𝑆) ∧ 𝐹:𝐵1-1-onto𝐶))

Theoremlmimf1o 19810 An isomorphism of left modules is a bijection. (Contributed by Stefan O'Rear, 21-Jan-2015.)
𝐵 = (Base‘𝑅)    &   𝐶 = (Base‘𝑆)       (𝐹 ∈ (𝑅 LMIso 𝑆) → 𝐹:𝐵1-1-onto𝐶)

Theoremlmimlmhm 19811 An isomorphism of modules is a homomorphism. (Contributed by Stefan O'Rear, 21-Jan-2015.)
(𝐹 ∈ (𝑅 LMIso 𝑆) → 𝐹 ∈ (𝑅 LMHom 𝑆))

Theoremlmimgim 19812 An isomorphism of modules is an isomorphism of groups. (Contributed by Stefan O'Rear, 21-Jan-2015.) (Revised by Mario Carneiro, 6-May-2015.)
(𝐹 ∈ (𝑅 LMIso 𝑆) → 𝐹 ∈ (𝑅 GrpIso 𝑆))

Theoremislmim2 19813 An isomorphism of left modules is a homomorphism whose converse is a homomorphism. (Contributed by Mario Carneiro, 6-May-2015.)
(𝐹 ∈ (𝑅 LMIso 𝑆) ↔ (𝐹 ∈ (𝑅 LMHom 𝑆) ∧ 𝐹 ∈ (𝑆 LMHom 𝑅)))

Theoremlmimcnv 19814 The converse of a bijective module homomorphism is a bijective module homomorphism. (Contributed by Stefan O'Rear, 25-Jan-2015.) (Revised by Mario Carneiro, 6-May-2015.)
(𝐹 ∈ (𝑆 LMIso 𝑇) → 𝐹 ∈ (𝑇 LMIso 𝑆))

Theorembrlmic 19815 The relation "is isomorphic to" for modules. (Contributed by Stefan O'Rear, 25-Jan-2015.)
(𝑅𝑚 𝑆 ↔ (𝑅 LMIso 𝑆) ≠ ∅)

Theorembrlmici 19816 Prove isomorphic by an explicit isomorphism. (Contributed by Stefan O'Rear, 25-Jan-2015.)
(𝐹 ∈ (𝑅 LMIso 𝑆) → 𝑅𝑚 𝑆)

Theoremlmiclcl 19817 Isomorphism implies the left side is a module. (Contributed by Stefan O'Rear, 25-Jan-2015.)
(𝑅𝑚 𝑆𝑅 ∈ LMod)

Theoremlmicrcl 19818 Isomorphism implies the right side is a module. (Contributed by Mario Carneiro, 6-May-2015.)
(𝑅𝑚 𝑆𝑆 ∈ LMod)

Theoremlmicsym 19819 Module isomorphism is symmetric. (Contributed by Stefan O'Rear, 26-Feb-2015.)
(𝑅𝑚 𝑆𝑆𝑚 𝑅)

Theoremlmhmpropd 19820* Module homomorphism depends only on the module attributes of structures. (Contributed by Mario Carneiro, 8-Oct-2015.)
(𝜑𝐵 = (Base‘𝐽))    &   (𝜑𝐶 = (Base‘𝐾))    &   (𝜑𝐵 = (Base‘𝐿))    &   (𝜑𝐶 = (Base‘𝑀))    &   (𝜑𝐹 = (Scalar‘𝐽))    &   (𝜑𝐺 = (Scalar‘𝐾))    &   (𝜑𝐹 = (Scalar‘𝐿))    &   (𝜑𝐺 = (Scalar‘𝑀))    &   𝑃 = (Base‘𝐹)    &   𝑄 = (Base‘𝐺)    &   ((𝜑 ∧ (𝑥𝐵𝑦𝐵)) → (𝑥(+g𝐽)𝑦) = (𝑥(+g𝐿)𝑦))    &   ((𝜑 ∧ (𝑥𝐶𝑦𝐶)) → (𝑥(+g𝐾)𝑦) = (𝑥(+g𝑀)𝑦))    &   ((𝜑 ∧ (𝑥𝑃𝑦𝐵)) → (𝑥( ·𝑠𝐽)𝑦) = (𝑥( ·𝑠𝐿)𝑦))    &   ((𝜑 ∧ (𝑥𝑄𝑦𝐶)) → (𝑥( ·𝑠𝐾)𝑦) = (𝑥( ·𝑠𝑀)𝑦))       (𝜑 → (𝐽 LMHom 𝐾) = (𝐿 LMHom 𝑀))

10.5.4  Subspace sum; bases for a left module

Syntaxclbs 19821 Extend class notation with the set of bases for a vector space.
class LBasis

Definitiondf-lbs 19822* Define the set of bases to a left module or left vector space. (Contributed by Mario Carneiro, 24-Jun-2014.)
LBasis = (𝑤 ∈ V ↦ {𝑏 ∈ 𝒫 (Base‘𝑤) ∣ [(LSpan‘𝑤) / 𝑛][(Scalar‘𝑤) / 𝑠]((𝑛𝑏) = (Base‘𝑤) ∧ ∀𝑥𝑏𝑦 ∈ ((Base‘𝑠) ∖ {(0g𝑠)}) ¬ (𝑦( ·𝑠𝑤)𝑥) ∈ (𝑛‘(𝑏 ∖ {𝑥})))})

