1
$\begingroup$

In Linear Algebra Done Right 3rd ed. Axler tries to prove a transformation $T\in \mathcal{L}(V,W)$ is surjective iff the dual map $T^*$ is injective. His reasoning starts with the following equation:$$\operatorname{range}(T)=W\iff (\operatorname{range}(T))^0=\{0\}.$$ How do I derive this? Is there something special about the fact that the only transformation that takes $\operatorname{range}(T)$ to the zero vector is the zero transformation?

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $\Bbbk$ be the field over which you are working. If $U$ is a subspace of $W$, then you have two possibilities:

  1. $U=W$: then the only linear form $\alpha\in W^*$ such that $\alpha(W)=\{0\}$ is the null form.
  2. $U\varsubsetneq W$: then, if $v\in W\setminus U$, if $U'$ is a subspace of $V$ such that $U\subset U'$ and that $U'\oplus\Bbbk v=W$, then if $\alpha\colon W\longrightarrow\Bbb k$ is defined by $\alpha(u+\lambda v)=\lambda$ ($u\in U'$ and $\lambda\in\Bbb k$), then $\alpha\in W^*$, $\alpha\neq0$, and $\alpha\in U^0$.

So, this proves that$$U=W\quad\text{if and only if}\quad U^0=\{0\}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.