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Exercise 1.6.33

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1.

Since

V = W_{1} \oplus W_{2}

means

W_{1} \cap W_{2} = {0}

and a basis is linearly independent and so contains no

0

, we have

β_{1} \cap β_{2} = \emptyset

. And it a special case of exercise 1.6.29(a) that

β_{1} \cup β_{2}

is a basis.

2.

Let

u_{i}

and

v_{j}

are vectors in

β_{1}

and

β_{2}

respectly. If there is a nonzero vector

u \in W_{1} \cap W_{2}

, we can write

u = \sum_{i = 1}^{n} a_{i} u_{i} = \sum_{j = 1}^{m} b_{j} v_{j}

. But it impossible since it will cause

\sum_{i = 1}^{n} a_{i} u_{i} - \sum_{j = 1}^{m} b_{j} v_{j} = 0

. On the other hand, for any $v \in V$ , we can write $v = \sum_{i = 1}^{n} c_{i} u_{i} + \sum_{j = 1}^{m} d_{j} v_{j} \in W_{1} + W_{2}$ . For any $x + y \in W_{1} + W_{2}$ with $x \in W_{1}$ and $y \in W_{2}$ , we have $x = \sum_{i = 1}^{n} e_{i} u_{i}$ and $y = \sum_{j = 1}^{m} f_{j} v_{j}$ . Thus we have $x + y = \sum_{i = 1}^{n} e_{i} u_{i} + \sum_{j = 1}^{m} f_{j} v_{j} \in V$ .

jephianlin

2011-06-27 00:00

Exercise 1.6.33

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