Exercise 7.A.5

Answers

Proof. We have

\begin{aligned} \dim null T^{∗} & = \dim {(range T)}^{⊥} \\ = \dim W - \dim range T \\ = \dim W + \dim null T - \dim V \end{aligned}

where the first line follows from 7.7 (a), the second from 6.50 and the third from 3.22. We alse have

\begin{aligned} \dim range T^{∗} & = \dim {(null T)}^{⊥} \\ = \dim V - \dim null T \\ = \dim range T \end{aligned}

where the first line follows from 7.7 (b), the second from 6.50 and the third from 3.22. □

celiopassos

2017-10-06 00:00