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In the above statement, $M=\ker E$.

I asked one question (here) like this, but now I added one hypothesys. Could someone help me again?

Notation:

$V$ is a infinite-dimensional inner product space;
$\langle\cdot,\cdot\rangle$ is the inner product of $V$;
$E:V\rightarrow V$ is a linear map;
$\text{Im}=\{E(v):v\in V\}$;
$\ker E=\{z\in V:E(z)=0\}$;
$(\ker E)^\perp=\{x\in V:\langle x,y\rangle=0 \;\forall y\in\ker E\}$.

Prove or give a couterexample: if $E^2=E$ and $\langle E(v),E(v)\rangle\leq\langle v,v\rangle$ for all $v\in V$, then $$\text{Im}\;E\subset\left(\ker E+(\ker E)^\perp\right)$$

Thanks!

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Are you sure the right hand side is $\big(\ker E +(\ker E)^\perp\big)$? Because we can also conclude $\def\im{{\rm im\,}} \im E=(\ker E)^\perp$.

If $E$ is idempotent ($E^2=E$), then $V=\ker E\,\oplus\,\im E\ $ (by $x=(x-Ex)\,+\,Ex$), and $E(u+w)=w$ for $u\in\ker E,\,w\in\im E$, so it is a projection onto $\im E$, in the direction of $\ker E$.

If in addition, we had that $\ker E\perp\im E$, then the statement would be obviously true.

Draw a picture, vectors of $\ker E$ give the direction of the projection, if any of these, say $u$, is oblique to the projected space ($\im E$), then there is a vector, orthogonal to $u$, which is shorter than its projected image.

To set it in algebra:$\ $ if $\langle u,w\rangle\ne 0$ for some $u\in\ker E,\,w\in\im E$, we can also assume $\|u\|=1$, for convenience. Then project $w$ orthogonally to $u$: $$v:=w-\langle u,w\rangle\,u\,.$$ Now the claim is that $\|v\|<\|Ev\|$. By the above description, we know that $Ev=w$. Since $v\perp u$, we can apply the Pythagorean theorem to conclude $$\|v\|^2 <\|v\|^2+\|(\langle u,w\rangle\,u)\|^2=\|w\|^2\,.$$ So, with the original assumption ($\langle Ev,Ev\rangle\le\langle v,v\rangle$ for all $v$) we get that $E$ must be orthogonal projection, so that $\im E=(\ker E)^\perp$.

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  • $\begingroup$ @julien Isn't this question also answered by your answer on the other question? $\endgroup$ – Potato Jun 14 '13 at 0:30
  • $\begingroup$ Ok. But, ${\rm im\,}E=(\ker E)^\perp$ is true, no? $\endgroup$ – Berci Jun 14 '13 at 0:35
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    $\begingroup$ If $\|E\|\leq 1$, we have that $\mbox{im E}\subseteq \ker E^\perp$. Whence indeed $V=\ker E\oplus (\ker E)^\perp$. And $\mbox{im E}= \ker E^\perp$ $\endgroup$ – Julien Jun 14 '13 at 0:42
  • $\begingroup$ @Potato In my counterexample 2, I give a bounded projection. But it is not possible to have it bounded by $1$ like here. $\endgroup$ – Julien Jun 14 '13 at 0:42

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