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Theorem List for Metamath Proof Explorer - 18801-18900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremislbs 18801* 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 18802 A basis is a set of vectors. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)       (𝐵𝐽𝐵𝑉)
 
Theoremlbsel 18803 An element of a basis is a vector. (Contributed by Mario Carneiro, 24-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)       ((𝐵𝐽𝐸𝐵) → 𝐸𝑉)
 
Theoremlbssp 18804 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 18805 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 18806 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 18807 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 18808 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 18809* 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 18810* 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 18811 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 18812 Subspace sum of spans of subsets is the span of their union. (spanuni 27575 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 18813 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 18814 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 18815 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 18816 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 18817 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 18818* 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 18819* 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 18820 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 18821 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 18822 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 18823 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 18824* 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 18825 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 18826 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 18827 Extend class notation with class of all left vector spaces.
class LVec
 
Definitiondf-lvec 18828 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 18829 The predicate "is a left vector space". (Contributed by NM, 11-Nov-2013.)
𝐹 = (Scalar‘𝑊)       (𝑊 ∈ LVec ↔ (𝑊 ∈ LMod ∧ 𝐹 ∈ DivRing))
 
Theoremlvecdrng 18830 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 18831 A left vector space is a left module. (Contributed by NM, 9-Dec-2013.)
(𝑊 ∈ LVec → 𝑊 ∈ LMod)
 
Theoremlsslvec 18832 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 18833 If a scalar product is zero, one of its factors must be zero. (hvmul0or 27054 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 18834 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 18835 If a scalar product belongs to a subspace, either the scalar component is zero or the vector component also belongs. (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 18836 Cancellation law for scalar multiplication. (hvmulcan 27101 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 18837 Cancellation law for scalar multiplication. (hvmulcan2 27102 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 18838 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 18839 A nonzero scalar product does not change the span of a singleton. (spansncol 27599 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 18840 Membership relation that implies equality of spans. (spansneleq 27601 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 18841 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 18842 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 18843 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 18844 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 18845 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 18846 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 18847* Equal spans of singletons must have proportional vectors. See lspsnss2 18730 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 18848* 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 18849 A member of the span of the singleton of a vector is a member of a subspace containing the vector. (elspansn4 27604 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 18850 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 18851 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 18852 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 18853* 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 18854 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 18855 vs. lspexchn2 18856); look for lspexch 18854 and prcom 4114 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 18855 Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch 18854 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 18856 Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch 18854 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 18857 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 18858 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 18859 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 18860 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 18861 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 18862 (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 18863 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 18864 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 18865 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 18866 Subspace sum has the covering property (using spans of singletons to represent atoms). Similar to Exercise 5 of [Kalmbach] p. 153. (spansncvi 27683 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 18867* Lemma for lspsolv 18868. (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 18868 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 18869* In a vector space, subspaces form an algebraic closure system whose closure operator has the exchange property. Strengthening of lssacs 18692 by lspsolv 18868. (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 18870 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 27612 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 18871* 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 18872 Lemma for lspprat 18878. 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 18868 (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 18873 Lemma for lspprat 18878. 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 18874 Lemma for lspprat 18878. In the first case of lsppratlem1 18872, 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 18875 Lemma for lspprat 18878. In the second case of lsppratlem1 18872, 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 18876 Lemma for lspprat 18878. 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 18877 Lemma for lspprat 18878. 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 18878* 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}))
 
Theoremislbs2 18879* An equivalent formulation of the basis predicate in a vector space: a subset is a basis iff no element is in the span of the rest of the set. (Contributed by Mario Carneiro, 14-Jan-2015.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)       (𝑊 ∈ LVec → (𝐵𝐽 ↔ (𝐵𝑉 ∧ (𝑁𝐵) = 𝑉 ∧ ∀𝑥𝐵 ¬ 𝑥 ∈ (𝑁‘(𝐵 ∖ {𝑥})))))
 
Theoremislbs3 18880* An equivalent formulation of the basis predicate: a subset is a basis iff it is a minimal spanning set. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)       (𝑊 ∈ LVec → (𝐵𝐽 ↔ (𝐵𝑉 ∧ (𝑁𝐵) = 𝑉 ∧ ∀𝑠(𝑠𝐵 → (𝑁𝑠) ⊊ 𝑉))))
 
Theoremlbsacsbs 18881 Being a basis in a vector space is equivalent to being a basis in the associated algebraic closure system. Equivalent to islbs2 18879. (Contributed by David Moews, 1-May-2017.)
𝐴 = (LSubSp‘𝑊)    &   𝑁 = (mrCls‘𝐴)    &   𝑋 = (Base‘𝑊)    &   𝐼 = (mrInd‘𝐴)    &   𝐽 = (LBasis‘𝑊)       (𝑊 ∈ LVec → (𝑆𝐽 ↔ (𝑆𝐼 ∧ (𝑁𝑆) = 𝑋)))
 
