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Exercise 1.4.15

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If we have both $a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n} = b_{1} v_{1} + b_{2} v_{2} + \dots b_{n} v_{n}$ then we have $(a_{1} - b_{1}) v_{1} + (a_{2} - b_{2}) v_{2} + \dots + (a_{n} - b_{n}) v_{n} = 0$ . By the property we can deduce that $a_{i} = b_{i}$ for all $i$ .

jephianlin

2011-06-27 00:00

Exercise 1.4.15

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