Hello, everyone!


Supposing that there is a unit vector in $n$-dimensional real space $\mathbf{x}\_1\in\mathbb{R}^n$, I want to get a group of $n-1$ vectors to form an orthogonal basis with $\mathbf{x}\_1$. One way to achieve this goal is to firstly randomly generate $n-1$ linear independent vectors and then orthogonalize them use the Gram-Schmidt process to obtain the orthogonality.


Then, I wonder if it is possible to achieve this goal by a group of orthogonal transformations, i.e., if there are $n-1$ orthogonal matrices $\{A\_i\}\\_{i=2}^n$ so that $\mathbf{x}\_i^\top\mathbf{x}\_j=0,\forall i\neq j$ where $\mathbf{x}\_i=A\_i\mathbf{x}\_1,i=2,\ldots,n$.


Is there any result about this question, please? Any suggestion will be welcome. Thank you very much!


Comment
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In the special case where $n=2$, it is possible by $A\_2=\begin{bmatrix} 0 & 1 \\\ -1 & 0 \end{bmatrix}$. It is straightforward to check that the orghotonality of $A\_2$ and $(A\_2\mathbf{x}\_1)^\top\mathbf{x}\_1=0,\forall\mathbf{x}\_1$. It is easy to comprehend because $A\_2$ is indeed a ration of 90 degree in a 2-dimensional plane.


However, it is not so simple in the case of $n=3$. First, to obatin the orthogonality by a transformation $A\_2$, i.e., $\mathbf{x}^\top A\\_2\mathbf{x}=0,\forall\mathbf{x}\in\mathbb{R}^n$, the following equation need to be satisfied.
\begin{align}
&a\\_{11}=a\\_{22}=a\\_{33}=0\\\
&a\\_{12}+a\\_{21}=a\\_{13}+a\\_{31}=a\\_{23}+a\\_{32}=0
\end{align}
where $a\_\{ij}$ is the $i,j$-th element of $A\_2$.


With the conditions above, it is straightforward to check that $|A\_2|=0$ and thus $A\_2$ cannot be an orthogonal matrix, which means there does not exist an orthogonal transformation to obtain an orthogonal vector to a given one.


In the case of $n=4$, it is possible to find an orthogonal transformation to obatin an orthogonal vector to a given one. But I did not find how to find 3 such transformations to form an orthogonal basis including the given vector.



Is it possible to obtain the vectors orthogonal to a given one by orthogonal transformations?
Adams gave the negative answer; in fact, his theorem is stronger, since there is no requirement that the operators $A\_i$ be given by orthonormal matrices.


The positive answer is given by the theorem of Radon-Hurwitz, which is usually described using the theory of Clifford algebras. Their problem is to classify collections of orthogonal $n\times n$ matrices $A\_1,A\_2,\dots,A\_k$ such that $A\_iA\_j=-A\_jA\_i$ for all $i\neq j$, and $A\_i^2=-I$.
Then it is automatic that the vectors $x,A\_1x,A\_2x,\dots,A\_kx$ are orthogonal. 


This kind of structure is called a module for the Clifford algebra $C\_k$. The Radon-Hurwitz result gives for each $n$ the largest $k=k(n)$ such that you can find a $C\_k$-module structure on $\mathbb{R}^n$; their lower bound turns out to be equal to Adams' upper bound.


You can easily find the explicit formula for $k(n)$ written down in various places: it has the form $k(n)=2d+\epsilon(d)$, where $d$ is the number of times $n$ is divided by $2$, and $0\leq \epsilon(d)\leq 2$ is a small number depending only on $d$ modulo $4$. From this, you can see that the only values of $n$ for which $k(n)=n$ are $1,2,4,8$.


