I initially had the matrix $A$, and through row reductions I found a matrix $H$.
$$A=\begin{bmatrix}
1 &2 & -3\\
2 & 4 &-6 \\
-1& -2& 1\\
3& 6 & -8
\end{bmatrix}$$
$$\text{Through row reductions:}$$
$$H=\begin{bmatrix}
1 &2 & 0\\
0 & 0 &1 \\
0& 0& 0\\
0& 0 & 0
\end{bmatrix}$$
I know that columns $1,3$ are linearly independent, and thus they form the basis.
$$\text {Basis for A: } = \{\begin{bmatrix}
1\\
2\\
-1\\
3
\end{bmatrix},\begin{bmatrix}
-3\\
-6\\
1\\
-8
\end{bmatrix}\}$$
Now I am asked to enlarge the basis for $\mathbb{R}^4$. So in this case would I have to find $2$ additional linearly independent vectors, and do the entire row reduction again, until I get $4$ vectors in the basis? If so, how do I find these vectors? Is it just guessting?
Best Answer
There are many pairs of vectors that will complete the basis, but an obvious choice is to use a basis of the orthogonal complement of this subspace. Since $N(A^T)=C(A)^\perp$, this amounts to finding a basis for the null space of the matrix that has the two basis vectors of $C(A)$ as its rows. (This is equivalent to the column-reduction suggested by Exodd in his comment to the question.)
In specific cases, you might be able to eyeball some set of standard basis vectors or simnple combinations thereof that are linearly independent of your basis for $C(A)$, but the above is a general procedure that will work in always work.