Determine the basis for a vector space $V$? Problem:
I have 5 vectors $v_1, v_2,...,v_5$ each of them having $5$ components:
$v_1 = \left[\begin{matrix} 5 \\ 4 \\ 3\\ 2 \\ 1  \\\end{matrix}\right] $ 
$v_2 = \left[\begin{matrix} -1 \\ 2 \\ 0 \\ -2 \\ 1  \\\end{matrix}\right] $ 
$v_3 = \left[\begin{matrix} 8 \\ 7 \\ 6 \\ 5 \\ 4  \\\end{matrix}\right] $ 
$v_4 = \left[\begin{matrix} 0 \\ 3 \\ 1\\ -1 \\ 2  \\\end{matrix}\right] $ 
$v_5 = \left[\begin{matrix} 10 \\ 8 \\ 6\\ 4 \\ 2  \\\end{matrix}\right] $ 
The question is determine a basis $B = \{b_1, b_2, ...\}$ for the vector space $V = span(v_1, v_2, ... , v_5)$.
What I know:
This is the way I understand the concept of basis: a set with the minimum number of vectors that combined can represent all other vectors in a vector space.
The concept of span is also familiar: all possible linear combinations of some vectors.
I have seen that to find the basis, I have basically to make the matrix created by putting next to each other each of the vectors in $RREF$.
We can observe from the problem that the $v_1$ is the double of $v_5$, and that all components of $v_2$ are smaller exactly one unit respect to the components of $v_4$.
Questions:
1 .Is $\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\\end{matrix}\right]$ a basis for the vector space $V$, no matter the values of the vectors? 


*If I use the RREF I can find a basis, what about if I want to find others?

 A: Q1. The columns of the identity matrix indeed form a basis of $\mathbb{R}^{5}$; the whole $5$-dimensional space. It is in fact the standard basis. 
However, we are interested in the subspace $V  = \text{span}(\lbrace v_{1}, \dots, v_{5}\rbrace)$. And as you mentioned, a basis is a minimal set. 
 $V$ is clearly a subspace of $\mathbb{R}^{5}$. If the span of these five vectors coincides with the entire $\mathbb{R}^{5}$, then indeed your proposed basis works. This happens if the $5$ vectors are linearly independent. If they are not, then we can describe their span using fewer vectors and the columns of the identity matrix are no longer a basis because they are not a minimal set.
Q2 There are infinitely many bases for a subspace of $\mathbb{R}^{n}$.
RREF allows us to identify a (maximal) subset of the vectors that are linearly independent and can hence be used to form a basis. A (nice) property of this approach is that it yields a basis that it consists of vectors among the original set. 
Another well known procedure to extract a basis is the Gram-Schmidt process. Starting from a vector of our original set, we end up with an orthogonal (or orthonormal) basis.
Once we have identified a basis $B = \lbrace b_{1}, \dots, b_{k} \rbrace$ for $V$, we can use that to produce many different bases.
Let $\mathbf{B} = \left[ b_{1}, \dots, b_{k}\right]$ be the $n \times k$ matrix formed by stacking the basis vectors (here, $n=5$).
Let $\mathbf{C}$ be an arbitrary $k \times k$ matrix.
Then, each column of the $n \times k$ matrix 
$$
\widehat{\mathbf{B}} = \mathbf{B}\mathbf{C}
$$
is a linear combination of the columns of $\mathbf{B}$ and hence lies in $V$.
If $\mathbf{C}$ is full rank (i.e., if its columns are linearly independent), then the columns of $\widehat{\mathbf{B}}$ will also be linearly independent (Note that the columns of $\mathbf{B}$ are by definition linearly independent). Hence, the columns of $\widehat{\mathbf{B}}$ also form a basis of $V$. Choosing different full-rank $k\times k$ matrices  $\mathbf{C}$ will yields new bases for $V$.
A additional remark: Interestingly, if we know how many vectors among $v_{1}, \dots, v_{5}$ are linearly independent (say $k$, here $1\le k \le 5$), then we could also take an equal number of random linear combinations of all vectors (say with coefficients selected i.i.d. according to the normal distribution). The $k$ created vectors will also be linearly independent (and hence, form a basis for $V$) with probability $1$.
