Exercise 3.D.14

Answers

Proof. Suppose there are $u, v \in V$ such that $Tv = Tu$ . We have that

M (v) = M (u) = (\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix})

For somo $c_{1}, \dots, c_{n} \in 𝔽$ . Thus

v = c_{1} v_{1} + \dots + c_{n} v_{n} = u

Hence $T$ is injective. We have

\begin{aligned} \dim 𝔽^{n, 1} & = \dim V \\ = \dim null T + \dim range T \\ = \dim range T \end{aligned}

Hence $T$ is surjective and, therefore, an isomorphism of $V$ onto $F^{n, 1}$ □

celiopassos

2017-10-06 00:00