Metamath Proof Explorer


Theorem lduallvec

Description: The dual of a left vector space is also a left vector space. Note that scalar multiplication is reversed by df-oppr ; otherwise, the dual would be a right vector space as is sometimes the case in the literature. (Contributed by NM, 22-Oct-2014)

Ref Expression
Hypotheses lduallvec.d 𝐷 = ( LDual ‘ 𝑊 )
lduallvec.w ( 𝜑𝑊 ∈ LVec )
Assertion lduallvec ( 𝜑𝐷 ∈ LVec )

Proof

Step Hyp Ref Expression
1 lduallvec.d 𝐷 = ( LDual ‘ 𝑊 )
2 lduallvec.w ( 𝜑𝑊 ∈ LVec )
3 lveclmod ( 𝑊 ∈ LVec → 𝑊 ∈ LMod )
4 2 3 syl ( 𝜑𝑊 ∈ LMod )
5 1 4 lduallmod ( 𝜑𝐷 ∈ LMod )
6 eqid ( Scalar ‘ 𝑊 ) = ( Scalar ‘ 𝑊 )
7 eqid ( oppr ‘ ( Scalar ‘ 𝑊 ) ) = ( oppr ‘ ( Scalar ‘ 𝑊 ) )
8 eqid ( Scalar ‘ 𝐷 ) = ( Scalar ‘ 𝐷 )
9 6 7 1 8 2 ldualsca ( 𝜑 → ( Scalar ‘ 𝐷 ) = ( oppr ‘ ( Scalar ‘ 𝑊 ) ) )
10 6 lvecdrng ( 𝑊 ∈ LVec → ( Scalar ‘ 𝑊 ) ∈ DivRing )
11 2 10 syl ( 𝜑 → ( Scalar ‘ 𝑊 ) ∈ DivRing )
12 7 opprdrng ( ( Scalar ‘ 𝑊 ) ∈ DivRing ↔ ( oppr ‘ ( Scalar ‘ 𝑊 ) ) ∈ DivRing )
13 11 12 sylib ( 𝜑 → ( oppr ‘ ( Scalar ‘ 𝑊 ) ) ∈ DivRing )
14 9 13 eqeltrd ( 𝜑 → ( Scalar ‘ 𝐷 ) ∈ DivRing )
15 8 islvec ( 𝐷 ∈ LVec ↔ ( 𝐷 ∈ LMod ∧ ( Scalar ‘ 𝐷 ) ∈ DivRing ) )
16 5 14 15 sylanbrc ( 𝜑𝐷 ∈ LVec )