I was working through a problem and was wondering if there was an easier way of finding the basis of the left null space of a given matrix.
(For a simple example) Suppose we have a matrix $A = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 4 & 8 \end{bmatrix}$ when reduced we can write it as $\text{rref}(A) = \begin{bmatrix} 1 & 2 & 4 \\ 0 & 0 & 0 \end{bmatrix} $
from rref(A) it is clear that:
Basis for $C(A) = \left\{ \begin{pmatrix} 1 \\ 2\end{pmatrix} \right\}$
Basis for $C(A^T) = \left\{ \begin{pmatrix} 1,&2, & 4 \end{pmatrix} \right\}$
Basis for $N(A) = \left\{ \begin{pmatrix} -2 \\ 1 \\ 0\end{pmatrix} , \begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}\right\}$
Now my question is am I able to deduce the left null space just from rref(A)?
Else, I would take the transpose of A, row reduce it and then find the left null space that way but I was wondering if there is an easier way?
Best Answer
You can’t really get the left null space directly from just the rref, but if you first augment the matrix with the appropriately-sized identity and then row-reduce it, the row vectors to the right of the zero rows of the rref constitute a basis for the left null space.
Using your example, row-reduce $$\left[\begin{array}{ccc|cc}1&2&4 & 1&0 \\ 2&4&8 & 0&1 \end{array}\right] \to \left[\begin{array}{ccc|rc} 1&2&4 & 1 &0 \\ 0&0&0 & -2&1 \end{array}\right].$$ The left null space is thus $\operatorname{span}\{(-2,1)\}$.
As for why this works, see this question. I’ll repeat a caveat from there: this method doesn’t often give you a “nice” basis, in that the vectors are often rather large multiples of what you would’ve computed by the more usual method of applying Gaussian elimination to the transpose.