The imaginary unit can be represented as real 2×2 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 3×3 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 4×4 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 2×2 representation is really the lowest dimensional representation of $i$ as a real matrix?
Best Answer
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.