# Changed the orthogonal basis, how to find the new coordinate of the same point?

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

$$e_1=(1,0,...)$$

$$e_2=(0,1,0,...)$$

...

$$e_n=(0,...,0,1)$$

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.

• 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 $(-2,1,1)$. Jun 2, 2019 at 16:07
• @LeeMosher Yes I need to mention that the solution is not unique. Jun 2, 2019 at 16:13

Of course there are $$\infty ^{n-1}$$ sets of unit vectors, which are mutually orthogonal and orthogonal to the $$n$$D vector $$(1,1, \cdots,1) / \sqrt {n}$$.
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.