Re: [線代] 證明題
※ 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
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◆ From: 112.104.169.48
※ 編輯: JohnMash 來自: 112.104.169.48 (07/14 21:46)
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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
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※ 編輯: JohnMash 來自: 112.104.98.253 (07/17 23:08)
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