推 Lpspace :This is the same. Your claim is not so easy. XD 01/02 20:55
→ Sfly :well...C-H is definitly harder than my claim 01/02 23:49
推 herstein :yes...二樓說的對 01/03 00:22
※ 引述《mqazz1 (無法顯示)》之銘言：
: 3*3 matrix A
: 4 vector: a, b, c, d
: Aa=b
: Ab=c
: Ac=d
: 請問這樣為什麼d會是a,b,c的linear combination呢?
: 該怎麼推呢?
: 謝謝!
Cayley-Hamilton is too powerful for this question.
It's easy to prove that V:=<a,Aa,A^2a,A^3a,...> = <a,Aa,A^2a>.
(for example, let k be the smallest integer such that
A^k(a) is in <a,Aa,..,A^(k-1)a> . Then V=<a,Aa,..,A^(k-1)a> and dimV=k.
Clearly dimV<=3, so k<=3 and V=<a,Aa,A^2a>.)
so A^3a = d is in <a,Aa,A^2a>=<a,b,c>.
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※ 編輯: Sfly 來自: 76.94.119.209 (01/02 19:45)