I need to prove that for all $n\in \mathbb{N}$ exists a matrix $A$ $\in \mathrm{SO}(\mathbb{R}^n)$ (orthogonal matrixes with determinant equal to 1) such that the inputs of the first row are equal to $1/\sqrt{n}$.
For example, for $n = 2$ we have that A must be:
$$A=\begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ a & b \end{pmatrix}$$
If $A$ has to be orthogonal it has to verify $AA^T = Id$ which gives the following equation system:
$$\left.\begin{matrix} a^2 + b^2 = 1\\ \frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}} = 0 \end{matrix}\right\} \Rightarrow \left.\begin{matrix} a = \pm \frac{1}{\sqrt{2}}\\ b = \mp \frac{1}{\sqrt{2}} \end{matrix}\right\}\overset{det(A)=1}{\Rightarrow }\left.\begin{matrix} a = - \frac{1}{\sqrt{2}}\\ b = + \frac{1}{\sqrt{2}} \end{matrix}\right\}$$
The problem is that I don't know how to do it for an arbitrary $n$. I tried to write the equations similarly to the example but that wasn't successful.
I would appreciate any help. Thanks!
by Gram–Schmidt process you can always create an orthonormal basis starting from a basis that contains the given vector $v_1=(\frac{1}{\sqrt{2}},...)$
https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process
putting the vectors of the basis in a matrix (as rows or columns) you obtain an orthogonal matrix