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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?

Tom
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    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. – JMoravitz Jul 04 '18 at 00:49

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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 $x \neq 0$.

Ben Grossmann
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    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. – Travis Willse Jul 04 '18 at 02:39
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    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. – Ben Grossmann Jul 04 '18 at 02:44
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    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. – Travis Willse Jul 04 '18 at 03:34
  • Can we say all $J$ has the form $J=O^T \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}O$ for some orthogonal matrices $O$? – khashayar Nov 24 '23 at 07:58
  • @khashayar We cannot. There are only two matrices of that form, but infinitely many valid $J$. We can say, however, that $$ J = S^{-1} \pmatrix{0&1\-1&0} S $$ for some invertible matrix $S$. – Ben Grossmann Nov 24 '23 at 14:05
  • @khashayar As an earlier comment mentions, the answer is unique up to conjugation (i.e. up to matrix similarity), but there is no requirement that this similarity matrix should be orthogonal. Using an orthogonal matrix would preserve the skew-symmetry of $J$. – Ben Grossmann Nov 24 '23 at 14:11
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    But in that case, the embedding from $\mathbb{C}$ to $2 \times 2 $ real matrices is not convolutive anymore. Am I right? I asked a somewhat similar question here https://math.stackexchange.com/questions/4813180/algebra-embedding-from-complex-matrices-into-real-matrices but no one answered yet. Would you take a look? – khashayar Nov 24 '23 at 19:59
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    @khashayar I haven't heard the term convolutive; I assume you mean "adjoint preserving" essentially. Nothing about this question requires that we satisfy this condition. But yes I suspect you're right – Ben Grossmann Nov 24 '23 at 20:45
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    Yes, I mean adjoint preserving or -algebra embedding when we consider to be conjugation in $\mathbb{C}$ and to be transpose in real matrices. Thank you for the explanation. – khashayar Nov 24 '23 at 21:25