Is an associative division algebra required for this phenomenon? 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 (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:
$$
R_1 = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\qquad
R_2 = \begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}\qquad\\
R_3 = \begin{pmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{pmatrix}\qquad
R_4 = \begin{pmatrix}0&0&0&1\\0&0&-1&0\\0&1&0&0\\-1&0&0&0\end{pmatrix}\qquad
$$


Is this possible for any other $d$? I.e., can we find $R_i$s even when there is no associative division algebra of dimension $d$?
 A: *

*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.

*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. 

*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.

*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 $i\neq j$, $i,j>1$. 

*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 $d=8$.

A: One more example: for $d=8$ based on octonions. And this are all possible dimensions. $R_1$ takes vectors of length one to vectors of length one hence it is an isometry. By multiplying from the left by $R_1^{-1}$ we may assume that $R_1=\mathop{\rm Id}$. Then for every $v\in S^{d-1}$, the unit sphere in $\mathbb{R}^d$, the map $v\mapsto(R_2v,R_3v,\ldots,R_dv)$ gives a parallelization of the tangent bundle of $S^{d-1}$. Then by a celebrated Bott-Kervaire-Milnor theorem such a parallelization exists only when $d\in\{1,2,4,8\}$.
Edit: More explicitly, the solution for $d = 8$ is given by
$$
\sum_{i=1}^8 a_i R_i =
\begin{pmatrix}
 a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8 \\
 -a_2 & a_1 & -a_4 & a_3 & -a_6 & a_5 & a_8 & -a_7 \\
 -a_3 & a_4 & a_1 & -a_2 & -a_7 & -a_8 & a_5 & a_6 \\
 -a_4 & -a_3 & a_2 & a_1 & -a_8 & a_7 & -a_6 & a_5 \\
 -a_5 & a_6 & a_7 & a_8 & a_1 & -a_2 & -a_3 & -a_4 \\
 -a_6 & -a_5 & a_8 & -a_7 & a_2 & a_1 & a_4 & -a_3 \\
 -a_7 & -a_8 & -a_5 & a_6 & a_3 & -a_4 & a_1 & a_2 \\
 -a_8 & a_7 & -a_6 & -a_5 & a_4 & a_3 & -a_2 & a_1 \\
\end{pmatrix}
$$
A: This is an alternative way to see this question.
By argument given by the current answer
 given by Ehud Meir we
may assume that $R_1$ is the identity matrix. 
Then the hypotheses imply that for $i=2,\cdots,d$ the maps
$v\rightarrow R_i v$ are $d-1$ independent vector fields on the unit
sphere $S^{d-1}$. 
But it is known that the maximal number of independent vector fields
on the unit sphere $S^{n-1}$ is $n-1$ if and only if $n=1, 2, 4 $ or
$8$. 
This follows from a much stronger result due to J. F. Adams (Annals of Mathematics, 1962)
