1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 !

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.