For which integers $d \geq 1$ can we find real matrices $R\_1, \dotsc, R\_d$ of size $d \times d$ such that for any unit vector $v \in \mathbb{R}^d$, $$R\_1 v, \dotsc, R\_d v$$ is an orthonormal basis? Note that the chosen set of $R\_i$s has to work simultaneously **for all** $v$. (Does this phenomenon have a name?)




---


It may not be obvious at first, but this question seems to be related to [associative division algebras](https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29) (i.e., reals, complex numbers, and quaternions). Indeed, I know that such matrices $R\_i$ can be found when $d=1,2,4$:


* $d=1$ is trivial—simply take $$R\_1 = \begin{pmatrix}1\end{pmatrix}$$
* $d=2$ is based on *complex numbers*—we represent $1$ and $i$ by $2 \times 2$ real matrices as follows:
$$
R\_1 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\qquad
R\_2 = \begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$
* $d=4$ is based on *quaternions*—we represent $1$, $i$, $j$, $k$ by $4 \times 4$ matrices [as follows](https://en.wikipedia.org/wiki/Quaternion#Matrix_representations):
$$
R\_1 = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\qquad
R\_2 = \begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}\qquad\\
R\_3 = \begin{pmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{pmatrix}\qquad
R\_4 = \begin{pmatrix}0&0&0&1\\0&0&-1&0\\0&1&0&0\\-1&0&0&0\end{pmatrix}\qquad
$$


Is this possible for any other $d$? I.e., can we find $R\_i$s even when there is no [associative division algebra](https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29) of dimension $d$?



Is an associative division algebra required for this phenomenon?
1. Consider the matrix $\sum\_i a\_iR\_i$. One can show that it sends every vector of length 1 to a vector of length $\sqrt{\sum\_i a\_i^2}$. It follows that if the norm of $a=(a\_i)$ is 1, then the matrix $\sum\_i a\_iR\_i$ is orthogonal. In particular, all the matrices $R\_i$ are orthogonal.
2. By multiplication from the left or from the right with an orthogonal matrix, we can assume without loss of generality that $R\_1=Id$ (the collection $\{SR\_iT\}$ where $S$ and $T$ are orthogonal matrices will still satisfy the condition.
3. Consider now $R\_2$. Then $R\_2$ cannot have a real eigenvalue $\lambda$, because then $R\_2-\lambda R\_1$ will have nontrivial kernel, which contradicts 1.
It follows that, since $R\_2$ is orthogonal, it can be written as the direct sum of $2\times 2$ rotation matrices. If one of these rotation angles is $\theta$, then by considering 1. for the matrix $R\_2+R\_1$ we reach the conclusion that $\cos(\theta)=0$. So $\theta=90$ or $270$. In any case, it follows that $R\_2^2 = -1$, and the same is true for $R\_i$ $i=3,\ldots, d$ of course. We also get that $d$ is even.
4. We can now multiply the collection $\{R\_i\}$ from the left by $-R\_2$. Again we will receive a collection of matrices which satisfies the condition, and contains the matrix $-R\_2^2=Id$. But then it follows that all the non-unital matrices in this collection square to -1. 
In other words, we get that $(R\_iR\_j)^2=-1$ for every $i\neq j$, $i,j>1$.
5. Consider now the algebra $\mathbb{R}\langle X\_1,\ldots X\_{d-1}\rangle/(X\_i^2=-1, X\_iX\_j + X\_jX\_i=0 (i\neq j))$. 
This algebra is a twisted group algebra with the group $G=(\mathbb{Z}/2)^{d-1}$.
This algebra is semisimple, and one can prove directly that its center is 2 dimensional and generated by $X:=X\_1\cdots X\_{d-1}$.   

If $r=2 \mod 4$, then $X^2=-1$. The center is then isomorphic with $\mathbb{C}$, so we get a central simple algebra of dimension $2^{d-2}$ over $\mathbb{C}$, which must be the matrix algebra $M\_n(\mathbb{C})$, 
where $n= 2^{(d-2)/2}$. This algebra has pnly one irreducible representation, and it is of dimension $2^{(d-2)/2}$ over $\mathbb{C}$, and therefore of dimension $2^{d/2}$ over $\mathbb{R}$. This rules out the possibility that $r=6,10,14...$.   

If $r=0 \mod 4$, then $X^2 = 1$, and the center is isomorphic with $\mathbb{R}\oplus\mathbb{R}$, and the algebra splits as the direct sum of two central simple $\mathbb{R}$-algebras, each of dimension $2^{(d-2)}$. 
Such an algebra will either be isomorphic with $M\_n(\mathbb{R})$ where $n = 2^{(d-2)/2}$, or with $M\_n(D)$ where $D$ is the quaternion algebra and $n= 2^{(d-6)/2}$.
In the first case the algebra has a unique simple representation of dimension $2^{(d-2)/2}$, and in the second case it has a unique irreducible representation of dimension $2^{(d+2)/2}$. For $d>8$ both dimensions are bigger than $d$ so there is no solution.
For $d=8$ we can study carefully the algebra and find a solution, the one given by Adam. In order to analyze the algebra we can, for example, consider the subalgebra generated by $X\_1,X\_2$ (in the proper quotient) which is isomorphic with the quaternion algebra. Then the algebra will be a tensor product of the quaternion algebra and its centralizer. This reduces the question to studying an algebra of a smaller dimension, and we can find out that at the end we get a matrix algebra over the reals, and so we have a solution for $d=8$.