Theoremislbs 19823* The predicate "𝐵 is a basis for the left module or vector space 𝑊". A subset of the base set is a basis if zero is not in the set, it spans the set, and no nonzero multiple of an element of the basis is in the span of the rest of the family. (Contributed by Mario Carneiro, 24-Jun-2014.) (Revised by Mario Carneiro, 14-Jan-2015.)
𝑉 = (Base‘𝑊)    &   𝐹 = (Scalar‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐾 = (Base‘𝐹)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    0 = (0g𝐹)       (𝑊𝑋 → (𝐵𝐽 ↔ (𝐵𝑉 ∧ (𝑁𝐵) = 𝑉 ∧ ∀𝑥𝐵𝑦 ∈ (𝐾 ∖ { 0 }) ¬ (𝑦 · 𝑥) ∈ (𝑁‘(𝐵 ∖ {𝑥})))))

Theoremlbsss 19824 A basis is a set of vectors. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)       (𝐵𝐽𝐵𝑉)

Theoremlbsel 19825 An element of a basis is a vector. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)       ((𝐵𝐽𝐸𝐵) → 𝐸𝑉)

Theoremlbssp 19826 The span of a basis is the whole space. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)       (𝐵𝐽 → (𝑁𝐵) = 𝑉)

Theoremlbsind 19827 A basis is linearly independent; that is, every element has a span which trivially intersects the span of the remainder of the basis. (Contributed by Mario Carneiro, 12-Jan-2015.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   𝐹 = (Scalar‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐾 = (Base‘𝐹)    &    0 = (0g𝐹)       (((𝐵𝐽𝐸𝐵) ∧ (𝐴𝐾𝐴0 )) → ¬ (𝐴 · 𝐸) ∈ (𝑁‘(𝐵 ∖ {𝐸})))

Theoremlbsind2 19828 A basis is linearly independent; that is, every element is not in the span of the remainder of the basis. (Contributed by Mario Carneiro, 25-Jun-2014.) (Revised by Mario Carneiro, 12-Jan-2015.)
𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   𝐹 = (Scalar‘𝑊)    &    1 = (1r𝐹)    &    0 = (0g𝐹)       (((𝑊 ∈ LMod ∧ 10 ) ∧ 𝐵𝐽𝐸𝐵) → ¬ 𝐸 ∈ (𝑁‘(𝐵 ∖ {𝐸})))

Theoremlbspss 19829 No proper subset of a basis spans the space. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   𝐹 = (Scalar‘𝑊)    &    1 = (1r𝐹)    &    0 = (0g𝐹)    &   𝑉 = (Base‘𝑊)       (((𝑊 ∈ LMod ∧ 10 ) ∧ 𝐵𝐽𝐶𝐵) → (𝑁𝐶) ≠ 𝑉)

Theoremlsmcl 19830 The sum of two subspaces is a subspace. (Contributed by NM, 4-Feb-2014.) (Revised by Mario Carneiro, 19-Apr-2016.)
𝑆 = (LSubSp‘𝑊)    &    = (LSSum‘𝑊)       ((𝑊 ∈ LMod ∧ 𝑇𝑆𝑈𝑆) → (𝑇 𝑈) ∈ 𝑆)

Theoremlsmspsn 19831* Member of subspace sum of spans of singletons. (Contributed by NM, 8-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    · = ( ·𝑠𝑊)    &    = (LSSum‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → (𝑈 ∈ ((𝑁‘{𝑋}) (𝑁‘{𝑌})) ↔ ∃𝑗𝐾𝑘𝐾 𝑈 = ((𝑗 · 𝑋) + (𝑘 · 𝑌))))

Theoremlsmelval2 19832* Subspace sum membership in terms of a sum of 1-dim subspaces (atoms), which can be useful for treating subspaces as projective lattice elements. (Contributed by NM, 9-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &    = (LSSum‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑇𝑆)    &   (𝜑𝑈𝑆)       (𝜑 → (𝑋 ∈ (𝑇 𝑈) ↔ (𝑋𝑉 ∧ ∃𝑦𝑇𝑧𝑈 (𝑁‘{𝑋}) ⊆ ((𝑁‘{𝑦}) (𝑁‘{𝑧})))))

Theoremlsmsp 19833 Subspace sum in terms of span. (Contributed by NM, 6-Feb-2014.) (Proof shortened by Mario Carneiro, 21-Jun-2014.)
𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)       ((𝑊 ∈ LMod ∧ 𝑇𝑆𝑈𝑆) → (𝑇 𝑈) = (𝑁‘(𝑇𝑈)))

Theoremlsmsp2 19834 Subspace sum of spans of subsets is the span of their union. (spanuni 29305 analog.) (Contributed by NM, 22-Feb-2014.) (Revised by Mario Carneiro, 21-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)       ((𝑊 ∈ LMod ∧ 𝑇𝑉𝑈𝑉) → ((𝑁𝑇) (𝑁𝑈)) = (𝑁‘(𝑇𝑈)))

Theoremlsmssspx 19835 Subspace sum (in its extended domain) is a subset of the span of the union of its arguments. (Contributed by NM, 6-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)    &   (𝜑𝑇𝑉)    &   (𝜑𝑈𝑉)    &   (𝜑𝑊 ∈ LMod)       (𝜑 → (𝑇 𝑈) ⊆ (𝑁‘(𝑇𝑈)))

Theoremlsmpr 19836 The span of a pair of vectors equals the sum of the spans of their singletons. (Contributed by NM, 13-Jan-2015.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → (𝑁‘{𝑋, 𝑌}) = ((𝑁‘{𝑋}) (𝑁‘{𝑌})))

