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Let $V$ be an inner product space and $T:V\rightarrow V$ be a projection on some subspace of $V$.

Let's assume [$\forall x\in V, \|T(x)\|≦\|x\|$] Assuming this, I have proved that $\ker(T)=\operatorname{ran}(T)^\perp$.

However, I don't know whether $\operatorname{ran}(T)=\ker(T)^\perp$. Is it true? Then how do i prove this?

egreg
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Jj-
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  • If $|Tx| = |x|$ for all $x\in V$, then $|Tx| =0$ implies $|x| =0$, so $\ker(T) = 0$. So $\ker(T) = 0$. As $T$ is a projection $T = {\rm Id}$? – martini Jun 06 '13 at 11:05
  • Sorry, there was a typo. It meant inequality, not equality. – Jj- Jun 06 '13 at 11:12
  • If you have already proved $\ker T=(\mbox{im} T)^\perp$, you are done, since $V=\mbox{im} T\oplus \ker T$ and $\mbox{im} T=\ker (I-T)={v\in V,;, Tv=v}$ for any idempotent. Note also that in finite dimension, $(F^\perp )^\perp=F$ for any subspace. – Julien Jun 06 '13 at 12:42
  • @julien Thank you and what does 'im' stand for? – Jj- Jun 06 '13 at 17:38
  • $\mbox{im}$ is for image, that is range. – Julien Jun 06 '13 at 18:45

3 Answers3

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Since $\ker(T)=rng(T)^\perp, rng(T)\subset \ker(T)^\perp$.

Let $x\in \ker(T)^\perp$

Then $\exists (u,v)\in rng(T)\times \ker(T)$ such that $x=u+v$.

Thus $\langle x,T(x) \rangle = \langle x,u \rangle= \langle x,u \rangle + \langle x,v \rangle = \langle x,u+v \rangle = \|x\|^2$.

Thus $\lVert T(x)-x \rVert = \|T(x)\|^2 - \|x\|^2 ≦ \|x\|^2 - \|x\|^2 = 0$.

Thus $T(x)=x$. Since $T$ is a projection, $x\in rng(T)$.

Consequently, $\ker(T)^\perp \subset rng(T)$.

Hence the equality holds.

Jj-
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Note that $\|Tx\|=\|x\|$ implies that $T$ is an isometry with trivial kernel. Also, $$2\langle Tx,Ty\rangle = \lVert T(x+y)\rVert^2-\|Tx\|^2-\|Ty\|^2=\lVert x+y\rVert^2-\|x\|^2-\|y\|^2=2\langle x,y\rangle$$

Hagen von Eitzen
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Let $M = \text{im} T.$ Given what you have already proven, it is enough to show that $M=(M^{\perp})^{\perp}.$ That $M\subseteq (M^{\perp})^{\perp}$ is clear, so we have left to prove that $(\ker T)^{\perp}\subseteq \text{im} T.$

Pick $x\in (\ker T)^{\perp}.$ By the Polarization identity and your condition, we have $\langle x,y\rangle \leq \langle Tx, Ty\rangle.$ Let $y\in \ker T.$ We have at least one of: $\langle Tx, y \rangle \geq 0$ or $\langle -Tx, y \rangle\geq 0.$

If the first, $\langle Tx, y \rangle \leq \langle T^2 x, Ty \rangle=0$ so $Tx\in (\ker T)^{\perp}.$ If the second, $\langle -Tx, y \rangle \leq \langle T^2(-x), Ty \rangle=0$ so $-Tx\in (\ker T)^{\perp}.$

Therefore $x-Tx \in (\ker T)^{\perp}$ but since $T$ is a projection, $x-Tx\in \ker T$ as well. Therefore $x-Tx=0$ so $x=Tx\in \text{im} T.$

Ragib Zaman
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