Theoremlvecdim 18882 The dimension theorem for vector spaces: any two bases of the same vector space are equinumerous. Proven by using lssacsex 18869 and lbsacsbs 18881 to show that being a basis for a vector space is equivalent to being a basis for the associated algebraic closure system, and then using acsexdimd 16898. (Contributed by David Moews, 1-May-2017.)
𝐽 = (LBasis‘𝑊)       ((𝑊 ∈ LVec ∧ 𝑆𝐽𝑇𝐽) → 𝑆𝑇)
 
Theoremlbsextlem1 18883* Lemma for lbsext 18888. The set 𝑆 is the set of all linearly independent sets containing 𝐶; we show here that it is nonempty. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐶𝑉)    &   (𝜑 → ∀𝑥𝐶 ¬ 𝑥 ∈ (𝑁‘(𝐶 ∖ {𝑥})))    &   𝑆 = {𝑧 ∈ 𝒫 𝑉 ∣ (𝐶𝑧 ∧ ∀𝑥𝑧 ¬ 𝑥 ∈ (𝑁‘(𝑧 ∖ {𝑥})))}       (𝜑𝑆 ≠ ∅)
 
Theoremlbsextlem2 18884* Lemma for lbsext 18888. Since 𝐴 is a chain (actually, we only need it to be closed under binary union), the union 𝑇 of the spans of each individual element of 𝐴 is a subspace, and it contains all of 𝐴 (except for our target vector 𝑥- we are trying to make 𝑥 a linear combination of all the other vectors in some set from 𝐴). (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐶𝑉)    &   (𝜑 → ∀𝑥𝐶 ¬ 𝑥 ∈ (𝑁‘(𝐶 ∖ {𝑥})))    &   𝑆 = {𝑧 ∈ 𝒫 𝑉 ∣ (𝐶𝑧 ∧ ∀𝑥𝑧 ¬ 𝑥 ∈ (𝑁‘(𝑧 ∖ {𝑥})))}    &   𝑃 = (LSubSp‘𝑊)    &   (𝜑𝐴𝑆)    &   (𝜑𝐴 ≠ ∅)    &   (𝜑 → [] Or 𝐴)    &   𝑇 = 𝑢𝐴 (𝑁‘(𝑢 ∖ {𝑥}))       (𝜑 → (𝑇𝑃 ∧ ( 𝐴 ∖ {𝑥}) ⊆ 𝑇))
 
Theoremlbsextlem3 18885* Lemma for lbsext 18888. A chain in 𝑆 has an upper bound in 𝑆. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐶𝑉)    &   (𝜑 → ∀𝑥𝐶 ¬ 𝑥 ∈ (𝑁‘(𝐶 ∖ {𝑥})))    &   𝑆 = {𝑧 ∈ 𝒫 𝑉 ∣ (𝐶𝑧 ∧ ∀𝑥𝑧 ¬ 𝑥 ∈ (𝑁‘(𝑧 ∖ {𝑥})))}    &   𝑃 = (LSubSp‘𝑊)    &   (𝜑𝐴𝑆)    &   (𝜑𝐴 ≠ ∅)    &   (𝜑 → [] Or 𝐴)    &   𝑇 = 𝑢𝐴 (𝑁‘(𝑢 ∖ {𝑥}))       (𝜑 𝐴𝑆)
 
Theoremlbsextlem4 18886* Lemma for lbsext 18888. lbsextlem3 18885 satisfies the conditions for the application of Zorn's lemma zorn 9088 (thus invoking AC), and so there is a maximal linearly independent set extending 𝐶. Here we prove that such a set is a basis. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝑉 = (Base‘𝑊)    &   𝐽 = (LBasis‘𝑊)    &   𝑁 = (LSpan‘𝑊)    &   (𝜑𝑊 ∈ LVec)    &   (𝜑𝐶𝑉)    &   (𝜑 → ∀𝑥𝐶 ¬ 𝑥 ∈ (𝑁‘(𝐶 ∖ {𝑥})))    &   𝑆 = {𝑧 ∈ 𝒫 𝑉 ∣ (𝐶𝑧 ∧ ∀𝑥𝑧 ¬ 𝑥 ∈ (𝑁‘(𝑧 ∖ {𝑥})))}    &   (𝜑 → 𝒫 𝑉 ∈ dom card)       (𝜑 → ∃𝑠𝐽 𝐶𝑠)
 
Theoremlbsextg 18887* For any linearly independent subset 𝐶 of 𝑉, there is a basis containing the vectors in 𝐶. (Contributed by Mario Carneiro, 17-May-2015.)
𝐽 = (LBasis‘𝑊)    &   𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)       (((𝑊 ∈ LVec ∧ 𝒫 𝑉 ∈ dom card) ∧ 𝐶𝑉 ∧ ∀𝑥𝐶 ¬ 𝑥 ∈ (𝑁‘(𝐶 ∖ {𝑥}))) → ∃𝑠𝐽 𝐶𝑠)
 
Theoremlbsext 18888* For any linearly independent subset 𝐶 of 𝑉, there is a basis containing the vectors in 𝐶. (Contributed by Mario Carneiro, 25-Jun-2014.) (Revised by Mario Carneiro, 17-May-2015.)
𝐽 = (LBasis‘𝑊)    &   𝑉 = (Base‘𝑊)    &   𝑁 = (LSpan‘𝑊)       ((𝑊 ∈ LVec ∧ 𝐶𝑉 ∧ ∀𝑥𝐶 ¬ 𝑥 ∈ (𝑁‘(𝐶 ∖ {𝑥}))) → ∃𝑠𝐽 𝐶𝑠)
 
