How to find a basis of $\mathbb R^n$ if we are given just one vector using the coordinated of the given vector Suppose we are given a non-zero vector
\begin{pmatrix}
           x_{1} \\
           x_{2} \\
           \vdots \\
           x_{n}
         \end{pmatrix}
How to find the a linearly independent basis for $\mathbb R^n$ using the coordinates of vector? Like for two dimensional case $$\begin{pmatrix} x_{1}\\ x_{2} \end{pmatrix}$$ $$\begin{pmatrix} x_{2} \\ -x_{1} \end{pmatrix}$$ are linearly independent and they span $\mathbb R^2$ and I want to see if I could do the same with $n$-dimensional case as well.
Maybe for simplicity we could assume the give vector is unit vector.
 A: An easy method: if $v$ is your non-zero vector, then $v_i$ is non-zero for some $i$. Form a basis by taking $v$ and every standard basis vector $e_j$, except $e_i$.
Why is this a basis? We have
$$v = a_1 e_1 + \ldots + a_n e_n,$$
where $a_i \neq 0$. We can write,
$$e_i = \frac{1}{a_i}v-\frac{a_1}{a_i} e_1 - \frac{a_2}{a_i} e_2 - \ldots - \frac{a_{i-1}}{a_i} e_{i-1} - \frac{a_{i+1}}{a_i} e_{i+1} - \ldots - \frac{a_n}{a_i} e_n.$$
This means $e_i$ is in the span of the other vectors. All the other standard basis vectors are in the set, so the set spans $\Bbb{R}^n$. As it contains $n$ vectors, it must be a basis.
A: Given what you specified in the comments under my other answer, about how the choice should be continuous, I have a definitive answer in $\Bbb{R}^3$ using the Hairy Ball Theorem. It states, if $f : S^2 \to \Bbb{R}^3$, where $S^2$ is the unit sphere in $\Bbb{R}^3$, is continuous and satisfies $f(\vec{x}) \cdot \vec{x} = 0$ for all $\vec{x} \in S^2$, then some $\vec{y} \in S^2$ exists such that $f(\vec{y}) = \vec{0}$.
Let's suppose such a function exists for vectors in $\Bbb{R}^3$. If we consider the second vector, then continuity dictates that this vector is a continuous function of the first. Let $f : \Bbb{R}^3 \setminus\{0\} \to \Bbb{R}^3 \setminus \{0\}$ be this continuous function.
Define a new function:
$$g : S^2 \to \Bbb{R}^3 : \vec{v} \mapsto f(\vec{v}) - (f(\vec{v}) \cdot \vec{v})\vec{v}.$$
Then $g$ is continuous, and
$$g(\vec{v}) \cdot \vec{v} = \vec{v} \cdot f(\vec{v}) - (f(\vec{v}) \cdot \vec{v})(\vec{v} \cdot \vec{v}) = 0,$$
as $\vec{v} \cdot \vec{v} = \|\vec{v}\|^2 = 1$. It follows therefore that $g(\vec{w}) = \vec{0}$ for some $\vec{w} \in S^2$. For this $\vec{w}$, we have
$$\vec{0} = g(\vec{w}) = f(\vec{w}) - (f(\vec{w}) \cdot \vec{w})\vec{w},$$
which is a non-trivial linear combination of $f(\vec{w})$ and $\vec{w}$ that equals $\vec{0}$. This means that $\vec{w}, f(\vec{w})$ is not linearly independent, hence cannot be the first two vectors of a basis.
The same proof works for $\Bbb{R}^n$ for odd $n \ge 3$.
A: Suppose all $x_i$ are not 0. Consider the matrix
$$
\begin{matrix}
x_1 & x_2 & \ldots x_n \\
x_1 & x_2 & \ldots -x_n \\
\ldots\\
x_1 & -x_2 & \ldots x_n \\
\end{matrix}
$$
Then the above determinant is not $0$.
Now if some $x_i $ are 0, for example for $n=4$ $x_3=x_4=0$ then consider the matrix with rows :
$$
\begin{matrix}
x_1 & x_2 & 0 & 0 \\
0 & x_1 & x_2 & 0 \\
0 & 0 & x_1 & x_2 \\
0 & 0 & x_1 & -x_2 \\
\end{matrix}
$$
