Let's say $B=(e_1, ..., e_n)$ is a standard basis of a space $\mathbb{R}^n$. If $x_1, x_2, ... x_n$ are vectors from $\mathbb{R}^n$ such that $e_i\in L(x_1, ..., x_n)$ where $i=1:n$. Check if set $A=(x_1, ..., x_n)$ is also a basis of $\mathbb{R}^n$.
I know that basis is the largest linearly independent set, also, basis is the smallest set that generates given space, since $A$ and $B$ have the same number of elements i have to prove that $A$ is linearly independent set in order to prove that it is a basis of $\mathbb{R}^n$, but i don't know how to do it. Any help appreciated!
Since each $e_i$ is in the span of $A$, we know that $A$ spans $\mathbb{R}^n$. This means $A$ is a basis of $\mathbb{R}^n$, since it is a spanning list of length $n= \dim \mathbb{R}^n$.
Expanding on my first statement, since each $e_i$ is in the span of $A$, we know there exist real numbers $C_{ij}$ with $1\leq i \leq n$ and $1 \leq j \leq n$ such that for each integer $1\leq i \leq n$, we have $$e_i = \sum_{j=1}^n C_{ij}x_j.$$ Since the $e_i$'s span $\mathbb{R}^n$, we have, for any $v\in\mathbb{R}^n$, real numbers $\alpha_i$ such that
\begin{align*} v &= \sum_{j=1}^n \alpha_ie_i \\ &= \sum_{i=1}^n \alpha_i \sum_{j=1}^n C_{ij}x_j, \end{align*}
which is definitely en element of $\text{span}(A)$, hence $\mathbb{R}^n \subseteq \text{span}(A)$. Since vector spaces are closed under linear combination, we also know $\text{span}(A) \subseteq \mathbb{R}^n$, and so $\mathbb{R}^n = \text{span}(A)$.