Theoremlbsexg 18889 Every vector space has a basis. This theorem is an AC equivalent; this is the forward implication. (Contributed by Mario Carneiro, 17-May-2015.)
𝐽 = (LBasis‘𝑊)       ((CHOICE𝑊 ∈ LVec) → 𝐽 ≠ ∅)
 
Theoremlbsex 18890 Every vector space has a basis. This theorem is an AC equivalent. (Contributed by Mario Carneiro, 25-Jun-2014.)
𝐽 = (LBasis‘𝑊)       (𝑊 ∈ LVec → 𝐽 ≠ ∅)
 
Theoremlvecprop2d 18891* If two structures have the same components (properties), one is a left vector space iff the other one is. This version of lvecpropd 18892 also breaks up the components of the scalar ring. (Contributed by Mario Carneiro, 27-Jun-2015.)
(𝜑𝐵 = (Base‘𝐾))    &   (𝜑𝐵 = (Base‘𝐿))    &   𝐹 = (Scalar‘𝐾)    &   𝐺 = (Scalar‘𝐿)    &   (𝜑𝑃 = (Base‘𝐹))    &   (𝜑𝑃 = (Base‘𝐺))    &   ((𝜑 ∧ (𝑥𝐵𝑦𝐵)) → (𝑥(+g𝐾)𝑦) = (𝑥(+g𝐿)𝑦))    &   ((𝜑 ∧ (𝑥𝑃𝑦𝑃)) → (𝑥(+g𝐹)𝑦) = (𝑥(+g𝐺)𝑦))    &   ((𝜑 ∧ (𝑥𝑃𝑦𝑃)) → (𝑥(.r𝐹)𝑦) = (𝑥(.r𝐺)𝑦))    &   ((𝜑 ∧ (𝑥𝑃𝑦𝐵)) → (𝑥( ·𝑠𝐾)𝑦) = (𝑥( ·𝑠𝐿)𝑦))       (𝜑 → (𝐾 ∈ LVec ↔ 𝐿 ∈ LVec))
 
Theoremlvecpropd 18892* If two structures have the same components (properties), one is a left vector space iff the other one is. (Contributed by Mario Carneiro, 27-Jun-2015.)
(𝜑𝐵 = (Base‘𝐾))    &   (𝜑𝐵 = (Base‘𝐿))    &   ((𝜑 ∧ (𝑥𝐵𝑦𝐵)) → (𝑥(+g𝐾)𝑦) = (𝑥(+g𝐿)𝑦))    &   (𝜑𝐹 = (Scalar‘𝐾))    &   (𝜑𝐹 = (Scalar‘𝐿))    &   𝑃 = (Base‘𝐹)    &   ((𝜑 ∧ (𝑥𝑃𝑦𝐵)) → (𝑥( ·𝑠𝐾)𝑦) = (𝑥( ·𝑠𝐿)𝑦))       (𝜑 → (𝐾 ∈ LVec ↔ 𝐿 ∈ LVec))
 
10.8  Ideals
 
10.8.1  The subring algebra; ideals
 
Syntaxcsra 18893 Extend class notation with the subring algebra generator.
class subringAlg
 
Syntaxcrglmod 18894 Extend class notation with the left module induced by a ring over itself.
class ringLMod
 
Syntaxclidl 18895 Ring left-ideal function.
class LIdeal
 
Syntaxcrsp 18896 Ring span function.
class RSpan
 
Definitiondf-sra 18897* Given any subring of a ring, we can construct a left-algebra by regarding the elements of the subring as scalars and the ring itself as a set of vectors. (Contributed by Mario Carneiro, 27-Nov-2014.) (Revised by Thierry Arnoux, 16-Jun-2019.)
subringAlg = (𝑤 ∈ V ↦ (𝑠 ∈ 𝒫 (Base‘𝑤) ↦ (((𝑤 sSet ⟨(Scalar‘ndx), (𝑤s 𝑠)⟩) sSet ⟨( ·𝑠 ‘ndx), (.r𝑤)⟩) sSet ⟨(·𝑖‘ndx), (.r𝑤)⟩)))
 
Definitiondf-rgmod 18898 Every ring can be viewed as a left module over itself. (Contributed by Stefan O'Rear, 6-Dec-2014.)
ringLMod = (𝑤 ∈ V ↦ ((subringAlg ‘𝑤)‘(Base‘𝑤)))
 
Definitiondf-lidl 18899 Define the class of left ideals of a given ring. An ideal is a submodule of the ring viewed as a module over itself. (Contributed by Stefan O'Rear, 31-Mar-2015.)
LIdeal = (LSubSp ∘ ringLMod)
 
Definitiondf-rsp 18900 Define the linear span function in a ring (Ideal generator). (Contributed by Stefan O'Rear, 4-Apr-2015.)
RSpan = (LSpan ∘ ringLMod)
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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 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