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1.Prove that if $T$ is an unitary operator, then $T^n$ is unitary.

$\langle T^n \alpha ,T^n \beta\rangle = \langle\alpha ,T^{*n}T^n \beta\rangle = \langle\alpha,I\beta\rangle=\langle \alpha,\beta\rangle$

2.Let $T$ and $U$ be Hermitian operators on a finite-dimensional complex inner product space $V$. Prove that $T+U$ is a hermitian operator and that if $TU=UT$ then $TU$ is Hermitian.

$\langle TU \alpha ,\beta\rangle = \langle\alpha ,(TU)^*\beta\rangle = \langle\alpha,U^*T^*\beta\rangle=\langle (UT)^*\alpha,\beta\rangle=\langle \alpha,UT\beta\rangle$

Thanks, for your help.

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    $\begingroup$ What have you tried? This isn't a forum for having people do your homework for you. $\endgroup$ Feb 20, 2013 at 20:36
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    $\begingroup$ Sorry, that was not my intention, I do this (edited my question) $\endgroup$
    – user63192
    Feb 20, 2013 at 20:47
  • $\begingroup$ @KevinCarlson,sorry I´m new, I edited my question $\endgroup$
    – user63192
    Feb 20, 2013 at 21:02

2 Answers 2

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1) Your proof looks good, but you have to elaborate on the reason why $\langle T^n \alpha ,T^n \beta\rangle = \langle\alpha ,T^{*n}T^n \beta\rangle$.

2) Your proof here also looks good, but you need to be more detailed. It's a bit unclear why $\langle\alpha,\,U^\ast T^\ast \beta\rangle = \langle (UT)^\ast \alpha,\,\beta\rangle$. If you add one more step: $\langle\alpha,\,U^\ast T^\ast \beta\rangle = \langle\alpha,\,UT \beta\rangle = \langle (UT)^\ast \alpha,\,\beta\rangle$, your reader will know that you have applied the condition that $T,U$ are Hermitian.

Also, you've proved that $\langle TU\alpha,\,\beta\rangle = \langle \alpha,\,UT\beta\rangle $, but this is not exactly what you are asked to prove. What you should prove is that $\langle TU\alpha,\,\beta\rangle =\langle \alpha,\,TU\beta\rangle $, so, one further step is needed.

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1) If $T$ is unitary then $||Tx||=||x||$ Use the fact that $T^n=TT^{n-1}$ and induction over $n$

2a) Recall that the scalar product in a complex inner space is linear in the first argument and antilinear in the second

2b) $\mathbb{I}=(UT)^{-1}UT$

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