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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 X. N ) )
Assertion matvsca 
|- ( ( N e. Fin /\ R e. V ) -> ( .s ` G ) = ( .s ` A ) )

Proof

Step Hyp Ref Expression
1 matbas.a 
 |-  A = ( N Mat R )
2 matbas.g 
 |-  G = ( R freeLMod ( N X. N ) )
3 vscaid 
 |-  .s = Slot ( .s ` ndx )
4 vscandxnmulrndx 
 |-  ( .s ` ndx ) =/= ( .r ` ndx )
5 3 4 setsnid 
 |-  ( .s ` G ) = ( .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 e. Fin /\ R e. V ) -> A = ( G sSet <. ( .r ` ndx ) , ( R maMul <. N , N , N >. ) >. ) )
8 7 fveq2d 
 |-  ( ( N e. Fin /\ R e. V ) -> ( .s ` A ) = ( .s ` ( G sSet <. ( .r ` ndx ) , ( R maMul <. N , N , N >. ) >. ) ) )
9 5 8 eqtr4id 
 |-  ( ( N e. Fin /\ R e. V ) -> ( .s ` G ) = ( .s ` A ) )