Given a vector equation with $n$ vectors, how can we determine if the span of the vectors is equal to $\mathbb{R}^n$ So I've recently started taking Linear Algebra, and I've been thinking about how to determine if the linear combinations of any n vectors can represent any vector in $\mathbb{R^n}$. More formally put, I am looking for a way to determine if, for n vectors $\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}$... $\vec{v_n}$ have the property that $\mathrm{Span}(\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}$... $\vec{v_{n}})$ = $\mathbb{R^n}$. 
So I started to approach this problem by looking at the basic unit vectors $\vec{i}, \vec{j}, \vec{k}$. I know that any vector in $\mathbb{R^3}$ can be represented by the span of these three vectors alone (it is quite intuitive, take some scalar times the first to get the first term of the vector, some scalar times the second, etc...).
If we write out the augmented matrix for these three vectors and some vector $\vec{v}$ with elements $a, b, c$ we see the following: 
\begin{bmatrix}
1 &  0&0& a \\ 
0 &  1& 0& b\\ 
0 &  0& 1& c
\end{bmatrix}
The first weight for the linear combinations is strictly set to be $a$, the second weight is $b$, and the third is $c$. Since, for any vector, these three vectors $i, j, k$ will always have solutions for the weights, namely a, b, and c. 
So, for any vector to be in the span of n vectors, we need a specific solution for each of the weights. (no free variables) In order to accomplish this, we would need the reduced row echelon form for any n vectors to be of the above "triangular" form, with $1's$ going down the diagonal. 
Is the above statement correct, or are there other cases in which this is true? Thanks!
 A: Yes, the span of the columns of a matrix is the entire $\mathbb R^n$ if and only if the row echelon form has a leading one in each row. For a square matrix this is exactly when the reduced row echelon form is the identity matrix!
In your argument you can note that the $(a\,b\,c)$ column is never actually used for anything during Gaussian elimination, except if you have already found that the row-echelon form of the non-augmented matrix has too few nonzero rows. And if it does, you can always at that point choose some $a,b,c$ that makes the bottom right element nonzero, such that the system doesn't have a solution. So you don't actually need to augment the matrix with a constants column -- doing Gaussian elimination of the non-augmented matrix will tell you all you need to know in this particular case.
A: It can be shown that $n$ vectors in $\mathbb R^n$ span $\mathbb R^n$ if and only if the reduced row echelon form is the identity. Another equivalent condition is that the matrix formed by the vectors is invertible. 
A: There are quite a number of equivalent statements which would say that n vectors span Rn.  Two of those have been stated above.  You could also say that the triangular matrix produced by row elimination has no zero rows,  or that the vectors are linearly independent, or that the null-space of the matrix A they form is {0}, or that all the eigenvalues of A are non-zero, or that the vectors are a basis for the space Rn, or that Ax =b has a unique solutions for every b, that A-1 exists.  With a little time I can probably think of others. 
All these statements are equivalent, and it's very useful to have so many ways of looking at it.  Depending on what problem you are trying to deal with, you get to choose which definition you will use to show that your n vectors span Rn. 
