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Let $V$ complex inner product space of finite dimension and $T$ an operator over $V$. Show that the transformation $$U = (I-iT) (I + iT) ^ {-1}$$ is a unitary operator

My Attempt:

$$\langle U \alpha, U\alpha \rangle = \langle (I−iT)(I+iT)^{-1} \alpha , (I−iT)(I+iT)^{-1} \alpha \rangle = \langle \alpha , (I−iT)^{-1}(I+iT)(I−iT)(I+iT)^{-1} \alpha \rangle $$

and I need proof this $(I−iT)^{-1}(I+iT)(I−iT)(I+iT)^{-1} = I $

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  • $\begingroup$ This is straight from the definition of a unitary operator. What have you tried so far? $\endgroup$ – EuYu Apr 18 '13 at 23:45
  • $\begingroup$ @EuYu This <Ux,Ux>=<$(I-ix)(I+iX)^{-1}$ x , $(I-ix)(I+iX)^{-1}1$ x > = < x , $(I-ix)^{-1}(I+iX)(I-ix)(I+iX)^{-1}$ x > and I need proof this $(I-ix)^{-1}(I+iX)(I-ix)(I+iX)^{-1}$ = I $\endgroup$ – user63192 Apr 18 '13 at 23:49
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    $\begingroup$ Let $T = -iI$. Then $iT = I$, and $U$ will be zero. $\endgroup$ – Zach L. Apr 18 '13 at 23:50
  • $\begingroup$ @user63192 Can you please add your attempt to the question instead of as a comment? Also, you're using $x$ as both a vector and as the operator. $\endgroup$ – EuYu Apr 19 '13 at 0:00
  • $\begingroup$ @EuYu Yes I'm sorry :D $\endgroup$ – user63192 Apr 19 '13 at 0:02
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This is the operator-valued version of the fact that the map $$x\mapsto \frac{1-ix}{1+ix}$$ transforms the real line onto the unit circle.

In operator terms, the real line becomes the set of self-adjoint operators. And the unit circle is the group of unitary operators.

So, let $T$ be self-adjoint. Then $(I\pm iT)^*=(I\mp iT)$, which implies $U^*=(I-iT)^{-1}(I+iT)$. Since all operators of the form $aI+bT$ commute with each other, we have $$U^*U = (I-iT)^{-1}(I+iT) (I-iT)(I+iT)^{-1} =(I-iT)^{-1}(I-iT)(I+iT) (I+iT)^{-1} =I $$ and $$UU^* = (I-iT)(I+iT)^{-1} (I-iT)^{-1}(I+iT) = (I-iT) (I-iT)^{-1}(I+iT)^{-1}(I+iT) =I $$

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