Metamath Proof Explorer


Theorem matdim

Description: Dimension of the space of square matrices. (Contributed by Thierry Arnoux, 20-May-2023)

Ref Expression
Hypotheses matdim.a A = I Mat R
matdim.n N = I
Assertion matdim I Fin R DivRing dim A = N N

Proof

Step Hyp Ref Expression
1 matdim.a A = I Mat R
2 matdim.n N = I
3 simpr I Fin R DivRing R DivRing
4 simpl I Fin R DivRing I Fin
5 xpfi I Fin I Fin I × I Fin
6 4 4 5 syl2anc I Fin R DivRing I × I Fin
7 eqid R freeLMod I × I = R freeLMod I × I
8 7 frlmdim R DivRing I × I Fin dim R freeLMod I × I = I × I
9 3 6 8 syl2anc I Fin R DivRing dim R freeLMod I × I = I × I
10 1 7 matbas I Fin R DivRing Base R freeLMod I × I = Base A
11 10 eqcomd I Fin R DivRing Base A = Base R freeLMod I × I
12 eqidd I Fin R DivRing Base A = Base A
13 ssidd I Fin R DivRing Base A Base A
14 1 7 matplusg I Fin R DivRing + R freeLMod I × I = + A
15 14 oveqdr I Fin R DivRing x Base A y Base A x + R freeLMod I × I y = x + A y
16 7 frlmlvec R DivRing I × I Fin R freeLMod I × I LVec
17 3 6 16 syl2anc I Fin R DivRing R freeLMod I × I LVec
18 lveclmod R freeLMod I × I LVec R freeLMod I × I LMod
19 17 18 syl I Fin R DivRing R freeLMod I × I LMod
20 19 adantr I Fin R DivRing x Base Scalar A y Base A R freeLMod I × I LMod
21 simprl I Fin R DivRing x Base Scalar A y Base A x Base Scalar A
22 1 7 matsca I Fin R DivRing Scalar R freeLMod I × I = Scalar A
23 22 fveq2d I Fin R DivRing Base Scalar R freeLMod I × I = Base Scalar A
24 23 eqcomd I Fin R DivRing Base Scalar A = Base Scalar R freeLMod I × I
25 24 adantr I Fin R DivRing x Base Scalar A y Base A Base Scalar A = Base Scalar R freeLMod I × I
26 21 25 eleqtrd I Fin R DivRing x Base Scalar A y Base A x Base Scalar R freeLMod I × I
27 simprr I Fin R DivRing x Base Scalar A y Base A y Base A
28 11 adantr I Fin R DivRing x Base Scalar A y Base A Base A = Base R freeLMod I × I
29 27 28 eleqtrd I Fin R DivRing x Base Scalar A y Base A y Base R freeLMod I × I
30 eqid Base R freeLMod I × I = Base R freeLMod I × I
31 eqid Scalar R freeLMod I × I = Scalar R freeLMod I × I
32 eqid R freeLMod I × I = R freeLMod I × I
33 eqid Base Scalar R freeLMod I × I = Base Scalar R freeLMod I × I
34 30 31 32 33 lmodvscl R freeLMod I × I LMod x Base Scalar R freeLMod I × I y Base R freeLMod I × I x R freeLMod I × I y Base R freeLMod I × I
35 20 26 29 34 syl3anc I Fin R DivRing x Base Scalar A y Base A x R freeLMod I × I y Base R freeLMod I × I
36 35 28 eleqtrrd I Fin R DivRing x Base Scalar A y Base A x R freeLMod I × I y Base A
37 1 7 matvsca I Fin R DivRing R freeLMod I × I = A
38 37 oveqdr I Fin R DivRing x Base Scalar A y Base A x R freeLMod I × I y = x A y
39 eqid Scalar A = Scalar A
40 eqidd I Fin R DivRing Base Scalar A = Base Scalar A
41 22 fveq2d I Fin R DivRing + Scalar R freeLMod I × I = + Scalar A
42 41 oveqdr I Fin R DivRing x Base Scalar A y Base Scalar A x + Scalar R freeLMod I × I y = x + Scalar A y
43 drngring R DivRing R Ring
44 1 matlmod I Fin R Ring A LMod
45 43 44 sylan2 I Fin R DivRing A LMod
46 1 matsca2 I Fin R DivRing R = Scalar A
47 46 3 eqeltrrd I Fin R DivRing Scalar A DivRing
48 39 islvec A LVec A LMod Scalar A DivRing
49 45 47 48 sylanbrc I Fin R DivRing A LVec
50 11 12 13 15 36 38 31 39 24 40 42 17 49 dimpropd I Fin R DivRing dim R freeLMod I × I = dim A
51 hashxp I Fin I Fin I × I = I I
52 4 4 51 syl2anc I Fin R DivRing I × I = I I
53 9 50 52 3eqtr3d I Fin R DivRing dim A = I I
54 2 2 oveq12i N N = I I
55 53 54 eqtr4di I Fin R DivRing dim A = N N