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In the book I'm reading, it says:

Observe that the $1{\times}1$ complex matrices correspond precisely to the set of all complex numbers. Furthermore, in this correspondence the $1{\times}1$ Hermitian matrices correspond to the real numbers.

Is it me, or something is wrong here? Perhaps the dimension of the matrix representing $\mathbb{C}$? Shouldn't it be $2{\times}2$?

How would you define $\mathbb{R}$ and $\mathbb{C}$ as hermitian matrices after all?

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up vote 2 down vote accepted

We have a bijective correspondence between $\Bbb C$ and $\cal M_1(\cal C)$ by the map $z\mapsto \pmatrix{z}$. As $\pmatrix{z}^t=\pmatrix{z}$, we have $\pmatrix z^H=\pmatrix{\bar z}$. So the Hermitian matrices are in bijection with real numbers.

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I can see how it represents $\mathbb{C}$, but I still can't see any $\mathbb{R}$ :-( –  Flavius Nov 19 '12 at 12:59
    
Oh, perhaps if the imaginary part was always $0$? –  Flavius Nov 19 '12 at 12:59
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@Flavius The author simply means to identify a complex number $z=a+bi\ (a,b\in\mathbb{R})$ with the $1\times1$ matrix $(z)=(a+bi)$. When $z$ is real, the matrix $(z)=(a)$ is Hermitian because $(a)^\ast = a$. When $z=bi$ is an imaginary number, $(z)=(bi)$ is skew-Hermitian: $(bi)^\ast = -(bi)$. –  user1551 Nov 19 '12 at 13:32
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