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Exercise 8.A.4

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Proof. Suppose $v \in G (α, T) \cap G (β, T)$ and suppose by contradiction that $v \neq 0$ . Then $v, v$ are generalized eigenvectors corresponding to distinct generalized eigenvalues of $T$ . Now 8.13 implies that $v, v$ is linearly independent, which is clearly a contradiction. Therefore $v$ must be $0$ . □

celiopassos

2017-10-06 00:00

Exercise 8.A.4

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