The dimension of the column space is equal to that of the row space. But the column space is not equal to the row space in this problem.
I know that. But I don't know how the textbook has found Nullspace of $\displaystyle A^{T}$ in this particular problem.
Are you sure the column space of A^T isn't the basis of the row space of A? I would think it is that.
It is stated on http://www.ltcconline.net/greenl/cou...ctors/rank.htm that that the column space of A^T is the row space of A.
I don't see why it wouldn't be the same the other way around.
I cannot see the "A" that apparently was posted so I don't know if we are talking about matrices or general linear transformations or what dimensions. It is true that the "row space" of a linear transformation A is the same as (strictly speaking "is isomorphic to") the "column space" for finite dimensions. For infinite dimensions (such as function spaces) two spaces can have the same dimension but not be isomorphic.
I think he's using rank-nullity and then just making sure he gets the correct number of independent vectors. He knows the column space of $\displaystyle A$ is two, and the first and third columns of $\displaystyle A$ are the ones he lists and they're independent. I'm not sure if this is obvious from $\displaystyle E^{-1}$, but I suspect it might be because of the nice factorization. I forget all the details about the relationships between these bases and how they change when you do elementary operations.
As for the nullspace of $\displaystyle A^T$, it looks like he just interprets matrix multiplication as acting on the rows of $\displaystyle A$ and then finds a vector in the nullspace of $\displaystyle A^T$. It's in this nullspace because he considers left multiplication on $\displaystyle A$. Since this nullspace has dimension 1, there's a basis.
Is this Strang? I know he talks about this in one of his MIT lectures online. That should answer everything.