Theoremlsppreli 19837 A vector expressed as a sum belongs to the span of its components. (Contributed by NM, 9-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝐴𝐾)    &   (𝜑𝐵𝐾)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → ((𝐴 · 𝑋) + (𝐵 · 𝑌)) ∈ (𝑁‘{𝑋, 𝑌}))

Theoremlsmelpr 19838 Two ways to say that a vector belongs to the span of a pair of vectors. (Contributed by NM, 14-Jan-2015.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)       (𝜑 → (𝑋 ∈ (𝑁‘{𝑌, 𝑍}) ↔ (𝑁‘{𝑋}) ⊆ ((𝑁‘{𝑌}) (𝑁‘{𝑍}))))

Theoremlsppr0 19839 The span of a vector paired with zero equals the span of the singleton of the vector. (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)       (𝜑 → (𝑁‘{𝑋, 0 }) = (𝑁‘{𝑋}))

Theoremlsppr 19840* Span of a pair of vectors. (Contributed by NM, 22-Aug-2014.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    · = ( ·𝑠𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → (𝑁‘{𝑋, 𝑌}) = {𝑣 ∣ ∃𝑘𝐾𝑙𝐾 𝑣 = ((𝑘 · 𝑋) + (𝑙 · 𝑌))})

Theoremlspprel 19841* Member of the span of a pair of vectors. (Contributed by NM, 10-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    · = ( ·𝑠𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → (𝑍 ∈ (𝑁‘{𝑋, 𝑌}) ↔ ∃𝑘𝐾𝑙𝐾 𝑍 = ((𝑘 · 𝑋) + (𝑙 · 𝑌))))

Theoremlspprabs 19842 Absorption of vector sum into span of pair. (Contributed by NM, 27-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → (𝑁‘{𝑋, (𝑋 + 𝑌)}) = (𝑁‘{𝑋, 𝑌}))

Theoremlspvadd 19843 The span of a vector sum is included in the span of its arguments. (Contributed by NM, 22-Feb-2014.) (Proof shortened by Mario Carneiro, 21-Jun-2014.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝑁 = (LSpan‘𝑊)       ((𝑊 ∈ LMod ∧ 𝑋𝑉𝑌𝑉) → (𝑁‘{(𝑋 + 𝑌)}) ⊆ (𝑁‘{𝑋, 𝑌}))

Theoremlspsntri 19844 Triangle-type inequality for span of a singleton. (Contributed by NM, 24-Feb-2014.) (Revised by Mario Carneiro, 21-Jun-2014.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)       ((𝑊 ∈ LMod ∧ 𝑋𝑉𝑌𝑉) → (𝑁‘{(𝑋 + 𝑌)}) ⊆ ((𝑁‘{𝑋}) (𝑁‘{𝑌})))

Theoremlspsntrim 19845 Triangle-type inequality for span of a singleton of vector difference. (Contributed by NM, 25-Apr-2014.) (Revised by Mario Carneiro, 21-Jun-2014.)
𝑉 = (Base‘𝑊)    &    = (-g𝑊)    &    = (LSSum‘𝑊)    &   𝑁 = (LSpan‘𝑊)       ((𝑊 ∈ LMod ∧ 𝑋𝑉𝑌𝑉) → (𝑁‘{(𝑋 𝑌)}) ⊆ ((𝑁‘{𝑋}) (𝑁‘{𝑌})))

Theoremlbspropd 19846* If two structures have the same components (properties), they have the same set of bases. (Contributed by Mario Carneiro, 9-Feb-2015.) (Revised by Mario Carneiro, 14-Jun-2015.) (Revised by AV, 24-Apr-2024.)
(𝜑𝐵 = (Base‘𝐾))    &   (𝜑𝐵 = (Base‘𝐿))    &   (𝜑𝐵𝑊)    &   ((𝜑 ∧ (𝑥𝑊𝑦𝑊)) → (𝑥(+g𝐾)𝑦) = (𝑥(+g𝐿)𝑦))    &   ((𝜑 ∧ (𝑥𝑃𝑦𝐵)) → (𝑥( ·𝑠𝐾)𝑦) ∈ 𝑊)    &   ((𝜑 ∧ (𝑥𝑃𝑦𝐵)) → (𝑥( ·𝑠𝐾)𝑦) = (𝑥( ·𝑠𝐿)𝑦))    &   𝐹 = (Scalar‘𝐾)    &   𝐺 = (Scalar‘𝐿)    &   (𝜑𝑃 = (Base‘𝐹))    &   (𝜑𝑃 = (Base‘𝐺))    &   ((𝜑 ∧ (𝑥𝑃𝑦𝑃)) → (𝑥(+g𝐹)𝑦) = (𝑥(+g𝐺)𝑦))    &   (𝜑𝐾𝑋)    &   (𝜑𝐿𝑌)       (𝜑 → (LBasis‘𝐾) = (LBasis‘𝐿))

Theorempj1lmhm 19847 The left projection function is a linear operator. (Contributed by Mario Carneiro, 15-Oct-2015.) (Revised by Mario Carneiro, 21-Apr-2016.)
𝐿 = (LSubSp‘𝑊)    &    = (LSSum‘𝑊)    &    0 = (0g𝑊)    &   𝑃 = (proj1𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑇𝐿)    &   (𝜑𝑈𝐿)    &   (𝜑 → (𝑇𝑈) = { 0 })       (𝜑 → (𝑇𝑃𝑈) ∈ ((𝑊s (𝑇 𝑈)) LMHom 𝑊))

