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The imaginary unit can be represented as real 2x2 matrix :

$ i = \begin{pmatrix} 0 & -1 \\ 1 & 0\\ \end{pmatrix},\, i^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1\\ \end{pmatrix} $

(see https://en.wikipedia.org/wiki/Imaginary_unit#Matrices). Now from my shallow knowledge of group representation theory I believe that it should also be possible to find higher dimensional real representations of $i$. I tried to find a 3x3 representation, which is basically asking if there is matrix square root like

$ \sqrt{\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & -1 \end{pmatrix}} \sqrt{\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & -1 \end{pmatrix}} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & -1 \end{pmatrix} $

but couldn't find any (by try and error). Nevertheless I was able to find a 4x4 matrix which can be used as representation of $i$:

$ \begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\0& 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\0& 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $

So can it be that $i$ can only be represented as real matrix of even dimensions $N=2n\, (n>0)$? Can this be proven and maybe a construction be given on how to construct higher dimensional real matrix representations of the imaginary unit? And can it be shown, that the 2x2 representation is really the lowest dimensional representation of $i$ as a real matrix?

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    $\begingroup$ If $A^2=-I_3$, then $(\det{A})^2 = -1$. $\endgroup$ – Catalin Zara Oct 7 '18 at 20:45
  • $\begingroup$ @CatalinZara Why not an official answer? $\endgroup$ – Paul Frost Oct 7 '18 at 22:56
  • $\begingroup$ @CatalinZara Ok, from math.stackexchange.com/questions/2408343/… I finally understand what you mean: "In odd dimensions however, there is no real matrices which are skew-symmetric and orthogonal. " $\endgroup$ – asmaier Oct 8 '18 at 20:35
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Suppose $T$ is an $n\times n$ matrix such that $T^2=-I$. Note then then $\mathbb{R}^n$ becomes a vector space over $\mathbb{C}$ (extending the $\mathbb{R}$-vector space structure), by letting $a+ib\in\mathbb{C}$ act by $aI+bT$. But then by picking a basis for this vector space over $\mathbb{C}$, we have $\mathbb{R}^n\cong \mathbb{C}^m$ for some $m$. As a real vector space, $\mathbb{C}^m\cong \mathbb{R}^{2m}$, and so it follows that $n=2m$ is even.

Conversely, for any $m$, $\mathbb{C}^m$ is a $2m$-dimensional real vector space, and multiplication by $i$ is an $\mathbb{R}$-linear map whose square is $-1$. So such a matrix does exist whenever $n$ is even. Explicitly, this is just the obvious generalization of your $4\times 4$ matrix, a block diagonal matrix with the $2\times 2$ diagonal matrix for $i$ on the diagonal blocks.

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I hope Catalin Zara's comment has convinced you that you can only find a $n\times n$ matrix $M$ with $M^2=-1$, if $n=2k$. Here I will construct such $M$ in full generality.

Let us make a few definitions. $$ \Sigma_{(p,q)} = \begin{pmatrix} 1_{p\times p} &0\\ 0 & -1_{q\times q} \end{pmatrix}, \qquad Y=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} $$ where $\Sigma_{(p,q)}$ is $(p+q)\times (p+q)$ , while $Y$ is $2\times 2$. Now, if you use Jordan normal form for real matrices, then with a little bit of work, you find that if $M$ satisfies $M^2=-1$, then its real Jordan normal form is necessarily of the form $J = \Sigma_{(p,q)}\otimes Y$ where $p+q=k$ and $\otimes$ is the Kronecker product. As a result, $M^2=-1$ if and only if $$ M = P^{-1}(\Sigma_{(p,k-p)}\otimes Y) P $$ where $P$ is an invertible real matrix and $0\leq p\leq k$. You can think of $p$ as "signature" of $M$, which classifies matrices with $M^2=-1$, up to a similarity relation.

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  • $\begingroup$ Could you elaborate on Catalin Zaras comment, because I didn‘t understand that. $\endgroup$ – asmaier Oct 8 '18 at 8:04

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