Re: [線代] 一題證明
※ 引述《tim8238818 (AAAAAAAAAAAAAAAAAAAAAAA)》之銘言:
: Let A, B and C be n x n matrices. We say that A is similar to B if there is
: an n x n non-singular matrix P, such that(P-1)AP = B. Prove each of the
: following statements. (P-1是P的inverse)
: a. If A is similar to B, then B is similar to A.
: b. If A is similar to B and B is similar to C, then A is similar to C
: 第一小題我寫
: A=B
: A=(p-1)AP
: PA=AP substitute to (P-1)AP=B
: (P-1)(PA)=B
: A=B
因為 A is similar to B,所以存在 non-singular 矩陣 P 使得 P^{-1}AP = B。
換言之,A = PBP^{-1}。
令 Q = P^{-1},我們得到 Q^{-1}BQ = A,因此 B is similar to A。
: 第二小題
: A=B and B=C=(P-1)AP
: A=(P-1)AP
: PA=AP substitute to (P-1)AP=C
: thus (P-1)PA=C
: A=C
因為 A is similar to B & B is similar to C,存在 non-singular 矩陣 P & Q
使得 P^{-1}AP = B & Q^{-1}BQ = C。
故 C = Q^{-1}BQ
= Q^{-1}(P^{-1}AP)Q
= (Q^{-1}P^{-1})A(PQ)
= (PQ)^{-1}A(PQ)
令 R = PQ,我們有 R^{-1}AR = C,因此 A is similar to C。
: 不知道這樣對不對,先謝謝板上神人賜教
大家的推文已經說明的很清楚了 :)
不過原PO可能對寫法不甚了解,因此還是回了這篇~
(把similar的定義弄清楚,similar和equal不一定一樣)
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