Question(s) regarding complex numbers and representation of quantities in general with matricies.

Question

A complex number is often simply cited to map to specific matricies:

$$ z = a+ib\mapsto\begin{pmatrix}a&b\\-b&a\end{pmatrix} \text{ where it is the case that } 1=e_1\mapsto\begin{pmatrix}1&0\\0&1\end{pmatrix} , i=e_2\mapsto\begin{pmatrix}0&1\\-1&0\end{pmatrix} $$

The response to questions regarding this representation of complex numbers are often met with the answer that it is sufficient that these matrix representations satisfy all the normal field axioms and whatever necessary properties of complex numbers such as associativity and commutativity.

This is fine, and technically true but it is helpful to understand the motivation for creating such representations as well as how it might be done in general.

Paul Garret states here that these representations may be made in general by considering how a complex number acts on the 'basis vectors' of $\mathbb{C}$.

$$ 1z = e_1*(e_1a+e_2b)=e_1a+e_2b= \begin{pmatrix}e_1&e_2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}= \begin{pmatrix}a&b\end{pmatrix}\begin{pmatrix}e_1\\e_2\end{pmatrix} \\ iz=e_2*(e_1a+e_2b)=e_2a-e_1b= \begin{pmatrix}e_1&e_2\end{pmatrix}\begin{pmatrix}-b\\a\end{pmatrix}= \begin{pmatrix}-b&a\end{pmatrix}\begin{pmatrix}e_1\\e_2\end{pmatrix} $$

For the sake of clarity I will specify that row-column matrix product above simple represents a linear combination of the basis $e_1$, $e_2$ with the components of the components $a$, $b$. In this notation, the operation between the matricies is simple matrix multiplication. I included the reversed form since each representation produces the same linear combination.

We may consider both equations at once by considering the tensor product:

$$ \begin{pmatrix}e_1\\e_2\end{pmatrix}\begin{pmatrix}e_1&e_2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}= \begin{pmatrix}e_1&e_2\\e_2&-e_1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}= \begin{pmatrix}e_1a+e_2b\\-e_1b+e_2a\end{pmatrix}= \begin{pmatrix}a&b\\-b&a\end{pmatrix}\begin{pmatrix}e_1\\e_2\end{pmatrix} $$

Question: What is the justification for removing explicit reference to the basis - the $\begin{pmatrix}e_1&e_2\end{pmatrix}^\intercal$ matrix - and making the identification:

$$ a+ib=\begin{pmatrix}a&b\\-b&a\end{pmatrix} $$

Below I have constructed a 'proof' to demonstrate my reasoning on the matter, but I feel that I am lacking sufficient justification for the seemingly arbitrary tensor product and commutation of the complex scalar quantity with the basis matrix.

In general scalars commute with matricies, and the tensor product of a column matrix and a row matrix is a well defined matrix multiplication. I am not confident in my understanding. Below is an attempt to algebraically prove that every complex number $z$ has a unique matrix representation.

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Suppose that $z\in\mathbb{C}$ is a scalar quantity and commutes with any given matrix. So potentially we could write:

$$ \begin{pmatrix}e_1\\e_2\end{pmatrix}z= \begin{pmatrix}e_1z\\e_2z\end{pmatrix}= \begin{pmatrix}ze_1\\ze_2\end{pmatrix}= z\begin{pmatrix}e_1\\e_2\end{pmatrix} $$

From above we have:

$$ z\begin{pmatrix}e_1\\e_2\end{pmatrix} =\begin{pmatrix}a&b\\-b&a\end{pmatrix}\begin{pmatrix}e_1\\e_2\end{pmatrix} $$

Then subtract the left side from the right side and factor the basis matrix to obtain:

$$ {\bf 0} =\begin{pmatrix}0\\0\end{pmatrix} =\begin{pmatrix}a&b\\-b&a\end{pmatrix}\begin{pmatrix}e_1\\e_2\end{pmatrix} -z\begin{pmatrix}e_1\\e_2\end{pmatrix} = \left[\begin{pmatrix}a&b\\-b&a\end{pmatrix} -z\right]\begin{pmatrix}e_1\\e_2\end{pmatrix} $$

Since $\begin{pmatrix}e_1&e_2\end{pmatrix}^\intercal\ne{\bf 0}$ then it must be the case that:

$$ 0=\begin{pmatrix}a&b\\-b&a\end{pmatrix} -z \Rightarrow z=\begin{pmatrix}a&b\\-b&a\end{pmatrix} $$

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Obviously the above is not a completely rigorous proof, but I was trying to fill in missing details from resources I lack and books I have not yet read.

Answer 1

In Representation Theory, one is interested in giving a correspondence: $$g\in G \leftrightarrow [\rho(g)]_{\beta}$$ between group elements and matrices (or linear operators in the absence of a basis).

This is done in such a way as to preserve the structure of the group (i.e. we make $\rho:G\to GL_n(V/F)$ a group-homomorphism). Alternatively, one may construct a group action: $$\theta: G\times V\to V$$ $$(g,v)\mapsto \theta(g,v)$$ that gives rise to the desired hom in the obvious way: $$\forall v\in V,\text{ }\text{ }\rho(g)v:= \theta(g,v).$$ When there is more structure involved (say field structure), one can require $\rho$ to be a field-homomorphism instead to make sure things play nicely in the image.

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Now, if we take $G:= \mathbb{C}$ and $V/F:= \mathbb{C}/\mathbb{R}$. We may give the values for the action $\theta$ as you have partially done above in the post.

Whereas for finite groups, we may list $\theta$ quickly by defining the action of conjugacy class representatives (e.g. cycle types for $G:= S_n$), complex multiplication is commutative, so each element defines its own conjugacy class (and hence is not theoretically useful here). Instead, we have the basis, $\beta:= \{1,i\}$, at our disposal. We define the actions of $1$ and $i$ (from $G$) on $1$ and $i$ (from $V$): $$\theta(1,1) := 1$$ $$\theta(1,i) := i$$ $$\theta(i,1) := i$$ $$\theta(i,i) := -1$$ and extend by linearity in both variables to get the desired $\theta$.

Then since $\rho(g)v = \theta(g,v)$, it should be clear with linear algebra intuition, that: $$\big[\rho(\color{blue}{a+bi})\big]_{\beta}[v]_{\beta} = \color{blue}{\begin{bmatrix}a&& -b\\ b&& a\end{bmatrix}}[v]_{\beta}$$

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To elaborate a little:

(1) Matrix representations for linear operators (a.k.a. transformations $T$) are given by listing the coordinate reps for its images of basis vectors--which here looks like: $$[T]_{\beta} = \bigg[[Te_1]_{\beta}\text{ }\text{ }[Te_2]_{\beta}\bigg].$$ (2) It should be further noted that $\theta$ satisfies: $$\theta(1,v) = v$$ $$\theta(g,\theta(h,v)) = \theta(gh,v)$$ so that it is a group action. The second line says complex multiplication is associative.

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The point of this answer was to phrase things in terms of Representation Theory so that the motivation is more clear. The actual construction of $\theta$ here is straight forward, but things complicate when you define other actions where $V$ is different.