Theorempj1lmhm2 19848 The left projection function is a linear operator. (Contributed by Mario Carneiro, 15-Oct-2015.) (Revised by Mario Carneiro, 21-Apr-2016.)
𝐿 = (LSubSp‘𝑊)    &    = (LSSum‘𝑊)    &    0 = (0g𝑊)    &   𝑃 = (proj1𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑇𝐿)    &   (𝜑𝑈𝐿)    &   (𝜑 → (𝑇𝑈) = { 0 })       (𝜑 → (𝑇𝑃𝑈) ∈ ((𝑊s (𝑇 𝑈)) LMHom (𝑊s 𝑇)))

10.6  Vector spaces

10.6.1  Definition and basic properties

Syntaxclvec 19849 Extend class notation with class of all left vector spaces.
class LVec

Definitiondf-lvec 19850 Define the class of all left vector spaces. A left vector space over a division ring is an Abelian group (vectors) together with a division ring (scalars) and a left scalar product connecting them. Some authors call this a "left module over a division ring", reserving "vector space" for those where the division ring is commutative, i.e., is a field. (Contributed by NM, 11-Nov-2013.)
LVec = {𝑓 ∈ LMod ∣ (Scalar‘𝑓) ∈ DivRing}

Theoremislvec 19851 The predicate "is a left vector space". (Contributed by NM, 11-Nov-2013.)
𝐹 = (Scalar‘𝑊)       (𝑊 ∈ LVec ↔ (𝑊 ∈ LMod ∧ 𝐹 ∈ DivRing))

Theoremlvecdrng 19852 The set of scalars of a left vector space is a division ring. (Contributed by NM, 17-Apr-2014.)
𝐹 = (Scalar‘𝑊)       (𝑊 ∈ LVec → 𝐹 ∈ DivRing)

Theoremlveclmod 19853 A left vector space is a left module. (Contributed by NM, 9-Dec-2013.)
(𝑊 ∈ LVec → 𝑊 ∈ LMod)

Theoremlsslvec 19854 A vector subspace is a vector space. (Contributed by NM, 14-Mar-2015.)
𝑋 = (𝑊s 𝑈)    &   𝑆 = (LSubSp‘𝑊)       ((𝑊 ∈ LVec ∧ 𝑈𝑆) → 𝑋 ∈ LVec)

Theoremlvecvs0or 19855 If a scalar product is zero, one of its factors must be zero. (hvmul0or 28786 analog.) (Contributed by NM, 2-Jul-2014.)
𝑉 = (Base‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &   𝑂 = (0g𝐹)    &    0 = (0g𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐴𝐾)    &   (𝜑𝑋𝑉)       (𝜑 → ((𝐴 · 𝑋) = 0 ↔ (𝐴 = 𝑂𝑋 = 0 )))

Theoremlvecvsn0 19856 A scalar product is nonzero iff both of its factors are nonzero. (Contributed by NM, 3-Jan-2015.)
𝑉 = (Base‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &   𝑂 = (0g𝐹)    &    0 = (0g𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐴𝐾)    &   (𝜑𝑋𝑉)       (𝜑 → ((𝐴 · 𝑋) ≠ 0 ↔ (𝐴𝑂𝑋0 )))

Theoremlssvs0or 19857 If a scalar product belongs to a subspace, either the scalar component is zero or the vector component also belongs to the subspace. (Contributed by NM, 5-Apr-2015.)
𝑉 = (Base‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    0 = (0g𝐹)    &   𝑆 = (LSubSp‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝐴𝐾)       (𝜑 → ((𝐴 · 𝑋) ∈ 𝑈 ↔ (𝐴 = 0𝑋𝑈)))

Theoremlvecvscan 19858 Cancellation law for scalar multiplication. (hvmulcan 28833 analog.) (Contributed by NM, 2-Jul-2014.)
𝑉 = (Base‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    0 = (0g𝐹)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐴𝐾)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝐴0 )       (𝜑 → ((𝐴 · 𝑋) = (𝐴 · 𝑌) ↔ 𝑋 = 𝑌))

Theoremlvecvscan2 19859 Cancellation law for scalar multiplication. (hvmulcan2 28834 analog.) (Contributed by NM, 2-Jul-2014.)
𝑉 = (Base‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    0 = (0g𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐴𝐾)    &   (𝜑𝐵𝐾)    &   (𝜑𝑋𝑉)    &   (𝜑𝑋0 )       (𝜑 → ((𝐴 · 𝑋) = (𝐵 · 𝑋) ↔ 𝐴 = 𝐵))

Theoremlvecinv 19860 Invert coefficient of scalar product. (Contributed by NM, 11-Apr-2015.)
𝑉 = (Base‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    0 = (0g𝐹)    &   𝐼 = (invr𝐹)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐴 ∈ (𝐾 ∖ { 0 }))    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → (𝑋 = (𝐴 · 𝑌) ↔ 𝑌 = ((𝐼𝐴) · 𝑋)))

Theoremlspsnvs 19861 A nonzero scalar product does not change the span of a singleton. (spansncol 29329 analog.) (Contributed by NM, 23-Apr-2014.)
𝑉 = (Base‘𝑊)    &   𝐹 = (Scalar‘𝑊)    &    · = ( ·𝑠𝑊)    &   𝐾 = (Base‘𝐹)    &    0 = (0g𝐹)    &   𝑁 = (LSpan‘𝑊)       ((𝑊 ∈ LVec ∧ (𝑅𝐾𝑅0 ) ∧ 𝑋𝑉) → (𝑁‘{(𝑅 · 𝑋)}) = (𝑁‘{𝑋}))

