Metamath Proof Explorer


Theorem matmulcell

Description: Multiplication in the matrix ring for a single cell of a matrix. (Contributed by AV, 17-Nov-2019) (Revised by AV, 3-Jul-2022)

Ref Expression
Hypotheses matmulcell.a ⊢ A = N Mat R
matmulcell.b ⊢ B = Base A
matmulcell.m ⊢ × ˙ = ⋅ A
Assertion matmulcell ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → I X × ˙ Y J = ∑ R j ∈ N I X j ⋅ R j Y J

Proof

Step Hyp Ref Expression
1 matmulcell.a ⊢ A = N Mat R
2 matmulcell.b ⊢ B = Base A
3 matmulcell.m ⊢ × ˙ = ⋅ A
4 1 2 matrcl ⊢ X ∈ B → N ∈ Fin ∧ R ∈ V
5 eqid ⊢ R maMul N N N = R maMul N N N
6 1 5 matmulr ⊢ N ∈ Fin ∧ R ∈ V → R maMul N N N = ⋅ A
7 3 6 eqtr4id ⊢ N ∈ Fin ∧ R ∈ V → × ˙ = R maMul N N N
8 7 a1d ⊢ N ∈ Fin ∧ R ∈ V → R ∈ Ring → × ˙ = R maMul N N N
9 4 8 syl ⊢ X ∈ B → R ∈ Ring → × ˙ = R maMul N N N
10 9 adantr ⊢ X ∈ B ∧ Y ∈ B → R ∈ Ring → × ˙ = R maMul N N N
11 10 impcom ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B → × ˙ = R maMul N N N
12 11 3adant3 ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → × ˙ = R maMul N N N
13 12 oveqd ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → X × ˙ Y = X R maMul N N N Y
14 13 oveqd ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → I X × ˙ Y J = I X R maMul N N N Y J
15 eqid ⊢ Base R = Base R
16 eqid ⊢ ⋅ R = ⋅ R
17 simp1 ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → R ∈ Ring
18 4 simpld ⊢ X ∈ B → N ∈ Fin
19 18 adantr ⊢ X ∈ B ∧ Y ∈ B → N ∈ Fin
20 19 3ad2ant2 ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → N ∈ Fin
21 1 15 2 matbas2i ⊢ X ∈ B → X ∈ Base R N × N
22 21 adantr ⊢ X ∈ B ∧ Y ∈ B → X ∈ Base R N × N
23 22 3ad2ant2 ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → X ∈ Base R N × N
24 1 15 2 matbas2i ⊢ Y ∈ B → Y ∈ Base R N × N
25 24 adantl ⊢ X ∈ B ∧ Y ∈ B → Y ∈ Base R N × N
26 25 3ad2ant2 ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → Y ∈ Base R N × N
27 simp3l ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → I ∈ N
28 simp3r ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → J ∈ N
29 5 15 16 17 20 20 20 23 26 27 28 mamufv ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → I X R maMul N N N Y J = ∑ R j ∈ N I X j ⋅ R j Y J
30 14 29 eqtrd ⊢ R ∈ Ring ∧ X ∈ B ∧ Y ∈ B ∧ I ∈ N ∧ J ∈ N → I X × ˙ Y J = ∑ R j ∈ N I X j ⋅ R j Y J