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.

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.
