Metamath Proof Explorer


Theorem matsc

Description: The identity matrix multiplied with a scalar. (Contributed by Stefan O'Rear, 16-Jul-2018)

Ref Expression
Hypotheses matsc.a A=NMatR
matsc.k K=BaseR
matsc.m ·˙=A
matsc.z 0˙=0R
Assertion matsc NFinRRingLKL·˙1A=iN,jNifi=jL0˙

Proof

Step Hyp Ref Expression
1 matsc.a A=NMatR
2 matsc.k K=BaseR
3 matsc.m ·˙=A
4 matsc.z 0˙=0R
5 simp3 NFinRRingLKLK
6 3simpa NFinRRingLKNFinRRing
7 1 matring NFinRRingARing
8 eqid BaseA=BaseA
9 eqid 1A=1A
10 8 9 ringidcl ARing1ABaseA
11 6 7 10 3syl NFinRRingLK1ABaseA
12 eqid R=R
13 eqid N×N=N×N
14 1 8 2 3 12 13 matvsca2 LK1ABaseAL·˙1A=N×N×LRf1A
15 5 11 14 syl2anc NFinRRingLKL·˙1A=N×N×LRf1A
16 simp1 NFinRRingLKNFin
17 simp13 NFinRRingLKiNjNLK
18 fvex 1RV
19 4 fvexi 0˙V
20 18 19 ifex ifi=j1R0˙V
21 20 a1i NFinRRingLKiNjNifi=j1R0˙V
22 fconstmpo N×N×L=iN,jNL
23 22 a1i NFinRRingLKN×N×L=iN,jNL
24 eqid 1R=1R
25 1 24 4 mat1 NFinRRing1A=iN,jNifi=j1R0˙
26 25 3adant3 NFinRRingLK1A=iN,jNifi=j1R0˙
27 16 16 17 21 23 26 offval22 NFinRRingLKN×N×LRf1A=iN,jNLRifi=j1R0˙
28 ovif2 LRifi=j1R0˙=ifi=jLR1RLR0˙
29 2 12 24 ringridm RRingLKLR1R=L
30 29 3adant1 NFinRRingLKLR1R=L
31 2 12 4 ringrz RRingLKLR0˙=0˙
32 31 3adant1 NFinRRingLKLR0˙=0˙
33 30 32 ifeq12d NFinRRingLKifi=jLR1RLR0˙=ifi=jL0˙
34 28 33 eqtrid NFinRRingLKLRifi=j1R0˙=ifi=jL0˙
35 34 mpoeq3dv NFinRRingLKiN,jNLRifi=j1R0˙=iN,jNifi=jL0˙
36 15 27 35 3eqtrd NFinRRingLKL·˙1A=iN,jNifi=jL0˙