Finding a basis of a set of vectors Find a basis for the subspace of $\mathbb R^4$ spanned by the vectors $(1,2,−1,0),(4,8,−4,−3),(0,1,3,4),(2,5,1,4)$.
I've tried reducing a matrix (with vectors as rows) to the triangular form and got $(1,2,-1,0), (0,1,3,4)\text{ and }(0,0,0,-3)$. A matrix with vectors as columns gives a different answer with four vectors, but two bases of the same vector space have the same number of elements. Is this correct?
 A: It helps to remember small silly examples that you should intuitively know the answer to immediately without any calculation required and check to see that your proposed method works for that silly small situation as well.
Here, think of the problem where we want to find a basis for the space spanned by the single vector $\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$
It should be painfully clear that the vector by itself is a basis (or more correctly a set containing the vector is).
Now, if we were to lay it sideways and row reduce., well, it is already in RREF, and we wind up with $\begin{bmatrix}1&1&1&1\end{bmatrix},$ which as noted before can be interpreted as a valid basis for our space.
If instead we were to lay it as a column and then row reduce, well, that gives us $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$, but this is not even an element of our space and so clearly is not a basis for it.

All of this is a long way of reminding you and giving you an easy way to remember what @TheoBendit mentioned above... that row reduction preserves the rowspace of a matrix, but can change the columnspace of a matrix.
That is to say, if you want to find a basis for a collection of vectors of $\Bbb R^n$, you may lay them out as rows in a matrix and then row reduce, the nonzero rows that remain after row reduction can then be interpreted as basis vectors for the space spanned by your original collection of vectors.
