Re: [線代] A*A=AA*, A^2u=0→Au=0
(A^2)u = 0 => A(Au) = 0
i.e. Au lies in the range of A, donote by R(A),
and lies in the null space of A, donote by N(A).
And N(A) = R(A)^⊥ (if A is normal) ...(*)
=> Au also lies in R(A)^⊥
=> Au = 0. Q.E.D.
---------------------------------------------------------
Proof of (*):
Suppose u lies in R(A)^⊥, i.e. <u,Av> = 0 for all v in V.
Then <(A^*)u,v> = 0 for all v in V.
=> (A^*)u = 0
=> Au = 0, i.e. u lies in N(A).
Conversely, suppose u lies in N(A), i.e. Au = 0
=> (A^*)u = 0 and <(A^*)u,v> = <u,Av> = 0 for all v in V.
Hence, u lies in R(A)^⊥. Q.E.D.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 1.34.27.65
推
03/02 00:11, , 1F
03/02 00:11, 1F
推
03/03 14:30, , 2F
03/03 14:30, 2F
討論串 (同標題文章)
完整討論串 (本文為第 2 之 2 篇)：