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.

  • $\begingroup$ what is the definition of dual mapping ? $\endgroup$ – Tsemo Aristide Jul 12 '17 at 12:28
  • $\begingroup$ @TsemoAristide See the page 67, section 2.24 in the book of Linear Algebra by Werner Greub $\endgroup$ – onurcanbektas Jul 12 '17 at 12:44
  • $\begingroup$ @TsemoAristide the link to the book: archive.org/details/springer_10.1007-978-1-4 $\endgroup$ – onurcanbektas Jul 12 '17 at 12:44
  • $\begingroup$ Does the superscript $\perp$ designate “annihilator?” If so, then the latter is a property of dual mappings in general (though it would make a bit more sense to me to write $\ker\pi^*=\operatorname{im}(\pi)^\perp$ in that case). $\endgroup$ – amd Jul 12 '17 at 16:48
  • 1
    $\begingroup$ Since you know the general case holds and the proof is just several lines, why you have to proove the special case? $\endgroup$ – C.Ding Jul 15 '17 at 11:58

$\newcommand{\im}{\operatorname{Im}}$ $\newcommand{\ang}[1]{\langle #1\rangle}$ 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.

  • $\begingroup$ If $E$ is not finite, then $(\im u)^\perp=\ker u^*;(\im u^*)^\perp=\ker u$ doesn't hold. $\endgroup$ – onurcanbektas Jul 15 '17 at 16:23
  • $\begingroup$ By the way, I have no idea what you mean by $(1-\pi)E$ in $E=\pi E\oplus (1-\pi)E$. $\endgroup$ – onurcanbektas Jul 15 '17 at 16:28
  • $\begingroup$ $\newcommand{\im}{\operatorname{Im}}$ $(1-\pi) E$ means $\im(1-\pi)$. $\endgroup$ – C.Ding Jul 15 '17 at 16:32
  • $\begingroup$ So, $1$ is the identity function then ? $\endgroup$ – onurcanbektas Jul 15 '17 at 16:34
  • $\begingroup$ yes, of course. $\endgroup$ – C.Ding Jul 15 '17 at 16:35

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 !

  • $\begingroup$ Thanks for you answer, but rest is still not clearer at all. $\endgroup$ – onurcanbektas Jul 12 '17 at 12:40

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