Let the standard orthogonal basis of $n$ dimensional space be written as:





Now create another orthogonal basis system by rotating the $x$ axis by $\arccos\frac{1}{\sqrt n}$ degrees, such that $e_1'=\frac{1}{\sqrt n}\sum_ie_i$.

How do we find out the direction of all other $e_i$ ?

For 2D case, the direction of $e_2'$ is $(-1,1)$.

For 3D case, it is a little bit tricky, for example, the direction of $e'_2$ is $(-2,1,1)$ and the direction of $e_3'$ is $(0,1,-1)$.

Are there any general formula to find out those basis $e_i'$ for any $n$-dimensions?

Now we have a point $p=(x_1,...,x_n)$ in the old coordinate, how do we find the new coordinate of the same point $p$ in the changed basis system, for the general $n$-dimensional case?

For example, the coordinate of point $(1,1)$ in 2D coordinate system, becomes $(\sqrt 2,0)$ in the new system.

  • 2
    $\begingroup$ Your question is unclear. In the 3D and higher cases, the requirement that the rotation takes $e_1$ to $e'_1$ does not uniquely specify the rotation. In particular, there are rotations satisfying that requirement that do not make the direction of $e'_2$ be $(-2,1,1)$. $\endgroup$
    – Lee Mosher
    Jun 2, 2019 at 16:07
  • $\begingroup$ @LeeMosher Yes I need to mention that the solution is not unique. $\endgroup$
    – High GPA
    Jun 2, 2019 at 16:13

1 Answer 1


Of course there are $\infty ^{n-1}$ sets of unit vectors, which are mutually orthogonal and orthogonal to the $n$D vector $(1,1, \cdots,1) / \sqrt {n}$.
They can be transformed into each other by a rotation around that base vector.

The quickest way I could find to build one of such sets starts from considering that:
- two vectors with a different number of null components are independent;
- two vectors which are orthogonal, remain orthogonal if multiplied by non-null scalars.

So, representing the vectors in column, the following matrix $$ \left( {\matrix{ 1 & 1 & 1 & 1 & \cdots \cr 1 & { - 1} & 1 & 1 & \cdots \cr 1 & 0 & { - 2} & 1 & \cdots \cr 1 & 0 & 0 & { - 3} & \cdots \cr \vdots & \vdots & \vdots & \vdots & \ddots \cr } } \right) $$ provides a set of vectors mutually orthogonal.
The next step is to normalize them, which is straightforward.

  • $\begingroup$ You are genius! $\endgroup$
    – High GPA
    Jun 3, 2019 at 10:10
  • $\begingroup$ @HighGPA; glad of the appreciation, wish it is what you are looking for $\endgroup$
    – G Cab
    Jun 3, 2019 at 10:43

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