Metamath Proof Explorer


Theorem matsca

Description: The matrix ring has the same scalars 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 matsca
|- ( ( N e. Fin /\ R e. V ) -> ( Scalar ` G ) = ( Scalar ` A ) )

Proof

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