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Given this set: $$ S=\left\{\begin{bmatrix}a&-b\\b&a\end{bmatrix}\middle|\,a,b\in\Bbb R\right\} $$ Part I:

Why is this set equivalent to the set of all complex numbers a+bi (when both are under multiplication?) There is one matrix that corresponds to a specific complex number. Can this example be found and how can it be demonstrated to give equivalent answers?

Part II:

What is a formula for the multiplicative inverse of the matrix shown in the set, using knowledge on inverses of complex numbers?

user112533
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    Take to complex numbers in their matrix representation, multiply them and check if whatever matrix you get still represents complex number. – Kaster May 19 '14 at 06:34
  • what equivalence relation are you referring to when you say "equivalent" –  May 19 '14 at 06:35
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    Define the function $f:\Bbb{C}\to S$ by the rule $$f(a+bi)=\pmatrix{a&-b\cr b&a\cr}.$$ Your task is to show that for all complex numbers $z_1$ and $z_2$ the following rules are obeyed: $$f(z_1+z_2)=f(z_1)+f(z_2),$$ $$f(z_1z_2)=f(z_1)f(z_2),$$ and furthermore that $f$ is a bijection. Look up isomorphism of rings/fields from a textbook on abstract algebra. The answer to Part II is "YES." – Jyrki Lahtonen May 19 '14 at 06:36

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Looking in the comments since I've posted, the following answer appears to be expanding on Jyrki's idea.

There exists a homomorphism $\phi:S \rightarrow \mathbb{C}$ defined as follows:

$$\begin{bmatrix} a & -b \\[0.3em] b & a \\[0.3em] \end{bmatrix} \mapsto (a + bi)$$

Of course, you will want to prove that this is indeed a homomorphism by checking the following conditions:

  • $\phi$ maps the multiplicative identity in $S$ to the multiplicative identity in $\mathbb{C}$.
  • $\phi(xy) = \phi(x)\phi(y)$ for any $x, y \in S$.
  • $\phi(x + y) = \phi(x) + \phi(y)$ for any $x, y \in S$.

Once you have done this, then show that $Im(\phi) = \mathbb{C}$, and $\ker(\phi) = \{0\}$, where $0$ is the additive identity in $S$. From here, you can apply the isomorphism theorem to show that $S$ is isomorphic to $\mathbb{C}$.

Kaj Hansen
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