Show that $T^*$ is one to one $\implies$ $T$ is onto

Question

Let $V$ and $W$ be $F$-vector space, and $V^*$ and $W^*$ be the dual space of $V$ and $W$, respectly. Let $T:V \to W$ be a linear transformation. Define $T^*:W^* \to V^*$ by $T^*(f)=f\circ T$ for all $f\in W^*$. Show that $T^*$ is one to one $\implies$ $T$ is onto.

My attempt: $T^*$ is one to one $\implies$ $\ker\{T^*\}=\{0\}$ $\implies$ $\dim(\text{Im}(T^*))=\dim(W^*)=\dim(W)$. I stop here, I think that I need to prove that $\dim(V)=\dim(W)$, but there are some gaps, thanks for help!

Answer 1

Try the contrapositive.

Note that if $S$ is a subspace of $W$ and $w_0 \notin S$, there is a linear functional $f$ such that $f(s) = 0$ for $s \in S$ and $f(w_0) = 1$.

Let $S = {\cal R} T$, and suppose $T$ is not surjective. Then there is some $w_0 \notin S$ and hence there is some $f$ such that $f(w_0) = 1$ and $f(T(x)) = 0$ for all $x$, or in other words, $T^* f = 0$.

Answer 2

Following the hint given above.

$Ker(T^{*})=\{ f: W \to \mathbb{K} \text{ such that } Im(T) \subset Ker(f)\}$

If $T^{*}$ is injective this precisely means that the only linear function whose kernel contains the image of $T$ must be the zero function. This says that $T$ is onto because we if could construct some $f$ setting $f(w)=0$ for every $w \in W$ and 1 in the rest, but the condition above would yield a contradiction.

Answer 3

Using some theorems from textbooks:

$dim(W) = dim(img(T)) + dim (img(T)^o)$ and $ker(T^*) = img(T)^o$.

$T^*$ is one to one, so $ker(T^*)$ is empty, so $img(T)^o$ is empty. $dim(img(T)^o) = 0$

Therefore $dim(W) = dim(img(T)$, so T is onto.