Theoremlspsneleq 19862 Membership relation that implies equality of spans. (spansneleq 29331 analog.) (Contributed by NM, 4-Jul-2014.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑁‘{𝑋}))    &   (𝜑𝑌0 )       (𝜑 → (𝑁‘{𝑌}) = (𝑁‘{𝑋}))

Theoremlspsncmp 19863 Comparable spans of nonzero singletons are equal. (Contributed by NM, 27-Apr-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑌𝑉)       (𝜑 → ((𝑁‘{𝑋}) ⊆ (𝑁‘{𝑌}) ↔ (𝑁‘{𝑋}) = (𝑁‘{𝑌})))

Theoremlspsnne1 19864 Two ways to express that vectors have different spans. (Contributed by NM, 28-May-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑌𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))       (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌}))

Theoremlspsnne2 19865 Two ways to express that vectors have different spans. (Contributed by NM, 20-May-2015.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌}))       (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))

Theoremlspsnnecom 19866 Swap two vectors with different spans. (Contributed by NM, 20-May-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑉 ∖ { 0 }))    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌}))       (𝜑 → ¬ 𝑌 ∈ (𝑁‘{𝑋}))

Theoremlspabs2 19867 Absorption law for span of vector sum. (Contributed by NM, 30-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑉 ∖ { 0 }))    &   (𝜑 → (𝑁‘{𝑋}) = (𝑁‘{(𝑋 + 𝑌)}))       (𝜑 → (𝑁‘{𝑋}) = (𝑁‘{𝑌}))

Theoremlspabs3 19868 Absorption law for span of vector sum. (Contributed by NM, 30-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑 → (𝑋 + 𝑌) ≠ 0 )    &   (𝜑 → (𝑁‘{𝑋}) = (𝑁‘{𝑌}))       (𝜑 → (𝑁‘{𝑋}) = (𝑁‘{(𝑋 + 𝑌)}))

Theoremlspsneq 19869* Equal spans of singletons must have proportional vectors. See lspsnss2 19752 for comparable span version. TODO: can proof be shortened? (Contributed by NM, 21-Mar-2015.)
𝑉 = (Base‘𝑊)    &   𝑆 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝑆)    &    0 = (0g𝑆)    &    · = ( ·𝑠𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)       (𝜑 → ((𝑁‘{𝑋}) = (𝑁‘{𝑌}) ↔ ∃𝑘 ∈ (𝐾 ∖ { 0 })𝑋 = (𝑘 · 𝑌)))

Theoremlspsneu 19870* Nonzero vectors with equal singleton spans have a unique proportionality constant. (Contributed by NM, 31-May-2015.)
𝑉 = (Base‘𝑊)    &   𝑆 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝑆)    &   𝑂 = (0g𝑆)    &    · = ( ·𝑠𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑉 ∖ { 0 }))       (𝜑 → ((𝑁‘{𝑋}) = (𝑁‘{𝑌}) ↔ ∃!𝑘 ∈ (𝐾 ∖ {𝑂})𝑋 = (𝑘 · 𝑌)))

Theoremlspsnel4 19871 A member of the span of the singleton of a vector is a member of a subspace containing the vector. (elspansn4 29334 analog.) (Contributed by NM, 4-Jul-2014.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑁‘{𝑋}))    &   (𝜑𝑌0 )       (𝜑 → (𝑋𝑈𝑌𝑈))

Theoremlspdisj 19872 The span of a vector not in a subspace is disjoint with the subspace. (Contributed by NM, 6-Apr-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑 → ¬ 𝑋𝑈)       (𝜑 → ((𝑁‘{𝑋}) ∩ 𝑈) = { 0 })

Theoremlspdisjb 19873 A nonzero vector is not in a subspace iff its span is disjoint with the subspace. (Contributed by NM, 23-Apr-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))       (𝜑 → (¬ 𝑋𝑈 ↔ ((𝑁‘{𝑋}) ∩ 𝑈) = { 0 }))

Theoremlspdisj2 19874 Unequal spans are disjoint (share only the zero vector). (Contributed by NM, 22-Mar-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))       (𝜑 → ((𝑁‘{𝑋}) ∩ (𝑁‘{𝑌})) = { 0 })

Theoremlspfixed 19875* Show membership in the span of the sum of two vectors, one of which (𝑌) is fixed in advance. (Contributed by NM, 27-May-2015.) (Revised by AV, 12-Jul-2022.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌}))    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑍}))    &   (𝜑𝑋 ∈ (𝑁‘{𝑌, 𝑍}))       (𝜑 → ∃𝑧 ∈ ((𝑁‘{𝑍}) ∖ { 0 })𝑋 ∈ (𝑁‘{(𝑌 + 𝑧)}))

Theoremlspexch 19876 Exchange property for span of a pair. TODO: see if a version with Y,Z and X,Z reversed will shorten proofs (analogous to lspexchn1 19877 versus lspexchn2 19878); look for lspexch 19876 and prcom 4641 in same proof. TODO: would a hypothesis of ¬ 𝑋 ∈ (𝑁‘{𝑍}) instead of (𝑁‘{𝑋}) ≠ (𝑁‘{𝑍}) be better overall? This would be shorter and also satisfy the 𝑋0 condition. Here and also lspindp* and all proofs affected by them (all in NM's mathbox); there are 58 hypotheses with the pattern as of 24-May-2015. (Contributed by NM, 11-Apr-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑍}))    &   (𝜑𝑋 ∈ (𝑁‘{𝑌, 𝑍}))       (𝜑𝑌 ∈ (𝑁‘{𝑋, 𝑍}))

