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Exercise 5.2.21

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Because $β$ is a basis, we have

\sum_{i = 1}^{k} span (β_{i}) = V .

Second, since the dimension of $span (β_{i})$ is the number of elements of $β_{i}$ , we know that the equality in the previous exercise about dimension holds. So $V$ is the direct sum of them.

jephianlin

2011-06-27 00:00

Exercise 5.2.21

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