Metamath Proof Explorer


Theorem kerlmhm

Description: The kernel of a vector space homomorphism is a vector space itself. (Contributed by Thierry Arnoux, 7-May-2023)

Ref Expression
Hypotheses kerlmhm.1 ⊢ 0 = ( 0g ‘ 𝑈 )
kerlmhm.k ⊢ 𝐾 = ( 𝑉 ↾s ( ◡ 𝐹 “ { 0 } ) )
Assertion kerlmhm ( ( 𝑉 ∈ LVec ∧ 𝐹 ∈ ( 𝑉 LMHom 𝑈 ) ) → 𝐾 ∈ LVec )

Proof

Step Hyp Ref Expression
1 kerlmhm.1 ⊢ 0 = ( 0g ‘ 𝑈 )
2 kerlmhm.k ⊢ 𝐾 = ( 𝑉 ↾s ( ◡ 𝐹 “ { 0 } ) )
3 eqid ⊢ ( ◡ 𝐹 “ { 0 } ) = ( ◡ 𝐹 “ { 0 } )
4 eqid ⊢ ( LSubSp ‘ 𝑉 ) = ( LSubSp ‘ 𝑉 )
5 3 1 4 lmhmkerlss ⊢ ( 𝐹 ∈ ( 𝑉 LMHom 𝑈 ) → ( ◡ 𝐹 “ { 0 } ) ∈ ( LSubSp ‘ 𝑉 ) )
6 2 4 lsslvec ⊢ ( ( 𝑉 ∈ LVec ∧ ( ◡ 𝐹 “ { 0 } ) ∈ ( LSubSp ‘ 𝑉 ) ) → 𝐾 ∈ LVec )
7 5 6 sylan2 ⊢ ( ( 𝑉 ∈ LVec ∧ 𝐹 ∈ ( 𝑉 LMHom 𝑈 ) ) → 𝐾 ∈ LVec )