推 s620555 :非常感謝~ 10/06 03:19
※ 引述《s620555 (小毛)》之銘言：
: 麻煩幫解
: 1) show that {A屬於GLn(R)∣det A = ±1 } is a subgroup of GLn(R) un-der
: multiplication
The definition of subgroup (by I.N. Herstein) (可以請教這裡的hersteinXD)
A subset H of the group G is a subgroup of G if and only if it is
nonempty and closed under products and inverses.
(The closure conditions mean the following: whenever a and b are in H, then
-1
ab and a are also in H)
GLn(R)其實可以寫成很多形式
http://en.wikipedia.org/wiki/General_linear_group
在Lie group 裡面 我們就是把 GLn(R) 當做是一種n*n維空間中的一般線性變換群
所以可以寫成一般的矩陣轉換形式且係數是實數且 det(A)≠0
A'_i = Σ a_ij A_j .where A and A'are n*n matrix 且 det(a_ij)= ±1
j
pf:
let A'_i and B'_i belong to H
n
A'_i = Σ a_ij A_j ; det(a_ij)= ±1 .where A and A'are n*n matrix
j
with matrix element a_ij;i,j=1...n
n
B'_i = Σ b_ik B_k ; det(b_jk)= ±1 .where B and B'are n*n matrix
k
with matrix element b_jk;j,k=1...n
First
n n
C'i ≡ A'_i*B'_i = Σ a_ij b_jk C_k =Σ c_ik C_k (這步實在讓我很掙扎想好久)
jk k BBS要寫整個矩陣很麻煩
應該純寫矩陣相乘就好了??
and det(a_ij b_jk) = det(a_ij) det(b_jk)= ±1 ,
so the multiplication is closed.
Second
-1 n -1 -1
A'_i =( Σ a_ij) A_j det (a_ij)= ±1
j
because det(a_ij)= ±1 ≠ 0 ,so the inverse n*n matrix exists
hence
{ A is belong to GLn(R)∣det A = ±1 } is a subgroup of GLn(R)
PS:
A屬於GLn(R)∣det A = ±1 } is a subgroup of GLn(R)
是有物理意義的,且det A = 1 是 SLn(R)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.28.113
※ 編輯: Lindemann 來自: 140.113.28.113 (10/06 03:04)