# Basis vectors as the elements of a linear combination.

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 $A$ spans $B$ and $B$ spans the whole space we get that $A$ also spans $\mathbb R^n$. If there was a linearly dependent vector, $A$ would have at most $n-1$ linearly independent vectors, and thus it cannot span the whole space. – Dog_69 May 23 '18 at 18:06
• Could you explain the first part of your comment please? – cdummie May 23 '18 at 19:33
• Yes. The pint is that any vector $v\in\mathbb R^n$ can be express as a linear combination of the $e_i's\in B$. And since each $e_i$ belongs to the span of $A$ we can write $e_i$ in terms of the $x_j$. So suppose $v=\sum_i v_ie_j$ and $e_i=\sum_j a_{ij} x_j$. Then $v=\sum_{i,j} v_i a_{ij} x_j$ which can be rewritten as $\sum_i v'_ix_i$, where $v'_i = \sum_j v_ja_{ji}$. Then my statement. – Dog_69 May 23 '18 at 19:39

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 $\mathbb{R}^n = \text{span}(A)$.

• you said that "Since each $e_i$ is in the span of $A$, we know that $A$ spans $\mathbb{R}^n$." Am i actually allowed to state that without any mathematics behind it? – cdummie May 23 '18 at 19:32
• @cdummie: The mathematics behind that statement are in my new comment. – Dog_69 May 23 '18 at 19:40
• @cdummie See the edit of my post. – Mapa May 23 '18 at 20:38
• Thanks for the help guys, i appreciate it! – cdummie May 24 '18 at 7:50

There are a number of equivalent ways to do this. The easiest to state is that the determinant of A must be nonzero. So just compute the determinant. If it is nonzero, then you're done.