Let $Bs=\{e_1,e_2,...,e_n\}$ is standard base $\mathbb R^{n}$. If $x_1,x_2,...,x_n$ vectors from space $R^{n}$ such that $e_i\in L(x_1,x_2,...,x_n)$, $i=1:n$ then set $\{x_1,x_2,...x_n\}$ is base of space $\mathbb R^{n}$?.
My answer is yes.
First I write $\alpha_1e_1+\alpha_2e_2+...+\alpha_ne_n=0$. Since $e_i\in L(x_1,x_2,...,x_n)$ we can write every vector in base $Bs$ as linear combination of $\{x_1,x_2,...,x_n\}$. For example:
$\begin{matrix} e_1=a_{11}x_1+a_{21}x_2+...+a_{n1}x_n\\ e_2=a_{12}x_1+a_{22}x_2+...+a_{n2}x_n\\ .........................\\ e_n=a_{1n}x_1+a_{2n}x_2+...+a_{nn}x_n \end{matrix}$
then $0=x1(\alpha_1a_{11}+\alpha_2a_{12}+...+\alpha_na_{1n})+x2(\alpha_1a_{21}+\alpha_2a_{22}+...+\alpha_na_{2n})+...+xn(\alpha_1a_{n1}+\alpha_2a_{n2}+...+\alpha_na_{nn})$
If we suppose opposite that this vectors is linear dependent, then one vector can be write as linear combination of other vectors, for example if I use $x_n$ to write as linear combination, then $(\alpha_1a_{n1}+\alpha_2a_{n2}+...+\alpha_{nn}a_{1n})\not=0$ because we can not divide something with zero, if we put $(\alpha_10+\alpha_20+...+\alpha_na_{nn})\not=0$ then $\alpha_n \not=0$,so now we have $0e_1+0e_2+...+\alpha_ne_n=0$, then $e_n=0$ but it is not true, so $ L(x_1,x_2,...,x_n)$ is base of $\mathbb R^n$, is this ok?
I don’t understand how you move from $(\alpha_1a_{n1}+\alpha_2a_{n2}+...+\alpha_{nn}a_{1n})\not=0 $ to $(\alpha_1 0+\alpha_2 0+...+\alpha_{nn}a_{1n})\not=0$?
However I would use the proof below.
Another proof
$(x_1,\dots,x_n)$ spans $(e_1,\dots,e_n)$ and therefore spans $\mathbb R^n$. As $(x_1,\dots,x_n)$ has $n$ elements, $(x_1,\dots,x_n)$ is a minimal family of vectors that spans $\mathbb R^n$. Hence $(x_1,\dots,x_n)$ is linearly independent and is a basis.