Orthonormal basis of $\mathbb{R}^n$ containing $v_1 = \frac{1}{\sqrt{n}}(1,1,\ldots,1)$

Question

I would like to construct the orthonormal basis $\{v_1,v_2,\ldots,v_n\}$ of $\mathbb{R}^n$ with $v_1 = \frac{1}{\sqrt{n}}(1,1,\ldots,1)$. I am looking for an analytic formulation of all vectors of the basis.

Any hints or idea are very welcome :). Thanks.

Answer 1

Due to a dire need to procrastinate I decided to work out explicitly the two suggestions that have been made so far.

Experimentally, it seems that applying the Gram-Schmidt process to the vectors $v_1, e_1, e_2, \ldots, e_{n-1}$ gives a very neat result: $$\begin{matrix} v_1 = \Big( & \frac1{\sqrt n}, & \frac1{\sqrt n}, & \frac1{\sqrt n}, & \ldots, & \frac1{\sqrt n} & \Big), \\ v_2 = \Big( & \sqrt{\frac{n-1}n}, & \frac{-1}{\sqrt{n(n-1)}}, & \frac{-1}{\sqrt{n(n-1)}}, & \ldots, & \frac{-1}{\sqrt{n(n-1)}} & \Big), \\ v_3 = \Big( & 0, & \sqrt{\frac{n-2}{n-1}}, & \frac{-1}{\sqrt{(n-1)(n-2)}}, & \ldots, & \frac{-1}{\sqrt{(n-1)(n-2)}} & \Big), \\ v_4 = \Big( & 0, & 0, & \sqrt{\frac{n-3}{n-2}}, & \ldots, & \frac{-1}{\sqrt{(n-2)(n-3)}} & \Big), \\ \vdots \end{matrix}$$ all the way down to $$\begin{matrix}v_n = \Big(0, & \ldots, & 0, & \sqrt{\frac12}, & \frac{-1}{\sqrt{2\cdot1}}\Big).\end{matrix}$$ So apparently the last $n-2$ vectors are the same as those for $\mathbb R^{n-1}$, just lifted to $\mathbb R^n$ by prepending a $0$. Presumably someone can prove it via induction.

On the other hand, inspired by MPW's answer we can take the unique reflection $T$ such that $Te_1=v_1$, and then let $v_i = Te_i$. That means $T = I - 2uu^T$, where $u=(e_1-v_1)/\|e_1-v_1\|$, which after some algebra gives $$\begin{align} T &= I-2\frac{(e_1-v_1)(e_1-v_1)^T}{(e_1-v_1)^T(e_1-v_1)} \\ &= I - \frac1{n-\sqrt n}\Big(nE_{11} - \sqrt nE_{1*} - \sqrt nE_{*1} + E_{**}\Big), \end{align}$$ where I'm using $E_{ab}$ to denote the matrix with ones in row(s) $a$ and column(s) $b$ and zeroes everywhere else. This turns out to be $$T=\begin{pmatrix} \frac1{\sqrt n} & \frac1{\sqrt n} & \frac1{\sqrt n} & \ldots & \frac1{\sqrt n} \\ \frac1{\sqrt n} & 1-\frac1{n-\sqrt n} & \frac{-1}{n-\sqrt n} & \ldots & \frac{-1}{n-\sqrt n} \\ \frac1{\sqrt n} & \frac{-1}{n-\sqrt n} & 1-\frac1{n-\sqrt n} & \ldots & \frac{-1}{n-\sqrt n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac1{\sqrt n} & \frac{-1}{n-\sqrt n} & \frac{-1}{n-\sqrt n} & \ldots & 1-\frac1{n-\sqrt n}, \end{pmatrix}$$ and the vectors $v_i$ are just the $n$ columns of $T$.

Both solutions are quite nice. The solution from reflection is more symmetrical, while the one from Gram-Schmidt has a more inductive structure.

Answer 2

You can use the Gram-Schmidt process.

Answer 3

Are the $n$ vectors given or not? If you already have a set of $n$ vectors containing $v_1$, you can employ Gram-Schmidt process. Otherwise, you need to obtain $n$ linearly independent vectors first.

Answer 4

Start with the canonical orthonormal basis $\{e_1,\ldots,e_n\}$. Let $T$ be the rotation that sends $e_1$ to your $v_1$. Then $\{Te_1,\ldots,Te_n\}$is what you want.