Re: [線代] linear combination

看板Math作者Sfly (topos)時間14年前 (2012/01/02 19:43)推噓2(2推 0噓 1→)

留言3則, 3人參與討論串5/5 (看更多)

※ 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>. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 76.94.119.209 ※ 編輯: Sfly 來自: 76.94.119.209 (01/02 19:45)

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Lpspace

01/02 20:55, , 1^F

01/02 20:55, 1^F

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Sfly

01/02 23:49, , 2^F

01/02 23:49, 2^F

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herstein

01/03 00:22, , 3^F

01/03 00:22, 3^F