Show $T$ is invertible if $T'$ is invertible where $T\in B(X)$, $T'\in B(X')$ Seems simple enough but I can't quite get it. $X$ is a complex Banach space, and $T\in B(X)$, $T'\in B(X')$ is its adjoint. Suppose $T'$ is invertible. How can we show that $T$ is invertible?
I have actually already shown $T$ to be 1-1, and that $T^{-1}$, were it to exist, would be bounded. So it actually only remains to show that $T$ is onto.
EDIT: I suspect the following simple result, which I have shown to be true, might help:
$\|Tx\| ≥{\|(T')^{-1}\|}^{-1} \|x\|$
This is how I know that $ker(T)=\{0\}$, hence 1-1. I think that were we to assume $T$ were not onto, we could somehow show $T'$ is not 1-1 (by showing it has non zero kernel), contradiction since $T'$ invertible. Can anyone see exactly how this might work?
 A: Let $T:X\longrightarrow Y$ be a bounded linear operator between two normed vector spaces, and let $T^*:Y^*\longrightarrow X^*$ denote its adjoint. By the open mapping theorem, $T^*$ is invertible if and only if it is bijective, as $X^*$ and $Y^*$ are automatically Banach spaces. This is not true for $T$ in general, but it is true if we assume $X,Y$ to be Banach spaces. 
Of course, if $T$ is invertible, $TS=ST=I$ implies $S^*T^*=T^*S^*=I$ hence $T^*$ is invertible with inverse $(T^{-1})^*$. You want the converse, which amounts to: does $T^*$ bijective imply $T$ bijective? Note that if $X,Y$ are supposed to be reflexive, then the result follows trivially via the identification $T^{**}\simeq T$ and the observation above.
Recall that the orthogonal of a set $S$ in $X$ is defined to be $S^\perp:=\{\phi\in X^*\;;\;\phi(x)=0\;\forall x\in S\}$. This is a weak* closed subspace of $X^*$. Now
$$
\mbox{Im}\:T^*\subseteq (\mbox{Ker}\;T)^\perp.
$$
Indeed, if $x^*=T^*y^*$, then $(x^*,x)=(T^*y^*,x)=(y^*,Tx)=0$ for every $x\in\mbox{Ker}\;T$.
As a consequence, if $T^*$ is onto, then $(\mbox{Ker}\;T)^\perp=X^*$. Now if we had $\mbox{Ker}\;T\neq\{0\}$, Hahn-Banach would provide us with a bounded linear functional $x^*\in X^*$ which does not vanish on $\mbox{Ker}\;T$, contradicting $(\mbox{Ker}\;T)^\perp=X^*$. Hence

$$T^*\;\mbox{surjective}\quad\Rightarrow\quad T\;\mbox{injective}.$$

Now observe that in general
$$
\mbox{Ker}\;T^*=(\mbox{Im}\;T)^\perp.
$$
Indeed, $T^*y^*=0$ if and only if $(T^*y^*,x)=0$ for all $x\in X$: that is $(y^*,Tx)=0$ for all $x\in X$, i.e. $(y^*,y)=0$ for all $y\in \mbox{Im}\;T$. So if $T^*$ is injective, we get 
$(\mbox{Im}\;T)^\perp=\{0\}$: i.e. every bounded linear functional $y^*\in Y^*$ which vanishes on $\mbox{Im}\;T$ must vanish on $Y^*$. By Hahn-Banach, this yields $\overline{\mbox{Im}\;T}=Y$. The good way to see this is to consider the annihilator: if $S\subseteq Y^*$, denote $S^o:=\{y\in Y\;;(y^*,y)=0\;\forall y^*\in S\;$. Then for any subspace $F$ in $Y$, $(F^\perp)^o=\overline{F}$. Hence
$$
(\mbox{Ker}\;T^*)^o=\overline{\mbox{Im}\;T}.
$$
By Hahn-Banach, $S^o=Y$ if and only if $S=\{0\}$. So we get an equivalence:

$$T^*\;\mbox{injective}\quad\iff\quad \overline{\mbox{Im}\;T}=Y.$$

Finally, and that's the tricky part of the whole thing, actually: when $X,Y$ are Banach spaces 

$$\mbox{Im}\;T\;\mbox{is closed}\quad\iff\quad \mbox{Im}\;T^*\;\mbox{is closed}.$$

Proof: see here, page 77.
Note: it can probably be simplified here given that we assume $T^*$ to be invertible, but I think it is good to remember this general fact.
Conclusion: these three pieces altogether show that when $X,Y$ are Banach spaces, $T^*$ is invertible if and only if $T$ is invertible.
