Vtyjkyuk

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 $e_iin L(x_1,x_2,...,x_n)$, $i=1:n$ 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 $e_iin L(x_1,x_2,...,x_n)$ we can write every vector in base $Bs$ as linear combination of $x_1,x_2,...,x_n$. For example:

$beginmatrix
e_1=a_11x_1+a_21x_2+...+a_n1x_n\
e_2=a_12x_1+a_22x_2+...+a_n2x_n\
.........................\
e_n=a_1nx_1+a_2nx_2+...+a_nnx_n
endmatrix$

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 $(alpha_1a_n1+alpha_2a_n2+...+alpha_nna_1n)not=0$ 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_nna_1n)not=0
$ to $(alpha_1 0+alpha_2 0+...+alpha_nna_1n)not=0$?

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.