For each given subspace W there is one and only one row-reduced echelon matrix that has W as its row space. Let m and n be positive integers and let F be a field. Suppose W is a subspace of $ {F ^{n}}$ and $ dim W \le m$. Prove that there is precisely one m x n row-reduced echelon matrix over F which has W as its row space.
 A: First, prove that if two $m\times n$ matrices in reduced row-echelon form are not identical, then they have different row spaces. Then it follows that all matrices whose rows have span $W$ have the same reduced row-echelon form. 
A: Ok I think I get what Gerry meant. 
First of all, the first non-zero entry of every row of an rref matrix must be 1. Also, all entries above and below the first non-zero entry of every row of an rref matrix must be 0. 
Suppose that the rref of two matrices $A$ and $B$ are $R$ and $R'$ respectively. Also suppose the $i$th rows of $R$ and $R'$ first start to differ. There are two ways they can differ: a) the first non-zero entries are in different columns or b)the first non-zero entries are in the same column but the rest of the entries differ in some ways. 
In the case of a, if the first non-zero entries are in different columns, suppose for $R$ it is at column $m$ and for $R'$ it is at column $n$ where without loss of generality $n > m$. Then, the row $i$ in $R$ cannot be spanned by any linear combinations of the row vectors in $R'$ because there is no non-zero entry in column $n$ in any of the rows in $R'$. Hence, $R$ and $R'$ do not have the same row space because a vector in $R$ is not in the row space of $R'$.
We have shown that the two rref matrices must have each row to have the first non-zero row to be in the same column for them to as least not be ruled out as not having the same row space. 
Given any rref matrix with $r$ non-zero rows, $r$ scalars can determine the vector in the row space it maps to. Given two non-identical rref matrices (with the same number of non-zero rows and the same columns where the first non-zero entries for each row appear), a vector in each of their row spaces is mapped to when $r$ scalars are used. This vector cannot be mapped to by the other rref matrix's rows because it will impose a relation on the $r$ scalars, and this cannot happen because the $r$ scalars are supposed to be independent.
I cannot explain the bold part very clearly, the reader can try to work it out himself. Basically by getting $r$ vectors to determine the vector that is been mapped to, the entries of the vector not in the column of the first non-zero entry of the rref's row vectors will be determined by the $r$ scalars. By using these $r$ scalars on another rref vector (that passed part a's test), another relation between those entries and the $r$ vectors can be determined. Equating these relations, a relationship can be found between the $r$ scalars, which cannot happen.
