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※ 引述《wakke (合理化勒索)》之銘言: : Let A be an n*n matrix over R such that A^5=I .Here, as usual, : I denotes the identity matrix. : Show that if the trace tr(A)=0 , then Av=v holds, : for some non-zero vector v Factorize x^5-1 in R[x] x^5-1=(x-1)(x^4+x^3+x^2+x+1) however, x^4+x^3+x^2+x+1 has NO real roots hence, x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d) where a=-e^{2πi/5}-e^{-2πi/5}=-2cos(2π/5) c=-e^{4πi/5}-e^{-4πi/5}=-2cos(4π/5)=2cos(π/5) and x^5-1=(x-1)(x^2+ax+b)(x^2+cx+d) Denote the characteristic polynomial of A as f(x) then f(x) should be (x-1)^k (x^2+ax+b)^p (x^2+cx+d)^q and k,p,q≧0, k+2p+2q=n and k,p,q are INTEGERS First, assume p≧q, p=q+r f(x)=(x-1)^k (x^4+x^3+x^2+x+1)^q (x^2+ax+b)^r =x^n + (-k+q+ra) x^{n-1} +... Because, tr(A)=0, -k+q+ra=0, and a is IRRATIONAL hence, r=0 k=q and k+4q=n then k≠0 Similarly, if q≧p we also have k≠0 Done ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 112.104.169.48 ※ 編輯: JohnMash 來自: 112.104.169.48 (07/14 21:46)
profyang :推~想了一下才懂 07/15 00:30
profyang :等等 所以你這樣等於也證明了n必須是五的倍數? 07/15 00:41
Vulpix :確實會有 5|n 這個結論沒錯 07/15 00:44
ppia :x^4+x^3+x^2+x+1 is irred. in |R[x] ( Eisenstein 07/17 19:40
x^4+x^3+x^2+x+1 is irreducible in Q[x] but reducible in R[x] x^5-1=(x-1)(x^2+ax+b)(x^2+cx+d) In mathematics, Eisenstein's criterion gives an easily checked sufficient condition for a polynomial with integer coefficients to be irreducible over the rational numbers. http://en.wikipedia.org/wiki/Eisenstein's_criterion
ppia :criterion ), so the minimal poly. of A must be 07/17 19:40
ppia :(i) x-1; (ii) x^4+x^3+x^2+x+1 (iii) x^5-1. 07/17 19:41
ppia :The char. poly. is : (i) (x-1)^n; 07/17 19:42
ppia :(ii) (x^4+x^3+x^2+x+1)^(n/4) 07/17 19:43
ppia :(iii) (x^4+x^3+x^2+x+1)^a (x-1)^b 07/17 19:44
ppia :a, b are positive integers with 4a+b = n 07/17 19:44
ppia :Since trA = 0, (iii) with a=b is the only 07/17 19:44
ppia :possibility. Therefore, 5|n and 1 is an eigenvle 07/17 19:45
※ 編輯: JohnMash 來自: 112.104.98.253 (07/17 23:08)