每週問題 January 11, 2016

證明秩─1線性算子的兩個性質。

Let T be a linear operator of rank one. Prove the following statements.
(a) There exists a unique scalar \lambda such that T^2=\lambda T.
(b) If \lambda\neq 1, then I-T is invertible, where I is the identity transformation.

 
參考解答:

(a) 考慮定義於向量空間 \mathcal{V} 的線性算子 T。因為 \hbox{rank}T=1,設 T 的值域為 \hbox{span}\{\mathbf{z}\},其中 \mathbf{z}\in\mathcal{V} 為一非零向量。因此,對於每一 \mathbf{x}\in\mathcal{V}

T(\mathbf{x})=\alpha(\mathbf{x})\mathbf{z}

其中 \alpha(\mathbf{x}) 為一純量。因為 T 是一個線性算子,\alpha 是一個線性泛函 (linear functional),滿足 \alpha(\mathbf{x}+\mathbf{y})=\alpha(\mathbf{x})+\alpha(\mathbf{y})\alpha(c\mathbf{x})=c\alpha(\mathbf{x}),其中 \mathbf{x},\mathbf{y}\in\mathcal{V}c 為一純量。對於每一個 \mathbf{x}\in\mathcal{V}

T^2(\mathbf{x})=T(\alpha(\mathbf{x})\mathbf{z})=\alpha(\mathbf{x})T(\mathbf{z})=\alpha(\mathbf{x})\alpha(\mathbf{z})\mathbf{z}=\alpha(\mathbf{z})T(\mathbf{x})

T^2=\alpha(\mathbf{z})T。令 \lambda=\alpha(\mathbf{z}) 即證明存在性。接著證明唯一性,假設兩個純量 \lambda_1\lambda_2 使得 T^2=\lambda_1TT^2=\lambda_2T,則 (\lambda_1-\lambda_2)T=0。但 T\neq 0 (因為 \hbox{rank}T\neq 0),推得 \lambda_1=\lambda_2

(b) 假設 \lambda\neq 1,我們要證明 (I-T)(\mathbf{x})=\mathbf{0} 蘊含 \mathbf{x}=\mathbf{0}。寫出 T(\mathbf{x})=I(\mathbf{x})=\mathbf{x},且 T^2(\mathbf{x})=T(\mathbf{x})。由 (a),T^2(\mathbf{x})=\lambda T(\mathbf{x})。令兩式相減,(\lambda-1)T(\mathbf{x})=\mathbf{0},故 T(\mathbf{x})=\mathbf{0},推得 \mathbf{x}=\mathbf{0}

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