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!


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.

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

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 $1,2,4,8$.

  • $\begingroup$ 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! $\endgroup$ – ppyang Feb 18 '13 at 9:32

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.

  • $\begingroup$ 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! $\endgroup$ – ppyang Feb 18 '13 at 8:12
  • $\begingroup$ 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! $\endgroup$ – ppyang Feb 18 '13 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.