Re: [線代] 線代(16)

看板Math作者 (澳仔金控台灣分行)時間20年前 (2005/09/13 14:03), 編輯推噓0(000)
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※ 引述《PttFund (批踢踢基金)》之銘言: : (1) For A in M_{n╳n}(C), let f: M_{n╳n}(C) ---> M_{n╳n}(C) : defined by f(X) = AX-XA for all X in M_{n╳n}(C). Prove : that f is a linear transformation satisfying : f(XY) = f(X)Y + Xf(Y), for all X, Y in M_{n╳n}(C). ︿︿ 之前有打錯||b f(XY) = A(XY) - (XY)A = AXY - XYA. f(X)Y + Xf(Y) = (AX-XA)Y + X(AY-YA) = AXY - XAY + XAY - XYA = AXY - XYA. Hence f(XY) = f(X)Y + Xf(Y). ~( ̄▽ ̄)~(_△_)~( ̄▽ ̄)~(_△_)~( ̄▽ ̄)~ 這邊我不得不說了, 這邊的 f 很像微分算子, f(X) = AX-XA = [A,X]. 如果繼續算下去的話, f(XYZ) = f(X)YZ + Xf(Y)Z + XYf(Z). : (2) Let f be a linear transformation of the vector space M_{n╳n}(C) : which satisfies : f(XY) = f(X)Y + Xf(Y), for all X, Y in M_{n╳n}(C). : Prove that there is an A in M_{n╳n}(C) such that f(X) = AX-XA. : Note: M_{n╳n}(C) = n 階複係數方陣. 這子題目前還在想|||b -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 219.68.227.219 題目來自: http://math.nuk.edu.tw/gui/images/92_linear_transfer_school.pdf ※ 編輯: TaiwanBank 來自: 219.68.227.219 (09/13 18:02)
文章代碼(AID): #139ckRcy (Math)
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文章代碼(AID): #139ckRcy (Math)