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The canonical way to represent the complex number $a+bi$ as a $2\times2$ matrix is with $\pmatrix{a &b\\-b&a}$, but I have also found that $\pmatrix{a&bx\\\frac{-b}{x}&a}$ will do for any non-zero $x$. Is there some error here or is this also a perfect representation? Furthermore, is this the only way to represent the complex numbers as matrices?

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    $\begingroup$ You'll have to consider norms as well. $\|a+bi\|=\sqrt{a^2+b^2}=\left\|\begin{pmatrix}a&b\\-b&a\end{pmatrix}\right\|$ using the usual operator norm on the right. If you were to let $x=-1$, yes, this would still work (and truly, there is very little distinction between $i$ and $-i$ beyond convention), but for other values of $x$, you'll have to define the norm differently. $\endgroup$ – JMoravitz Jul 4 '18 at 0:49
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If you want a representation in the sense of representation theory, then all we need is a matrix satisfying $J^2 = -I$, where $I$ denotes the identity matrix. With that, we can represent the complex number $a + bi$ as the matrix $aI + bJ$. In the usual representation of the complex numbers, we have chosen $$ J = \pmatrix{0&-1\\1&0} $$ (in fact, you have the transpose of the canonical representation in your question). And there are certainly reasons that this $J$ is a "natural" choice.

If we want to find all possible representations, however, we should look for all matrices which satisfy $J^2 = -I$. To that end, a nice approach for $2 \times 2$ matrices is to use the Cayley-Hamilton theorem.

In particular, if $J$ is a real matrix which satisfies $J^2 + I = 0$, then its characteristic polynomial must be $$p(x) = \det(xI - J) = x^2 + 1$$ Thus, we have $$ p(x) = x^2 + 1 = x^2 - \operatorname{tr}(J)x + \det(J) $$ Thus, we must have $\operatorname{tr}(J) = 0$ and $\det(J) = 1$. Indeed, the matrices $$ J = \pmatrix{0 & x\\-1/x & 0} $$ satisfy this condition, and lead to the representations which you have identified.


In general, $J$ must have the form $$ J = \pmatrix{y&x\\-(y^2 + 1)/x&-y} $$ with $x,y \in \Bbb R$ and $x \neq 0$.

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  • $\begingroup$ Since you mention representation theory, perhaps it's worth mentioning that all of these $J$'s are the same up to conjugation, and so from the rep. theory p.o.v. once you understand the canonical embedding, you understand all of them. $\endgroup$ – Travis Willse Jul 4 '18 at 2:39
  • $\begingroup$ Isomorphism of representations seems like it would be a bit of a confusing addition to the answer above, especially since this is geared towards an audience that has potentially no background in abstract algebra. I am glad that you made the comment, though. $\endgroup$ – Omnomnomnom Jul 4 '18 at 2:44
  • $\begingroup$ I don't disagree---but I don't mean to suggest even using the language "isomorphism of representations" is necessary/appropriate---just that, since the answer already references representation theory, one could point out that the fact that since the J are all similar to one another, representation theory implies that the embeddings are all "essentially" the same, and in that sense there's nothing new to gain by studying the "non-canonical" embeddings. $\endgroup$ – Travis Willse Jul 4 '18 at 3:34

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