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Relation of this antisymmetric matrix $r = \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}$ to $i$

Let $H$ be the subset of $M_2(\mathbb R)$ consisting of all matrices of the form $\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$ for $a, b \in \mathbb R$.

  • Show that $(\mathbb C,+)$ is isomorphic to $(H,+)$.
  • Show that $(\mathbb C, \times)$ is isomorphic to $(H, \times)$.

$H$ is said to be a matrix representation of the complex numbers.

I beg some help please. I fail even to define one to one functions mapping $\mathbb C$ onto $H$. All the best.


marked as duplicate by t.b., Jonas Meyer, Mike Spivey, Did, J. M. is a poor mathematician Dec 6 '11 at 9:52

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    $\begingroup$ Each element of $H$ is determined by a pair of real numbers. How are elements of $\mathbb C$ determined? $\endgroup$ – Jonas Meyer Dec 6 '11 at 5:21

Our matrices are of the form $$\left(\begin{smallmatrix} a & -b \\ b & a\end{smallmatrix}\right)$$ While our complex numbers are of the form $a + bi$

Both of these depend on $a, b$

Can you see a way to map from $\mathbb{C} \rightarrow H$ ?

Hint: For the harder one, multiplication $((a+bi)(c+di)) = (ac +adi + cbi -bd)$ so to have an isomorphism we will want $f(ac - bd + (ad + cb)i) = f((a+bi))f((c+di))$

we'll try the only choice that really makes sense $a + bi \mapsto \left(\begin{smallmatrix} a& -b \\ b & a\end{smallmatrix}\right)$

Then $$f(ac - bd + (ad + cb)i) = \left(\begin{smallmatrix} (ac - bd) & (-ad -cb) \\ (ad + cb) & (ac - bd)\end{smallmatrix}\right)= \left(\begin{smallmatrix} a & - b \\ b & a\end{smallmatrix}\right)\left(\begin{smallmatrix} c & -d \\ d & c\end{smallmatrix}\right) = f(a + bi)f(c + di)$$

Now of course, you have to show that this is one-to-one and onto (although that shouldn't be that hard) and I believe the addition should be similar.

  • $\begingroup$ @ Daven; thanks be blessed $\endgroup$ – neemy Dec 6 '11 at 5:56

Here's a hint : If you consider a complex number $z$, it can be written as $a+bi$ with $a,b \in \mathbb R$. (I haven't chosen the letters $a$ and $b$ for no reason.)

If you want a proof I don't mind showing. Just ask.

Hope that helps,

  • $\begingroup$ thanks , i would like to get the proof please $\endgroup$ – neemy Dec 6 '11 at 5:46
  • $\begingroup$ I think Deven Ware has already written parts of it down. You should try filling in the blanks ; it is really not a hard exercise for you. $\endgroup$ – Patrick Da Silva Dec 6 '11 at 6:22

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