proof of surjectivity of dualT implies injectivity of T and conversely

Question

This is from Axler's Linear Algebra book page 107 and 108. If $T$ is a linear map from $V$ to $W$, and $T^*$ is a dual map from $W^*$ to $V^*$,

Why is range $T$ = $W$ (surjectivity of $T$) a necessary and sufficient condition for (range $T$)$^0 = \{0\}$?

Also, why is null $T =\{0\}$ (injectivity of $T$) a necessary and sufficient condition for (null $T$)$^0 = V^*$?

Answer 1

Claim $\mathbf 1$. Let $V$ be a finite-dimensional vector space and $U$ a subspace of $V$. Then $U = V$ if and only if $U^0 = \{0\}$.

Proof. It is clear that $V^0 = \{f \in V^* : V \subseteq \ker f\} = \{0\}$. On the other hand, if $U$ a subspace of $V$ with $U^0 = \{0\}$, then using $$\dim V = \dim U + \dim U^0$$ we have $\dim V = \dim U$, meaning that $U = V$. $\blacksquare$

Claim $\mathbf 2$. Let $V$ be a finite-dimensional vector space and $U$ a subspace of $V$. Then $U = \{0\}$ if and only if $U^0 = V^*$.

Proof. It is clear that $\{0\}^0 = \{f \in V^* : \{0\} \subseteq \ker f\} = V^*$. On the other hand, if $U$ a subspace of $V$ with $U^0 = V^*$, then using $$\dim V^* = \dim V = \dim U + \dim U^0$$ we have $\dim U = 0$, meaning that $U = \{0\}$. $\blacksquare$