Given $\hat v\in V$ find $\hat v_1,\dots,\hat v_k$ orthonormals such that $\hat v\cdot\hat v_i\ge0$ for any $i=1,\dots,k$

Question

So let be $V$ a (real or complex) vector space equipped with an inner product $\langle\,\cdot\,,\,\cdot\,\rangle:V\times V\rightarrow \Bbb R$ and so let be $\hat v\in V$ a unit vector. So if $\langle\hat v,\hat v\rangle=1$ then given a (orthonormal) basis $\mathcal E:=\{\hat e_i:i=1,\dots,k\}$ surely for any $r\le k$ there exist $i_1,\dots,i_r=1,\dots,k$ such that $$ v^{i_s}:=\langle\hat v,\hat e_{i_s}\rangle\neq0 $$ for any $i_s=i_1,\dots,i_r$. So supposing it is $i_1<\dots<i_r$ then I define $$ \vec v_i:=\begin{cases}v^{i}\hat e_{i},\,\,\,\text{if}\,\,\,i=i_1,\dots,i_r\\ \hat e_i,\,\,\,\text{otherwise}\end{cases} $$ for any $i=1,\dots,k$ so that $$ \langle\hat v,\vec v_i\rangle=\begin{cases}(v^i)^2,\,\,\,\text{if}\,\,\,i=i_1,\dots,i_r\\ 1,\,\,\,\text{otherwise}\end{cases}\ge0 $$ for any $i=1,\dots,k$. However unfortunately $\vec v_1,\dots,\vec v_k$ are not generally orthogonal but using Gram–Schmidt process it is surely possible to orthonormalise them but then I do not able to understand if $$ \langle\hat v,\hat v_i\rangle\ge0 $$ for any $i=1,\dots,k$. So how make the vectors $\vec v_1,\dots,\vec v_k$? Could someone help me, please?

Answer 1

Here is a simple argument for the real case. Assume $V=\mathbb{R}^k$. Fix a unit vector $v\in V$. Choose any orthonormal basis in $V$, say, $u_1,\dots u_k$. There certainly exist unit vectors $u\in V$ such that $\langle u_i,u\rangle\geq 0$ for all $i$. Choose one such vector $u$. Since the orthogonal group acts transitively on the unit ball of an inner product space, there exists an orthogonal transformation $T$ such that $Tu=v$. Put $v_i=Tu_i$. Then $$0\leq \langle u_i,u\rangle=\langle Tu_i,Tu\rangle = \langle v_i,v\rangle\quad (i=1,2,\dots k)$$

Answer 2

So let be $\varepsilon ^1,\dots,\varepsilon^k$ scalars defined thorugh the equation $$ \varepsilon^i:=\text{sgn}{\big(\langle\hat v,\hat e_i\rangle\big)} $$ for any $i=1,\dots,k$ so that let be $\hat v_1,\dots,\hat v_k$ vectors defined through the equations $$ \vec v_i:=\underbrace{\varepsilon^i\hat e_i}_{\text{no sum}}$$ for any $i=1,\dots,k$. So clearly $$ \langle\hat v_i,\hat v_j\rangle=\langle\varepsilon^i\hat e_i,\varepsilon^j\hat e_j\rangle=\varepsilon^i\varepsilon^j\langle\hat e_i,\hat e_j\rangle=\delta_{i,j} $$ for any $i,j=1,\dots,k$ and moreover $$ \langle\hat v,\hat v_i\rangle=\langle\hat v,\varepsilon^i\hat e_i\rangle=\varepsilon^i\langle\hat v,\hat e_i\rangle=\text{sgn}{\big(\langle\hat v,\hat e_i\rangle\big)}\langle\hat v,\hat e_i\rangle=\text{abs}{\big(\langle\hat v,\hat e_i\rangle\big)}\ge0$$ for any $i=1,\dots, k$ so that the statement follows.