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.

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## 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 \( \
Transcribed Image Text:## 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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