Exercise 7.A.6

Answers

Proof. (a) If $T$ were self-adjoint, we would have

⟨ T p, q ⟩ = ⟨ p, T^{*} q ⟩ = ⟨ p, T q ⟩ .

However, let $p (x) = a_{0} + a_{1} x + a_{2} x^{2}$ and $q (x) = b_{0} + b_{1} x + b_{2} x^{2}$ . We have

\begin{aligned} ⟨ T p, q ⟩ & = ⟨ a_{1} x, q ⟩ \\ = a_{1} \int_{0}^{1} b_{0} x + b_{1} x^{2} + b_{2} x^{3} d x \\ = a_{1} (\frac{b_{0}}{2} x^{2} + \frac{b_{1}}{3} x^{3} + \frac{b_{2}}{4} x^{4}) |_{0}^{1} \\ = a_{1} (\frac{b_{0}}{2} + \frac{b_{1}}{3} + \frac{b_{2}}{4}) . \end{aligned}

Similarly

⟨ p, T q ⟩ = ⟨ p, b_{1} x ⟩ = b_{1} (\frac{a_{0}}{2} + \frac{a_{1}}{3} + \frac{a_{2}}{4}) .

Thus, taking $a_{1} = 0$ and $b_{1}, a_{0}, a_{2} > 0$ clearly shows $⟨ T p, q ⟩ \neq ⟨ p, T q ⟩$ .

(b) 7.10 requires the chosen basis to be orthonormal. □

celiopassos

2017-10-06 00:00