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Exercise 6.A.5

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Proof. Suppose $v \in V$ such that $(T - \sqrt{2} I) v = 0$ . Then $Tv = \sqrt{2} v$ . This implies that $| | v | | \geq | | Tv | | = | | \sqrt{2} v | | = \sqrt{2} | | v | |$ . Since $| | v | |$ cannot be negative, it follows that $| | v | | = 0$ . Hence $v = 0$ , showing that $T - \sqrt{2} I$ is injective and, therefore, invertible. □

celiopassos

2017-10-06 00:00

Exercise 6.A.5

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