Exercise 6.10.7

Answers

Follow the proof of Theorem 6.43. We have

R (x) = \frac{\sum_{i = 1}^{n} λ_{i} | a_{i} |^{2}}{∥ x ∥^{2}} \geq \frac{λ_{n} \sum_{i = 1}^{n} | a_{i} |^{2}}{∥ x ∥^{2}} = \frac{λ_{n} ∥ x ∥^{2}}{∥ x ∥^{2}} = λ_{n} .

And the value is attainable since $R (v_{n}) = λ_{n}$ .

jephianlin

2011-06-27 00:00