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Exercise 6.A.25

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Proof. If $S$ is not injective, then there exists a non-zero $v \in V$ , such that $Sv = 0$ . But ${⟨ v, v ⟩}_{1} = ⟨ Sv, Sv ⟩ = ⟨ 0, 0 ⟩ = 0$ , therefore ${⟨⋅,⋅⟩}_{1}$ cannot satisfy the definiteness property of inner products. □

celiopassos

2017-10-06 00:00

Exercise 6.A.25

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