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I understand that complex numbers have a real 2-by-2 matrix representation as per this discussion. This can be generalized: a complex $n$-by-$m$ matrix has a real (2$n$)-by-(2$m$) matrix representation. One simply constructs the larger matrix of real entries by replacing each entry of the smaller complex matrix with a 2-by-2 real matrix as per the first representation.

How can I prove that this construction preserves multiplication and inverses? That is, if $S$ is the transformation from the set of $n$-by-$m$ complex matrices to the set of (2$n$)-by-(2$m$) real matrices, I want to show that:

  1. $S(AB) = S(A)S(B)$
  2. $S(A^{-1}) = (S(A))^{-1}$, where we assume $n=m$.

This is easy to show by brute force for the case $n=m=2$. However, I'm quite lost on how to do this for general $n$ and $m$. The case of the inverse seems especially difficult to me. Any advice would be appreciated!

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Note first: It is enough to show $S(AB) = S(A)S(B)$ and that $S(I_n) = I_{2n}$ (where $I_n$ and $I_{2n}$ are the unit matrices of the corresponding dimensions), because the law about the inverses then follows directly: If $A \in \mathbb{C}^{n \times n}$ is invertible, we have $$I_{2n} = S(I_n) = S(A A^{-1}) = S(A) S(A^{-1}).$$ Thus, the matrix $S(A)$ is invertible with inverse $S(A^{-1})$, and that shows $S(A)^{-1} = S(A^{-1})$.

Now, the identity $S(I_{n}) = I_{2n}$ is clear from the definition of your representation $S$, so the only thing left to show is the multiplicativity of $S$. The magic for showing this lies in finding a good notation for this kind of representation. Namely, recall first the $2 \times 2$-representation of complex numbers $$S_{22} : \mathbb{C} \to \mathbb{R}^{2 \times 2}, a + ib \mapsto S_{22}(a + ib) = a I_2 + b J_2,$$ where $I_2$ is the $2 \times 2$ unit matrix and $J = \left( \begin{array}{clcr} 0 & -1\\1 & 0 \end{array} \right).$ For this case, the identity $S_{22}(AB) = S_{22}(A)S_{22}(B)$ is just a small calculation. Now, let us use block matrix notation for the higher representations: your representation $S$ can be written as the following function:

$$S : \mathbb{C}^{m \times n} \to \mathbb{R}^{2m \times 2n}, \, \left(\begin{matrix} z_{11} & \dots & z_{1n} \\ \vdots & \ddots & \vdots \\ z_{m1} & \dots & z_{mn} \\ \end{matrix} \right) \mapsto \left[\begin{matrix} S_{22}(z_{11}) & \dots & S_{22}(z_{1n}) \\ \vdots & \ddots & \vdots \\ S_{22}(z_{m1}) & \dots & S_{22}(z_{mn}) \\ \end{matrix} \right]. $$

The square brackets here mean that the entries are not numbers, but they are themselves matrices (I do not know how familiar you are with block matrices). Now, you only need to know that you can carry out multiplication of block matrices basically in the same way as if the entries were numbers: you can just write out the usual formula for multiplication of matrices by summing up certain products of entries, just that you now have to pay attention to the order of the products since the entries are matrices, so in general non-commutative (see also the description on wikipedia, for example). This means that if you two matrices $A = (z_{ij})_{ij} \in \mathbb{C}^{m \times n}$ and $B = (w_{ij})_{ij} \in \mathbb{C}^{n \times r},$ we can calculate

$$S(A)S(B) = [S_{22}(z_{ij})]_{ij} \cdot [S_{22}(w_{jk})]_{jk} = \left[ \sum_{j = 1}^n S_{22}(z_{ij}) \cdot S_{22}(w_{jk}) \right]_{ik} = \left[ \sum_{j = 1}^n S_{22}(z_{ij}w_{jk}) \right]_{ik} = \left[ S_{22}(\sum_{j = 1}^n z_{ij}w_{jk}) \right]_{ik} = S(AB).$$

Here, we used the property $S_{22}(AB) = S_{22}(A) S_{22}(B)$ that we can prove by hand.

Now, this is a bit of cheating, because I now basically just used the magical property of block matrices that I can use their "nice" way of multiplying subblocks with one another. But that property is easy to convince yourself of, and not strictly difficult to prove, it just gets notationally ugly. And being able to mess around with block matrices is really handy, so I suggest playing around with those a bit until the above calculations seem believable to you.

Lukas Miaskiwskyi
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  • Thanks for explaining so carefully! I learnt a lot by going through your solution. –  Oct 06 '19 at 21:21
  • I do have one follow-up question, though. It's clear that if $A$ is invertible, then so is $S(A)$. But if $S(A)$ is invertible, does it necessarily follow that $A$ is invertible? –  Oct 07 '19 at 00:59
  • Yup, it does. One way to see this is, for example, by proceeding by contraposition: if $A$ is not invertible, then, for example, its columns $a_1, \dots, a_n \in \mathbb{C}^n$ would be linearly dependent. But then, if you apply $S_{22}$ componentwise to the $a_i$, so that $S_{22}(a_i) \in \mathbb{R}^{2n \times 2}$, because of linearity of the $S_{22}$, also the set of matrices $S_{22}(a_1), \dots, S_{22}(a_n)$ would be linearly dependent. But the matrix $S(A)$ is formed out of these matrices, and by that one can follow that $S(A)$ has linearly dependent columns, so it can't be invertible. – Lukas Miaskiwskyi Oct 07 '19 at 20:37