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Exercise 5.2.8

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We know that $\dim (E_{λ_{2}}) \geq 1$ . So pick a nonzero vector $v \in E_{λ_{2}}$ . Also pick a basis $β$ for $E_{λ_{2}}$ . Then $α = β \cup {v}$ forms a basis consisting of eigenvectors. It’s a basis because the cardinality is $n$ and the help of Theorem 5.8.

jephianlin

2011-06-27 00:00

Exercise 5.2.8

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