作者TaiwanBank (澳仔金控台湾分行)
看板Math
标题Re: [线代] 线代(16)
时间Tue Sep 13 14:17:51 2005
※ 引述《PttFund (批踢踢基金)》之铭言：
: (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).
︿︿
之前有打错||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).
~(￣▽￣)~(＿△＿)~(￣▽￣)~(＿△＿)~(￣▽￣)~
这边我不得不说了, 这边的 f 很像微分算子, f(X) = AX-XA = [A,X].
如果继续算下去的话, 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 阶复系数方阵.
这子题目前还在想|||b
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题目来自:
http://math.nuk.edu.tw/gui/images/92_linear_transfer_school.pdf
※ 编辑: TaiwanBank 来自: 219.68.227.219 (09/13 18:02)