If $\pi$ and $\pi^*$ are dual projection operators in $E, E^*$, show that $Im ( \pi^*) = (ker\pi)^\perp$. 
Let $E,E^*$ be dual spaces($E^*$ doesn't have to be $L(E)$).Suppose $\pi: E \to E$ and $\pi^*: E^* \to E^*$ are dual
  mappings.Assume that $\pi$ is a projection operator in E. Prove that
  $\pi^*$ is a projection operator in $E^*$ and that
  $$Im ( \pi^*) = (ker\pi)^\perp$$ 
  and  $$Im(\pi) = (ker \pi^*)^\perp$$

I have shown that $\pi^*$ is also a projection operator, but I have now idea how to show the following relations.
I would appreciate any help.
Note: The sign $A^\perp$ means the orthogonal space of A, i.e
$$(ker\pi)^\perp = \{ x^* \in E^* | <x^*, x> = 0 \quad \forall x \in ker\pi  \}$$
Note: We are not assuming neither $E$ nor $E^*$ is finite.
 A: Let $y^* \in Im ( \pi^*)$. Then $y^*=\pi^*(x^*)$ for some $x^* \in E^*$. Then we have for $x \in ker \pi$:
$y^*(x)=x^*( \pi(x))=0$, hence $y^* \in (ker\pi)^\perp$, therefore
$Im ( \pi^*) \subseteq (ker\pi)^\perp$.
The rest is your turn !
A: $\newcommand{\im}{\operatorname{Im}}$ $\newcommand{\ang}[1]{\langle #1\rangle}$
First, for a general linear operator $u:E\to E$, 

$$(\im u)^\perp=\ker u^*;(\im u^*)^\perp=\ker u.$$

In fact,
$$\begin{align*}
(\im u)^\perp &=\{x^*\in E^*|\ang{x^*,ux} = 0,\forall x\in E\}\\
&=\{x^*\in E^*|\ang{u^*x^*,x}=0,\forall x\in E\}\\
&=\{x^*\in E^*|u^*x^*=0\}=\ker u^*.
\end{align*}
$$
Second, for a projection $\pi$,

$$(\im\pi)^{\perp\perp}=\im \pi.$$

Indeed, $E=\pi E\oplus (1-\pi)E,$
so $$\forall x\in (\pi E)^{\perp\perp},\quad x=y+z,$$ where $y\in \pi E$ and $z\in (1-\pi) E,$ $$0=(1-\pi)x=(1-\pi)y+(1-\pi)z=0+z=z$$
since  $1-\pi\in (\pi E)^\perp$.
Thus $x=y\in \pi E$, another direction is just by definition.
Therefore, $\im \pi=(\im\pi)^{\perp\perp}=(\ker u^*)^\perp.$
And the other is similary.
