Why is the reduced echelon form of a set of independent vectors, the identity matrix? 
If a matrix has linearly independent rows, then its  reduced echelon form is  the identity matrix.  

I haven't found a concise explanation for this... I have the whole notion in my head but I cannot express this in words.  Can someone explain it?
 A: An $n\times n$ matrix whose rows (or columns) form a linearly independent set has rank $n$.  The elementary row (or column) operations are rank preserving, so the reduced row echelon form of the matrix will have rank $n$.
Now, using the properties of a matrix in reduced row echelon form (i.e., first nonzero term in each row is $1$, and no nonzero terms in any column with a leading $1$) see if you can show that the only $n\times n$ matrix of rank $n$ in reduced row echelon form is the identity matrix.
A: This might help you:
Suppose we have a matrix $A$ whose columns $v_1,\dots,v_k$ are vectors in $\Bbb R^n$.
The reduced echelon form of $A$ will have a pivot in every column if and only if the vectors $v_1,\dots,v_k$ are linearly independent.  The reduced row echelon form of $A$ will have a pivot in every row if and only if the vectors $v_1,\dots,v_k$ span $\Bbb R^n$.
The only reduced matrix with a pivot in every row and every column is the identity matrix.
A: The contrapositive of this is: If the reduced row echelon form is not the identity matrix, then the rows are linearly dependent.
This might be easier to understand. First, if the RREF of a square matrix is not the identity matrix, then one row will not have a leading one. That row must be made all of zeroes. Now track how you got this row of zeroes: if you track back to your original matrix, this row of zeroes is obtained as a linear combinations of rows in the original matrix, and the corresponding linear combination gives you a linear dependency of some rows in the matrix. 
As you get a linear dependency, the rows are linearly dependent.
