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Exercise 5.2.20

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Since we can check that

W_{1} \oplus W_{2} \oplus W_{3} = (W_{1} \oplus W_{2}) \oplus W_{3},

if $V$ is direct sum of $W_{1}, W_{2}, \dots W_{k}$ , the dimension of $V$ would be the sum of the dimension of each $W_{i}$ by using Exercise 1.6.29(a) inductively.

Conversely, we first prove that if we have

\sum_{i = 1}^{k} W_{i} = V,

then we must have

\dim (V) \leq \sum_{i = 1}^{k} \dim (W_{i}) .

We induct on $k$ . For $k = 2$ , we may use the formula in Exercise 1.6.29(a). Suppose it holds for $k = m$ . We know that

(\sum_{i = 1}^{m - 1} W_{i}) + W_{m} = \sum_{i = 1}^{m} W_{i}

and

\dim (\sum_{i = 1}^{m} W_{i}) \leq \dim (\sum_{i = 1}^{m - 1} W_{i}) + \dim (W_{m}) \leq \sum_{i = 1}^{m} \dim (W_{i})

by induction hypothesis and the case for $k = 2$ .

To prove the original statement, suppose, by contradiction, that

W_{1} \cap \sum {i = 2}^{k} W_{i}

has nonzero element. By the formula in Exercise 1.6.29(a) we know that

\dim (\sum {i = 2}^{k} W_{i}) > \dim (V) - \dim (W_{1}) = \sum_{i = 2}^{k} \dim (W_{i}) .

This is impossible, so we get the desired result.

jephianlin

2011-06-27 00:00

Exercise 5.2.20

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