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?

  • 1
    $\begingroup$ 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

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 $x \neq 0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.