## Inner Product Space ProblemsConsider an inner product space $ V $ over a field $ F $.### Problem (a)Let $ u, w \in V $. Prove the following inequality:\[ \|u + \langle u, w \rangle w \|^2 \geq \|u\|^2 + \| \langle u, w \rangle w \|^2 \]where equality holds if and only if $ u $ and $ w $ are orthogonal.### Problem (b)Let $ e_1, \ldots, e_m \in V $. Suppose that:\[ \| a_1 e_1 + \cdots + a_m e_m \|^2 = |a_1|^2 + \cdots + |a_m|^2 \]for every $ a_1, \ldots, a_m \in F $. Prove that $ e_1, \ldots, e_m $ form an orthonormal list of vectors in $ V $.---### Detailed Explanation#### Part (a)To tackle part (a), we need to show that the inequality holds and determine the conditions for equality.1. **Initial Setup**: - Start by applying the properties of inner products and norms in the space $ V $. - Explore cases where the inner product $ \langle u, w \rangle $ affects the outcome.2. **Orthogonality Condition**: - Recall that two vectors $ u $ and $ w $ are orthogonal if $ \langle u, w \rangle = 0 $. - Analyze the scenario when $ \langle u, w \rangle \neq 0 $ to see why the inequality is strict.#### Part (b)In part (b), the goal is to prove that the given vectors form an orthonormal set based on the given equation.1. **Orthogonality Proof**: - Use the given norm equation to infer that $ e_i $ must pairwise orthogonal. - Show that $ |a_1|^2 + \cdots + |a_m|^2 $ being equal to $ \| a_1 e_1 + \cdots + a_m e_m \|^2 $ implies pairwise inner products \( \

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Answered: Let V be an inner product space over F.…

Let V be an inner product space over F. (a) Let u, w e V. Prove that ||2+ (u, w)w|> |u+||(u, w)w|2, where equality holds if and only if u and w are orthogonal.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

## Inner Product Space Problems

Consider an inner product space $ V $ over a field $ F $.

### Problem (a)

Let $ u, w \in V $. Prove the following inequality:

\[ \|u + \langle u, w \rangle w \|^2 \geq \|u\|^2 + \| \langle u, w \rangle w \|^2 \]

where equality holds if and only if $ u $ and $ w $ are orthogonal.

### Problem (b)

Let $ e_1, \ldots, e_m \in V $. Suppose that:

\[ \| a_1 e_1 + \cdots + a_m e_m \|^2 = |a_1|^2 + \cdots + |a_m|^2 \]

for every $ a_1, \ldots, a_m \in F $. Prove that $ e_1, \ldots, e_m $ form an orthonormal list of vectors in $ V $.

---

### Detailed Explanation

#### Part (a)

To tackle part (a), we need to show that the inequality holds and determine the conditions for equality.

1. **Initial Setup**:
- Start by applying the properties of inner products and norms in the space $ V $.
- Explore cases where the inner product $ \langle u, w \rangle $ affects the outcome.

2. **Orthogonality Condition**:
- Recall that two vectors $ u $ and $ w $ are orthogonal if $ \langle u, w \rangle = 0 $.
- Analyze the scenario when $ \langle u, w \rangle \neq 0 $ to see why the inequality is strict.

#### Part (b)

In part (b), the goal is to prove that the given vectors form an orthonormal set based on the given equation.

1. **Orthogonality Proof**:
- Use the given norm equation to infer that $ e_i $ must pairwise orthogonal.
- Show that $ |a_1|^2 + \cdots + |a_m|^2 $ being equal to $ \| a_1 e_1 + \cdots + a_m e_m \|^2 $ implies pairwise inner products \( \

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