# A proproblem of dual space

Given problem:

Let $V$ and $W$ be nonzero vector spaces over the same field and let $T:V->W$ be a linear transformation. (a)Prove that $T$ is onto iff $T^t$ is one to one. (b)Prove that $T^t$ is onto iff $T$ is one to one. ($T^t$ is the mapping $T^t:W'->V'$defined by $T^t (g)=gT$ for all $g\in W'.$, $V'$ and $W'$ are the spaces of functionals from $V$ to $F$ and $W$ to $F$. My try : If $V$ and $W$ are finite dimensional

$dim (KerT)+dim (ImT)=dim (V)---(1)$

$dim (KerT^t)+dim (ImT^t)=dim (W')---(2)$

$dim (Im(T))=dim(Im (T^t))----(3)$(I have proved it)

(a) $T$ is onto implies $dim (Im(T))=dim W$ and we know $dim (W)=dim (W')$.Then using (2) and (3) we get $dim (KerT^t)=0$ ie. $T^t$ is one to one.

(b) $T^t$ is on to iimplies $dim (ImT^t)=dim (V')$ and we know $dim (V')=dim (V)$ then by using (1)and (3) $dim (KerT )=0$ ie. $T$ is one to one.

But I am STUCKED IN INFINITE DIMENSIONAL CASE. How can I solve it?Another approach in finite dimensional case will also help me.

I'll sketch a solution for the first part and leave the second to you. Assume that $T$ is onto. Let us show that $T^t$ is one-to-one. Let $g \in W'$ such that $T^t(g) = g \circ T = 0$. Then $g(T(v)) = 0$ for all $v \in V$. Given $w \in W$, choose $v \in V$ such that $Tv = w$ and deduce that $g(T(v)) = g(w) = 0$. Since $w$ was arbitrary, $g = 0$ which shows that $T^t$ is one-to-one.

Assume now that $T^t$ is one-to-one. Let us show that $T$ is onto. Assume by contradiction that $\operatorname{Im}(T) \subsetneq W$ and choose a complement $U$ such that $\operatorname{Im}(T) \oplus U = W$. Since $U \neq \{ 0 \}$, there exists a linear functional $g \in W'$ such that $g|_{\operatorname{Im}(T)} = 0$. But then $T^t(g) = 0$ while $g \neq 0$, a contradiction.