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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!

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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.

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It seems there is an orthogonal/skew-symmetric mix-up in the wording of this question, possibly from $SO_n$/$\mathfrak{so}_n$. –  Paul Reynolds Feb 17 '13 at 15:48
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2 Answers

up vote 4 down vote accepted

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 $1,2,4,8$.

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It's helpful of your answer. However there are some points in your answer that I cannot understand clearly since I do not major in mathematics. Could you please recommend some references or papers to explain your answer in more detail, please? Thank you very much! –  ppyang Feb 18 '13 at 9:32
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The kind of transformation that you want, i.e., a linear transformation $A$ such that $Ax\cdot x = 0$ for all $x$ and $|Ax|=|x|$, is possible only when the dimension of the space is even. The reason is that, in odd dimensions $A$ will always have a real eigenvector.

When the dimension of the space is divisible by $4$ you can have three orthogonal transformations $A_i$ such that, whenever $x$ is a unit vector, $x$, $A_1x$, $A_2x$, and $A_3x$ are orthonormal, and when the dimension is divisible by $8$, you can have $7$ such transformations. However, when the dimension is $16$, you can't get $15$ such transformations.

This (and the corresponding higher dimensional statements that say how many such transformations exist in any given dimension $n$) is a now-classical result about division algebras and vector fields that was long conjectured and finally proved by Adams in the 1950s.

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It's very kind of you for your help! Could you tell me the title of Adams's paper mentioned in your answer, please? I would like to understand how the result is obtained. Thank you! –  ppyang Feb 18 '13 at 8:12
    
I think both of your answers are very helpful. However, since I cannot accept both of your answers in the system and the answer given by Charles Rezk is in more detail, I choose to set his answer as the accepted answer. Anyway, it's very kind of you and thank you very much! –  ppyang Feb 18 '13 at 9:17
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