Basis of row space equals the basis of a subspace in $\mathbb{R}^n$? Question: Find a basis for the subspace of $\mathbb{R}^4$ spanned by the given vectors:
$(-1,1,-2,0),(3,3,6,0),(9,0,0,3)$
The solution to this problem is the basis for the row space of these vectors. Is the basis for row space and subspace the same thing? Also, when solving this problem, the solution manual solved it by placing the vectors like this:
$\begin{bmatrix}-1& 1& -2& 0\\3& 3& 6& 0\\9& 0& 0& 3\end{bmatrix}$
But when finding a basis, shouldn't the vectors be placed as columns of a matrix?
Thanks!
 A: The row space of a matrix is the subspace spanned by the row vectors of that matrix.  Hence, if we wish to find a basis for the subspace spanned by those vectors, then we need to make those vectors rows of a matrix and find its row space.
Alternately, we could put the vectors as columns in a matrix, but then we'd need to find the column space instead.  Remember that the column space of a matrix is the subspace spanned by its column vectors.
A: If you make the vectors to form the columns of a matrix, so
$$
\begin{bmatrix}
-1 & 3 & 9 \\
1  & 3 & 0 \\
-2 & 6 & 0 \\
0  & 0 & 3
\end{bmatrix}
$$
and use elementary row operation on it, to get
$$
\begin{bmatrix}
1 & -3 & -9 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0  & 0 & 0
\end{bmatrix}
$$
you can see that the three vectors form a linearly independent set. If, instead, some column turned out to be non dominant (that is, without a pivot in it), the corresponding vector could be eliminated from the set and the vectors corresponding to dominant column form a basis of the subspace.
This information is lost if you do the row operations on
$$\begin{bmatrix}-1& 1& -2& 0\\3& 3& 6& 0\\9& 0& 0& 3\end{bmatrix}$$
but you still find a basis of the subspace by considering the non zero rows in the row echelon form.
So, if you want to extract a basis from the given set, you need to use column vectors (or do elementary column operation on the row vectors); if you need any basis, you can use row vectors.
