Question

Let \( \langle \cdot, \cdot \rangle : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \) be an inner product on the vector space \( \mathbb{R}^n \) over \( \mathbb{R} \). Consider the following statements: P: \( |\langle u, v \rangle| \leq \frac{1}{2} \left( \langle u, u \rangle + \langle v, v \rangle \right) \) for all \( u, v \in \mathbb{R}^n \). Q: If \( \langle u, v \rangle = \langle 2u, -v \rangle \) for all \( v \in \mathbb{R}^n \), then \( u = 0 \). Then, which of the following is correct?

• A. both P and Q are TRUE
• B. P is TRUE and Q is FALSE
• C. P is FALSE and Q is TRUE
• D. both P and Q are FALSE

Hint:

The Cauchy-Schwarz inequality is a fundamental result in inner product spaces and is used to prove many important inequalities.

Answer 1

The Correct Option is A

- Statement P: This is a statement of the Cauchy-Schwarz inequality, which is a well-known result in the theory of inner products. The inequality is true for all inner products, so this statement is TRUE.
- Statement Q: If \( \langle u, v \rangle = \langle 2u, -v \rangle \), then we can simplify the equation to: \[ \langle u, v \rangle = -2 \langle u, v \rangle. \] This implies that \( \langle u, v \rangle = 0 \) for all \( v \), and hence \( u = 0 \) (since the inner product of \( u \) with any vector \( v \) is zero only if \( u = 0 \)). Thus, this statement is also TRUE. Thus, both statements P and Q are TRUE. Final Answer: (A) both P and Q are TRUE.