Theoremlspexchn1 19877 Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch 19876 to see if this will shorten proofs. (Contributed by NM, 20-May-2015.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → ¬ 𝑌 ∈ (𝑁‘{𝑍}))    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌, 𝑍}))       (𝜑 → ¬ 𝑌 ∈ (𝑁‘{𝑋, 𝑍}))

Theoremlspexchn2 19878 Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch 19876 to see if this will shorten proofs. (Contributed by NM, 24-May-2015.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → ¬ 𝑌 ∈ (𝑁‘{𝑍}))    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑍, 𝑌}))       (𝜑 → ¬ 𝑌 ∈ (𝑁‘{𝑍, 𝑋}))

Theoremlspindpi 19879 Partial independence property. (Contributed by NM, 23-Apr-2015.)
𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌, 𝑍}))       (𝜑 → ((𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}) ∧ (𝑁‘{𝑋}) ≠ (𝑁‘{𝑍})))

Theoremlspindp1 19880 Alternate way to say 3 vectors are mutually independent (swap 1st and 2nd). (Contributed by NM, 11-Apr-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))    &   (𝜑 → ¬ 𝑍 ∈ (𝑁‘{𝑋, 𝑌}))       (𝜑 → ((𝑁‘{𝑍}) ≠ (𝑁‘{𝑌}) ∧ ¬ 𝑋 ∈ (𝑁‘{𝑍, 𝑌})))

Theoremlspindp2l 19881 Alternate way to say 3 vectors are mutually independent (rotate left). (Contributed by NM, 10-May-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))    &   (𝜑 → ¬ 𝑍 ∈ (𝑁‘{𝑋, 𝑌}))       (𝜑 → ((𝑁‘{𝑌}) ≠ (𝑁‘{𝑍}) ∧ ¬ 𝑋 ∈ (𝑁‘{𝑌, 𝑍})))

Theoremlspindp2 19882 Alternate way to say 3 vectors are mutually independent (rotate right). (Contributed by NM, 12-Apr-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑍𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))    &   (𝜑 → ¬ 𝑍 ∈ (𝑁‘{𝑋, 𝑌}))       (𝜑 → ((𝑁‘{𝑍}) ≠ (𝑁‘{𝑋}) ∧ ¬ 𝑌 ∈ (𝑁‘{𝑍, 𝑋})))

Theoremlspindp3 19883 Independence of 2 vectors is preserved by vector sum. (Contributed by NM, 26-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌 ∈ (𝑉 ∖ { 0 }))    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))       (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{(𝑋 + 𝑌)}))

Theoremlspindp4 19884 (Partial) independence of 3 vectors is preserved by vector sum. (Contributed by NM, 26-Apr-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → ¬ 𝑍 ∈ (𝑁‘{𝑋, 𝑌}))       (𝜑 → ¬ 𝑍 ∈ (𝑁‘{𝑋, (𝑋 + 𝑌)}))

Theoremlvecindp 19885 Compute the 𝑋 coefficient in a sum with an independent vector 𝑋 (first conjunct), which can then be removed to continue with the remaining vectors summed in expressions 𝑌 and 𝑍 (second conjunct). Typically, 𝑈 is the span of the remaining vectors. (Contributed by NM, 5-Apr-2015.) (Revised by Mario Carneiro, 21-Apr-2016.) (Proof shortened by AV, 19-Jul-2022.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    · = ( ·𝑠𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑 → ¬ 𝑋𝑈)    &   (𝜑𝑌𝑈)    &   (𝜑𝑍𝑈)    &   (𝜑𝐴𝐾)    &   (𝜑𝐵𝐾)    &   (𝜑 → ((𝐴 · 𝑋) + 𝑌) = ((𝐵 · 𝑋) + 𝑍))       (𝜑 → (𝐴 = 𝐵𝑌 = 𝑍))

Theoremlvecindp2 19886 Sums of independent vectors must have equal coefficients. (Contributed by NM, 22-Mar-2015.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐾 = (Base‘𝐹)    &    · = ( ·𝑠𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝑌 ∈ (𝑉 ∖ { 0 }))    &   (𝜑𝐴𝐾)    &   (𝜑𝐵𝐾)    &   (𝜑𝐶𝐾)    &   (𝜑𝐷𝐾)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))    &   (𝜑 → ((𝐴 · 𝑋) + (𝐵 · 𝑌)) = ((𝐶 · 𝑋) + (𝐷 · 𝑌)))       (𝜑 → (𝐴 = 𝐶𝐵 = 𝐷))

Theoremlspsnsubn0 19887 Unequal singleton spans imply nonzero vector subtraction. (Contributed by NM, 19-Mar-2015.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &    = (-g𝑊)    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑 → (𝑁‘{𝑋}) ≠ (𝑁‘{𝑌}))       (𝜑 → (𝑋 𝑌) ≠ 0 )

Theoremlsmcv 19888 Subspace sum has the covering property (using spans of singletons to represent atoms). Similar to Exercise 5 of [Kalmbach] p. 153. (spansncvi 29413 analog.) TODO: ugly proof; can it be shortened? (Contributed by NM, 2-Oct-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &    = (LSSum‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑇𝑆)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)       ((𝜑𝑇𝑈𝑈 ⊆ (𝑇 (𝑁‘{𝑋}))) → 𝑈 = (𝑇 (𝑁‘{𝑋})))

