Exercise 6.B.2

Answers

Proof. The backward direction follows directly from 6.30.

Suppose

| | v | |^{2} = | ⟨ v, e_{1} ⟩ |^{2} + \dots + | ⟨ v, e_{m} ⟩ |^{2} .

By 6.35 we can extend $e_{1}, \dots, e_{m}$ to an orthonormal basis $e_{1}, \dots, e_{m}, f_{1}, \dots, f_{n}$ of $V$ . Then, by 6.30, we have

| | v | |^{2} = | ⟨ v, e_{1} ⟩ |^{2} + \dots + | ⟨ v, e_{m} ⟩ |^{2} + | ⟨ v, f_{1} ⟩ |^{2} + \dots + | ⟨ v, f_{n} ⟩ |^{2} .

Subtracting (1) from (2) yields

0 = | ⟨ v, f_{1} ⟩ |^{2} + \dots + | ⟨ v, f_{n} ⟩ |^{2},

showing that $⟨ v, f_{j} ⟩ = 0$ for each $j$ . From 6.30 again, we have

v = ⟨ v, e_{1} ⟩ e_{1} + \dots + ⟨ v, e_{m} ⟩ e_{m} + ⟨ v, f_{1} ⟩ f_{1} + \dots + ⟨ v, f_{n} ⟩ f_{n} = ⟨ v, e_{1} ⟩ e_{1} + \dots + ⟨ v, e_{m} ⟩ e_{m},

which shows that $v \in span (e_{1}, \dots, e_{m})$ . □

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2017-10-06 00:00