Metamath Proof Explorer


Theorem matvsca

Description: The matrix ring has the same scalar multiplication as its underlying linear structure. (Contributed by Stefan O'Rear, 4-Sep-2015) (Proof shortened by AV, 12-Nov-2024)

Ref Expression
Hypotheses matbas.a 𝐴 = ( 𝑁 Mat 𝑅 )
matbas.g 𝐺 = ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) )
Assertion matvsca ( ( 𝑁 ∈ Fin ∧ 𝑅𝑉 ) → ( ·𝑠𝐺 ) = ( ·𝑠𝐴 ) )

Proof

Step Hyp Ref Expression
1 matbas.a 𝐴 = ( 𝑁 Mat 𝑅 )
2 matbas.g 𝐺 = ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) )
3 vscaid ·𝑠 = Slot ( ·𝑠 ‘ ndx )
4 vscandxnmulrndx ( ·𝑠 ‘ ndx ) ≠ ( .r ‘ ndx )
5 3 4 setsnid ( ·𝑠𝐺 ) = ( ·𝑠 ‘ ( 𝐺 sSet ⟨ ( .r ‘ ndx ) , ( 𝑅 maMul ⟨ 𝑁 , 𝑁 , 𝑁 ⟩ ) ⟩ ) )
6 eqid ( 𝑅 maMul ⟨ 𝑁 , 𝑁 , 𝑁 ⟩ ) = ( 𝑅 maMul ⟨ 𝑁 , 𝑁 , 𝑁 ⟩ )
7 1 2 6 matval ( ( 𝑁 ∈ Fin ∧ 𝑅𝑉 ) → 𝐴 = ( 𝐺 sSet ⟨ ( .r ‘ ndx ) , ( 𝑅 maMul ⟨ 𝑁 , 𝑁 , 𝑁 ⟩ ) ⟩ ) )
8 7 fveq2d ( ( 𝑁 ∈ Fin ∧ 𝑅𝑉 ) → ( ·𝑠𝐴 ) = ( ·𝑠 ‘ ( 𝐺 sSet ⟨ ( .r ‘ ndx ) , ( 𝑅 maMul ⟨ 𝑁 , 𝑁 , 𝑁 ⟩ ) ⟩ ) ) )
9 5 8 eqtr4id ( ( 𝑁 ∈ Fin ∧ 𝑅𝑉 ) → ( ·𝑠𝐺 ) = ( ·𝑠𝐴 ) )