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Is a contraction idempotent operator self-adjoint?

In the other words, if $T:H\to H$ is a bounded linear operator such that $||T||\leq1$ and $T^{2}=T$, can we conclude $T=T^*$?

Yan kai
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morapi
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  • Any attempts? Your thoughts? – 5xum Dec 02 '14 at 09:42
  • I tried to solve this problem so much, but all of my attempts yielded nothing. I solve it when T is one-to-one operator, but probably all contraction idempotent operators are not one-to-one. – morapi Dec 02 '14 at 09:50
  • It seems that $T=T^*$ is in general false. See the canonical decomposition in http://en.wikipedia.org/wiki/Contraction_%28operator_theory%29 where the unitary operator $U$ need not be self-adjoint. – Urgje Dec 02 '14 at 10:45
  • $T^2 = T$ means $T$ is a projection. A projection is self-adjoint if and only if it is an orthogonal projection. $T = 0$ is trivial, and otherwise, one has $\lVert P\rVert \geqslant 1$ for all projections $P\neq 0$. So the question is, can you show that a projection with norm $1$ is an orthogonal projection? – Daniel Fischer Dec 02 '14 at 12:44
  • In my textbooks a projection is self-adjoint by the definition. Also by a theorem in Tsoy-Wo if T is a projection we have either $||T||=0$ or $||T||=1$ – morapi Dec 02 '14 at 13:10

1 Answers1

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Here is a proof.

Note that $\|T^*\|=\|T\|\leq1$. For an $x\in H$, $\langle Tx,T^*x\rangle=\langle T^2x,x\rangle=\langle Tx,x\rangle$. Then \begin{align} \|Tx-T^*x\|^2&=\|Tx\|^2+\|T^*x\|^2-2\text{Re}\,\langle Tx,T^*x\rangle\\ &\leq\|Tx\|^2+\|x\|^2-2\text{Re}\,\langle Tx,x\rangle\\ &=\|Tx-x\|^2. \end{align} If we now take $x=Ty$ for some $y\in H$, we have $Tx=x$, and so $T^*x=Tx$, which translates to $$T^*Ty=TTy=Ty.$$ This shows that $T^*T=T$. Then $T^*=(T^*T)^*=T^*T=T$. So $T$ is selfadjoint (actually, it is positive).

Martin Argerami
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