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Is it correct that set ${\mathbb C}$ is isomorphic to the set of following 2x2 matrices: $$\left( \begin{array}{cc} a &-b\\ b &a \end{array}\right) $$ $a \in {\mathbb R}$ and $b \in {\mathbb R} $?

In other words: are these two sets identical?

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  • $\begingroup$ Yes, these are isomorphic fields, but no, that does not mean the two sets are equal. $\endgroup$ – Ittay Weiss Nov 17 '13 at 18:27
  • $\begingroup$ Actually a meant that they are "identical" (with quotes). Sometimes when there is isomorphism between two sets it is said that they are alike. $\endgroup$ – mechanician Nov 17 '13 at 18:29
  • $\begingroup$ the term is 'essentially the same'. $\endgroup$ – Ittay Weiss Nov 17 '13 at 18:30
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Yes. Provided that $a^2+b^2 \neq 0$, that means the determinant is non-zero.

the map $\varphi: (\mathbb{C}-\{0\},.) \to (M, \times)$ given by $(a+bi) \mapsto \left( \begin{array}{cc} a &-b\\ b &a \end{array}\right)$ is an isomorphism. You can check it easily ($M$ is the set of all such matrices and $\times$ is matrix multiplication)

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  • $\begingroup$ What is the reason for $a^2 + b^2 \neq 0$? I think that zero matrix is absolutely equivalent to zero complex number, isn't it? $\endgroup$ – mechanician Nov 17 '13 at 18:12
  • $\begingroup$ @mechanician: Well, if you look at complex numbers as a ring, then you don't need that, but if you look at complex numbers as a group under multiplication, you do need that condition to ensure the existence of inverses. I assumed you considered them isomorphic as 'groups'. In what category you're talking about these two being isomorphic? $\endgroup$ – user66733 Nov 17 '13 at 18:15
  • $\begingroup$ Ok, I understand it. $\endgroup$ – mechanician Nov 17 '13 at 18:26
  • $\begingroup$ Should we not have $a^2+b^2 \ge 0$, because the absolute value of a complex number cannot be smaller than zero? $\endgroup$ – asmaier Apr 10 '14 at 20:14

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