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Exercise 1.4.12

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To prove $span (S_{1}) \subset span (S_{2})$ we may let $v \in S_{1}$ . Then we can write $v = a_{1} x_{1} + a_{2} x_{2} + \dots + a_{3} x_{3}$ where $x_{i}$ is an element of $S_{1}$ and so is $S_{2}$ for all $n = 1, 2, \dots, n$ . But this means $v$ is a linear combination of $S_{2}$ and we complete the proof. If $span (S_{1}) = V$ , we know $span (S_{2})$ is a subspace containing $span (S_{1})$ . So it must be $V$ .

jephianlin

2011-06-27 00:00

Exercise 1.4.12

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