每週問題 September 23, 2013

這是通過零空間證明一特殊分塊矩陣的可逆性。

Let A=\begin{bmatrix}  P\\  Q  \end{bmatrix} be an n\times n real matrix. If N(P)=C(Q^T), show that A is nonsingular. Note that N(P) is the nullspace of P and C(Q^T) is the column space of Q^T.

 
參考解答:

對於 \mathbf{x}\in N(A),若能證明 \mathbf{x}=\mathbf{0},即可推論 A 是一可逆矩陣。假設 A\mathbf{x}=\mathbf{0},則 P\mathbf{x}=\mathbf{0}Q\mathbf{x}=\mathbf{0},也就是說,\mathbf{x}\in N(P)\mathbf{x}\in N(Q)。因為 N(P)=C(Q^T),可知存在 \mathbf{y} 使得 \mathbf{x}=Q^T\mathbf{y},即有 \mathbf{x}^T=\mathbf{y}^TQ。上式右乘 \mathbf{x},可得 \mathbf{x}^T\mathbf{x}=\mathbf{y}^TQ\mathbf{x}=\mathbf{y}^T\mathbf{0}=0,故 \Vert\mathbf{x}\Vert^2=0,推論 \mathbf{x}=\mathbf{0}

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