每週問題 October 8, 2012

A, B 是實矩陣,證明:若 N(A^T)\subseteq N(B^T),則 C(B)\subseteq C(A)

Let A and B be an m\times n and m\times p real matrices, respectively. If N(A^T)\subseteq N(B^T), prove that C(B)\subseteq C(A).

 
參考解答:

因為 N(A^T)\subseteq N(A^T)+N(B^T),利用集合容斥性質即得 (N(A^T)+N(B^T))^\perp\subseteq N(A^T)^\perp,其中 \perp 代表正交補餘。同樣地,(N(A^T)+N(B^T))^\perp\subseteq N(B^T)^\perp,合併以上結果可得

(N(A^T)+N(B^T))^\perp\subseteq N(A^T)^\perp\cap N(B^T)^\perp

已知 N(A^T)\subseteq N(B^T),就有 N(A^T)+N(B^T)=N(B^T),因此 N(B^T)^\perp \subseteq N(A^T)^\perp\cap N(B^T)^\perp。最後將子空間正交補餘關係 N(A^T)^\perp = C(A)N(B^T)^\perp = C(B) 代入上式,即得 C(B)\subseteq C(A) \cap C(B),故可推論 C(B)\subseteq C(A)

PowSol-Oct-8-12

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