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I know the * operator means the conjugate transpose but I am not sure how to get the conjugate transpose of $\mathbb R$

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  • $\begingroup$ What do you mean with R? $\endgroup$
    – Wuestenfux
    Feb 17 '19 at 7:17
  • $\begingroup$ If by R you mean the set of real numbers, $\mathbb{R}^{*}$ typically refers to the set of non-zero real numbers. $\endgroup$ Feb 17 '19 at 7:18
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$\mathbb{R^*} = \mathbb{R} - \{0\} $

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In general if $R$ is an arbitrary ring with $1$, then $R^\ast$ denotes the set of units, i.e. of invertible elements. Moreover, $R^\ast$ forms a group with respect to the ring multiplication, the so called group of units. In particular, if $R$ is a field, then $R^\ast=R \setminus \{0\}$.

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Just because $z^*$ for a complex number $z\in\mathbb C$ and $R^*$ for a ring $R$ use the same symbol, you can't conclude it is the same "* operator". One is complex conjugation, the other is taking the group of units. These two operations share nothing apart from their notation.

To answer the question: $\mathbb R^* = \mathbb R\setminus\{0\}$.

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