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作者  hunter (.) 站內  ALGEBRA
標題  Re: [問題] 請教一題線代
時間  2009/02/10 Tue 22:58:50

※ 引述《hunter (.)》之銘言:
> Let V be a finite dimensional vector space over R,
> and T is a linear transformation of V into itself.
> Suppose that the characteristic polynomial p_T(x)
> of T is written as p_T(x)=p_1(x)p_2(x),where p_1(x)
> and p_2(x) are two relatively prime polynomials with
> real coefficients. Show that every vector v in V can
> be written in a unique way as v=v_1+v_2, where v_1,v_2
> in V, p_1(T)(v_1)=0 and p_2(T)(v_2)=0
> 覺得滿有趣的問題,我猜可能要用rational canonical form
> 的概念去解,可是還是想不太出來:(

跟別人討論出來了,好像沒這麼難:)

"existence":
Since (p_1(x),p_2(x))=1, there exists m(x),n(x) such that
1 = m(x)p_1(x)+n(x)p_2(x).
v = m(T)p_1(T)v + n(T)p_2(T)v for any v in V.
Let v_1=n(T)p_2(T)v, v_2=m(T)p_1(T)v,
p_T(x) is characteristic polynomial, so p_T(T)v=0 by Caley-Hamilton theorem.
Hence p_1(T)v_1=0 and p_2(T)v_2=0.

"uniqueness":
It suffices to show that if v_1+v_2=0, where p_1(T)(v_1)=0,p_2(T)(v_2)=0,
then v_1=v_2=0.
In fact, p_1(T)(v_1+v_2) = 0 implies p_1(T)(v_2)=0.
Since (p_1(x),p_2(x))=1, there exists m(x),n(x) such that
1 = m(x)p_1(x)+n(x)p_2(x)
v_2 = m(T)p_1(T)(v_2)+n(T)p_2(T)(v_2) = 0 + 0 = 0
Therefore v_1=v_2=0.
                        #



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