Exercise 7.A.19

Answers

Proof. We saw in Exercise 16 that $null T = null T^{∗}$ (for $T$ normal). Thus $(z_{1}, z_{2}, z_{3}) \in null T$ and we have

\begin{aligned} 0 & = ⟨ T^{∗} (z_{1}, z_{2}, z_{3}), (1, 1, 1) ⟩ \\ = ⟨ (z_{1}, z_{2}, z_{3}), T (1, 1, 1) ⟩ \\ = ⟨ (z_{1}, z_{2}, z_{3}), (2, 2, 2) ⟩ \\ = 2 z_{1} + 2 z_{2} + 3 z_{3} . \end{aligned}

Dividing by $2$ yields the desired result. □

celiopassos

2017-10-06 00:00