Theoremlspsolvlem 19889* Lemma for lspsolv 19890. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   𝐹 = (Scalar‘𝑊)    &   𝐵 = (Base‘𝐹)    &    + = (+g𝑊)    &    · = ( ·𝑠𝑊)    &   𝑄 = {𝑧𝑉 ∣ ∃𝑟𝐵 (𝑧 + (𝑟 · 𝑌)) ∈ (𝑁𝐴)}    &   (𝜑𝑊 ∈ LMod)    &   (𝜑𝐴𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑋 ∈ (𝑁‘(𝐴 ∪ {𝑌})))       (𝜑 → ∃𝑟𝐵 (𝑋 + (𝑟 · 𝑌)) ∈ (𝑁𝐴))

Theoremlspsolv 19890 If 𝑋 is in the span of 𝐴 ∪ {𝑌} but not 𝐴, then 𝑌 is in the span of 𝐴 ∪ {𝑋}. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)       ((𝑊 ∈ LVec ∧ (𝐴𝑉𝑌𝑉𝑋 ∈ ((𝑁‘(𝐴 ∪ {𝑌})) ∖ (𝑁𝐴)))) → 𝑌 ∈ (𝑁‘(𝐴 ∪ {𝑋})))

Theoremlssacsex 19891* In a vector space, subspaces form an algebraic closure system whose closure operator has the exchange property. Strengthening of lssacs 19714 by lspsolv 19890. (Contributed by David Moews, 1-May-2017.)
𝐴 = (LSubSp‘𝑊)    &   𝑁 = (mrCls‘𝐴)    &   𝑋 = (Base‘𝑊)       (𝑊 ∈ LVec → (𝐴 ∈ (ACS‘𝑋) ∧ ∀𝑠 ∈ 𝒫 𝑋𝑦𝑋𝑧 ∈ ((𝑁‘(𝑠 ∪ {𝑦})) ∖ (𝑁𝑠))𝑦 ∈ (𝑁‘(𝑠 ∪ {𝑧}))))

Theoremlspsnat 19892 There is no subspace strictly between the zero subspace and the span of a vector (i.e. a 1-dimensional subspace is an atom). (h1datomi 29342 analog.) (Contributed by NM, 20-Apr-2014.) (Proof shortened by Mario Carneiro, 22-Jun-2014.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)       (((𝑊 ∈ LVec ∧ 𝑈𝑆𝑋𝑉) ∧ 𝑈 ⊆ (𝑁‘{𝑋})) → (𝑈 = (𝑁‘{𝑋}) ∨ 𝑈 = { 0 }))

Theoremlspsncv0 19893* The span of a singleton covers the zero subspace, using Definition 3.2.18 of [PtakPulmannova] p. 68 for "covers".) (Contributed by NM, 12-Aug-2014.) (Revised by AV, 13-Jul-2022.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)       (𝜑 → ¬ ∃𝑦𝑆 ({ 0 } ⊊ 𝑦𝑦 ⊊ (𝑁‘{𝑋})))

Theoremlsppratlem1 19894 Lemma for lspprat 19900. Let 𝑥 ∈ (𝑈 ∖ {0}) (if there is no such 𝑥 then 𝑈 is the zero subspace), and let 𝑦 ∈ (𝑈 ∖ (𝑁‘{𝑥})) (assuming the conclusion is false). The goal is to write 𝑋, 𝑌 in terms of 𝑥, 𝑦, which would normally be done by solving the system of linear equations. The span equivalent of this process is lspsolv 19890 (hence the name), which we use extensively below. In this lemma, we show that since 𝑥 ∈ (𝑁‘{𝑋, 𝑌}), either 𝑥 ∈ (𝑁‘{𝑌}) or 𝑋 ∈ (𝑁‘{𝑥, 𝑌}). (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))    &    0 = (0g𝑊)    &   (𝜑𝑥 ∈ (𝑈 ∖ { 0 }))    &   (𝜑𝑦 ∈ (𝑈 ∖ (𝑁‘{𝑥})))       (𝜑 → (𝑥 ∈ (𝑁‘{𝑌}) ∨ 𝑋 ∈ (𝑁‘{𝑥, 𝑌})))

Theoremlsppratlem2 19895 Lemma for lspprat 19900. Show that if 𝑋 and 𝑌 are both in (𝑁‘{𝑥, 𝑦}) (which will be our goal for each of the two cases above), then (𝑁‘{𝑋, 𝑌}) ⊆ 𝑈, contradicting the hypothesis for 𝑈. (Contributed by NM, 29-Aug-2014.) (Revised by Mario Carneiro, 5-Sep-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))    &    0 = (0g𝑊)    &   (𝜑𝑥 ∈ (𝑈 ∖ { 0 }))    &   (𝜑𝑦 ∈ (𝑈 ∖ (𝑁‘{𝑥})))    &   (𝜑𝑋 ∈ (𝑁‘{𝑥, 𝑦}))    &   (𝜑𝑌 ∈ (𝑁‘{𝑥, 𝑦}))       (𝜑 → (𝑁‘{𝑋, 𝑌}) ⊆ 𝑈)

