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 × N
Assertion matsca ⊢ N ∈ Fin ∧ R ∈ V → Scalar ⁡ G = Scalar ⁡ A

Proof

Step Hyp Ref Expression
1 matbas.a ⊢ A = N Mat R
2 matbas.g ⊢ G = R freeLMod N × N
3 scaid ⊢ Scalar = Slot Scalar ⁡ ndx
4 scandxnmulrndx ⊢ Scalar ⁡ ndx ≠ ⋅ ndx
5 3 4 setsnid ⊢ Scalar ⁡ G = Scalar ⁡ G sSet ⋅ 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 ∧ R ∈ V → A = G sSet ⋅ ndx R maMul N N N
8 7 fveq2d ⊢ N ∈ Fin ∧ R ∈ V → Scalar ⁡ A = Scalar ⁡ G sSet ⋅ ndx R maMul N N N
9 5 8 eqtr4id ⊢ N ∈ Fin ∧ R ∈ V → Scalar ⁡ G = Scalar ⁡ A