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Exercise 7.C.13

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Proof. It is false. Let $e_{1}, \dots, e_{n}$ be an orthonormal basis of $V$ and define $S e_{j} = e_{1}$ for $j = 1, \dots, n$ . Then $| | S e_{j} | | = 1$ for each $e_{j}$ , but obviously $S$ is not invertible, therefore $S$ is not an isometry (7.42 requires isometries to be invertible). □

celiopassos

2017-10-06 00:00

Exercise 7.C.13

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