Theoremlsppratlem3 19896 Lemma for lspprat 19900. In the first case of lsppratlem1 19894, since 𝑥 ∉ (𝑁‘∅), also 𝑌 ∈ (𝑁‘{𝑥}), and since 𝑦 ∈ (𝑁‘{𝑋, 𝑌}) ⊆ (𝑁‘{𝑋, 𝑥}) and 𝑦 ∉ (𝑁‘{𝑥}), we have 𝑋 ∈ (𝑁‘{𝑥, 𝑦}) as desired. (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))    &    0 = (0g𝑊)    &   (𝜑𝑥 ∈ (𝑈 ∖ { 0 }))    &   (𝜑𝑦 ∈ (𝑈 ∖ (𝑁‘{𝑥})))    &   (𝜑𝑥 ∈ (𝑁‘{𝑌}))       (𝜑 → (𝑋 ∈ (𝑁‘{𝑥, 𝑦}) ∧ 𝑌 ∈ (𝑁‘{𝑥, 𝑦})))

Theoremlsppratlem4 19897 Lemma for lspprat 19900. In the second case of lsppratlem1 19894, 𝑦 ∈ (𝑁‘{𝑋, 𝑌}) ⊆ (𝑁‘{𝑥, 𝑌}) and 𝑦 ∉ (𝑁‘{𝑥}) implies 𝑌 ∈ (𝑁‘{𝑥, 𝑦}) and thus 𝑋 ∈ (𝑁‘{𝑥, 𝑌}) ⊆ (𝑁‘{𝑥, 𝑦}) as well. (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))    &    0 = (0g𝑊)    &   (𝜑𝑥 ∈ (𝑈 ∖ { 0 }))    &   (𝜑𝑦 ∈ (𝑈 ∖ (𝑁‘{𝑥})))    &   (𝜑𝑋 ∈ (𝑁‘{𝑥, 𝑌}))       (𝜑 → (𝑋 ∈ (𝑁‘{𝑥, 𝑦}) ∧ 𝑌 ∈ (𝑁‘{𝑥, 𝑦})))

Theoremlsppratlem5 19898 Lemma for lspprat 19900. Combine the two cases and show a contradiction to 𝑈 ⊊ (𝑁‘{𝑋, 𝑌}) under the assumptions on 𝑥 and 𝑦. (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))    &    0 = (0g𝑊)    &   (𝜑𝑥 ∈ (𝑈 ∖ { 0 }))    &   (𝜑𝑦 ∈ (𝑈 ∖ (𝑁‘{𝑥})))       (𝜑 → (𝑁‘{𝑋, 𝑌}) ⊆ 𝑈)

Theoremlsppratlem6 19899 Lemma for lspprat 19900. Negating the assumption on 𝑦, we arrive close to the desired conclusion. (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))    &    0 = (0g𝑊)       (𝜑 → (𝑥 ∈ (𝑈 ∖ { 0 }) → 𝑈 = (𝑁‘{𝑥})))

Theoremlspprat 19900* A proper subspace of the span of a pair of vectors is the span of a singleton (an atom) or the zero subspace (if 𝑧 is zero). Proof suggested by Mario Carneiro, 28-Aug-2014. (Contributed by NM, 29-Aug-2014.)
𝑉 = (Base‘𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑈𝑆)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑈 ⊊ (𝑁‘{𝑋, 𝑌}))       (𝜑 → ∃𝑧𝑉 𝑈 = (𝑁‘{𝑧}))

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144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 206 20501-20600 207 20601-20700 208 20701-20800 209 20801-20900 210 20901-21000 211 21001-21100 212 21101-21200 213 21201-21300 214 21301-21400 215 21401-21500 216 21501-21600 217 21601-21700 218 21701-21800 219 21801-21900 220 21901-22000 221 22001-22100 222 22101-22200 223 22201-22300 224 22301-22400 225 22401-22500 226 22501-22600 227 22601-22700 228 22701-22800 229 22801-22900 230 22901-23000 231 23001-23100 232 23101-23200 233 23201-23300 234 23301-23400 235 23401-23500 236 23501-23600 237 23601-23700 238 23701-23800 239 23801-23900 240 23901-24000 241 24001-24100 242 24101-24200 243 24201-24300 244 24301-24400 245 24401-24500 246 24501-24600 247 24601-24700 248 24701-24800 249 24801-24900 250 24901-25000 251 25001-25100 252 25101-25200 253 25201-25300 254 25301-25400 255 25401-25500 256 25501-25600 257 25601-25700 258 25701-25800 259 25801-25900 260 25901-26000 261 26001-26100 262 26101-26200 263 26201-26300 264 26301-26400 265 26401-26500 266 26501-26600 267 26601-26700 268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 392 39101-39200 393 39201-39300 394 39301-39400 395 39401-39500 396 39501-39600 397 39601-39700 398 39701-39800 399 39801-39900 400 39901-40000 401 40001-40100 402 40101-40200 403 40201-40300 404 40301-40400 405 40401-40500 406 40501-40600 407 40601-40700 408 40701-40800 409 40801-40900 410 40901-41000 411 41001-41100 412 41101-41200 413 41201-41300 414 41301-41400 415 41401-41500 416 41501-41600 417 41601-41700 418 41701-41800 419 41801-41900 420 41901-42000 421 42001-42100 422 42101-42200 423 42201-42300 424 42301-42400 425 42401-42500 426 42501-42600 427 42601-42700 428 42701-42800 429 42801-42900 430 42901-43000 431 43001-43100 432 43101-43200 433 43201-43300 434 43301-43400 435 43401-43500 436 43501-43600 437 43601-43700 438 43701-43800 439 43801-43900 440 43901-44000 441 44001-44100 442 44101-44200 443 44201-44300 444 44301-44400 445 44401-44500 446 44501-44600 447 44601-44700 448 44701-44800 449 44801-44900 450 44901-45000 451 45001-45093
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