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Theorem List for Metamath Proof Explorer - 19101-19200   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremreslmhm2 19101 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)       ((𝐹 ∈ (𝑆 LMHom 𝑈) ∧ 𝑇 ∈ LMod ∧ 𝑋𝐿) → 𝐹 ∈ (𝑆 LMHom 𝑇))
 
Theoremreslmhm2b 19102 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 19103 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 19104* 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 19105* 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 19106* 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 19107* 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 19108* 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 19109* 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 19110 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 19111 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 19112 An isomorphism of modules is a homomorphism. (Contributed by Stefan O'Rear, 21-Jan-2015.)
(𝐹 ∈ (𝑅 LMIso 𝑆) → 𝐹 ∈ (𝑅 LMHom 𝑆))
 
Theoremlmimgim 19113 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 19114 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 19115 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 19116 The relation "is isomorphic to" for modules. (Contributed by Stefan O'Rear, 25-Jan-2015.)
(𝑅𝑚 𝑆 ↔ (𝑅 LMIso 𝑆) ≠ ∅)
 
Theorembrlmici 19117 Prove isomorphic by an explicit isomorphism. (Contributed by Stefan O'Rear, 25-Jan-2015.)
(𝐹 ∈ (𝑅 LMIso 𝑆) → 𝑅𝑚 𝑆)
 
Theoremlmiclcl 19118 Isomorphism implies the left side is a module. (Contributed by Stefan O'Rear, 25-Jan-2015.)
(𝑅𝑚 𝑆𝑅 ∈ LMod)
 
Theoremlmicrcl 19119 Isomorphism implies the right side is a module. (Contributed by Mario Carneiro, 6-May-2015.)
(𝑅𝑚 𝑆𝑆 ∈ LMod)
 
Theoremlmicsym 19120 Module isomorphism is symmetric. (Contributed by Stefan O'Rear, 26-Feb-2015.)
(𝑅𝑚 𝑆𝑆𝑚 𝑅)
 
Theoremlmhmpropd 19121* 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))
 
10.6.4  Subspace sum; bases for a left module
 
Syntaxclbs 19122 Extend class notation with the set of bases for a vector space.
class LBasis
 
Definitiondf-lbs 19123* 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 19124* 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 19125 A basis is a set of vectors. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)       (𝐵𝐽𝐵𝑉)
 
Theoremlbsel 19126 An element of a basis is a vector. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)       ((𝐵𝐽𝐸𝐵) → 𝐸𝑉)
 
Theoremlbssp 19127 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 19128 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 19129 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 19130 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 19131 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 19132* 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 19133* 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 19134 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 19135 Subspace sum of spans of subsets is the span of their union. (spanuni 28531 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 19136 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 19137 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 19138 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 19139 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 19140 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 19141* 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 19142* 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 19143 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 19144 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 19145 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 19146 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 19147* 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.)
(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))    &   (𝜑𝐾 ∈ V)    &   (𝜑𝐿 ∈ V)       (𝜑 → (LBasis‘𝐾) = (LBasis‘𝐿))
 
Theorempj1lmhm 19148 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 19149 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)))
 
10.7  Vector spaces
 
10.7.1  Definition and basic properties
 
Syntaxclvec 19150 Extend class notation with class of all left vector spaces.
class LVec
 
Definitiondf-lvec 19151 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 multiplication is commutative i.e. a field. (Contributed by NM, 11-Nov-2013.)
LVec = {𝑓 ∈ LMod ∣ (Scalar‘𝑓) ∈ DivRing}
 
Theoremislvec 19152 The predicate "is a left vector space". (Contributed by NM, 11-Nov-2013.)
𝐹 = (Scalar‘𝑊)       (𝑊 ∈ LVec ↔ (𝑊 ∈ LMod ∧ 𝐹 ∈ DivRing))
 
Theoremlvecdrng 19153 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 19154 A left vector space is a left module. (Contributed by NM, 9-Dec-2013.)
(𝑊 ∈ LVec → 𝑊 ∈ LMod)
 
Theoremlsslvec 19155 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 19156 If a scalar product is zero, one of its factors must be zero. (hvmul0or 28010 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 19157 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 19158 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 19159 Cancellation law for scalar multiplication. (hvmulcan 28057 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 19160 Cancellation law for scalar multiplication. (hvmulcan2 28058 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 19161 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 19162 A nonzero scalar product does not change the span of a singleton. (spansncol 28555 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 19163 Membership relation that implies equality of spans. (spansneleq 28557 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 19164 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 19165 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 19166 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 19167 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 19168 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 19169 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 19170* Equal spans of singletons must have proportional vectors. See lspsnss2 19053 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 19171* 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 19172 A member of the span of the singleton of a vector is a member of a subspace containing the vector. (elspansn4 28560 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 19173 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 19174 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 19175 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 19176* 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.)
𝑉 = (Base‘𝑊)    &    + = (+g𝑊)    &    0 = (0g𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑌𝑉)    &   (𝜑𝑍𝑉)    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑌}))    &   (𝜑 → ¬ 𝑋 ∈ (𝑁‘{𝑍}))    &   (𝜑𝑋 ∈ (𝑁‘{𝑌, 𝑍}))       (𝜑 → ∃𝑧 ∈ ((𝑁‘{𝑍}) ∖ { 0 })𝑋 ∈ (𝑁‘{(𝑌 + 𝑧)}))
 
Theoremlspexch 19177 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 19178 vs. lspexchn2 19179); look for lspexch 19177 and prcom 4299 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 19178 Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch 19177 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 19179 Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch 19177 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 19180 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 19181 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 19182 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 19183 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 19184 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 19185 (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 19186 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.)
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 19187 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 19188 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 19189 Subspace sum has the covering property (using spans of singletons to represent atoms). Similar to Exercise 5 of [Kalmbach] p. 153. (spansncvi 28639 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 19190* Lemma for lspsolv 19191. (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 19191 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 19192* In a vector space, subspaces form an algebraic closure system whose closure operator has the exchange property. Strengthening of lssacs 19015 by lspsolv 19191. (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 19193 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 28568 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 19194* 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.)
𝑉 = (Base‘𝑊)    &    0 = (0g𝑊)    &   𝑆 = (LSubSp‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝑋𝑉)    &   (𝜑𝑋0 )       (𝜑 → ¬ ∃𝑦𝑆 ({ 0 } ⊊ 𝑦𝑦 ⊊ (𝑁‘{𝑋})))
 
Theoremlsppratlem1 19195 Lemma for lspprat 19201. 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 19191 (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 19196 Lemma for lspprat 19201. 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 19197 Lemma for lspprat 19201. In the first case of lsppratlem1 19195, 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 19198 Lemma for lspprat 19201. In the second case of lsppratlem1 19195, 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 19199 Lemma for lspprat 19201. 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 19200 Lemma for lspprat 19201. 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})))
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