Fundamental theorem of linear algebra: The rows and columns forms a basis for a matrix $A$ In my course literature, the author claims the following:

Because the column vectors of the matrix $A^T$ is equivalent to the row vectors of $A$, the following statements must be equivalent

*

*$A$'s column vectors forms a basis


*$A$'s row vectors forms a basis

I e-mailed my tutor in class for an explanation, asking if the author had made a mistake in his book. My tutor simply reiterated what I thought the author meant.
I already understand that the row vectors of $A^T$ is equivalent to the column vectors of $A$, and the other way around. As such, I thought the author meant to say:
$A$'s column vectors forms a basis $\iff A^T$'s row vectors forms a basis
However, I don't see the author justifying such made statements with the quote I gave above.

I expected my tutor to have misunderstood me, and well, checking for more sources; they claim the same. How is is that the quote above is true?
 A: Suppose $A$'s columns form a basis, then $Aw = 0$ if and only if $w = 0$.
Now, suppose the columns of $A^T$ do not form a basis. Then there exists non-zero $v$ so that $A^Tv = 0$.
Then, for any $w$, we have
$$
\langle v, A w\rangle = \langle A^Tv,w\rangle = 0
$$
so that $v \in (\textrm{Im }A)^\perp$ (i.e. $\textrm{ ker }A^T \subset (\textrm{Im }A)^\perp$). But because $A$'s columns form a basis, $(\textrm{Im }A)^\perp = \{0\}$, and you have a contradiction.
(In fact, in proving the full theorem, you will show that in general, $\textrm{ ker }A^T = (\textrm{Im }A)^\perp$ etc.)
A: It certainly appears that your version is the one intended. 
The statement made by the author is actually true, but the "because" statement doesn't justify it at all. 
Roughly: if the column vectors form a basis of $\mathbb R^n$, then there must be $n$ rows (so that the columns are in the right space), and there must be $n$ columns, because every basis of $\mathbb R^n$ has $n$ vectors. So if the column vectors form a basis, the matrix is square. 
That means that the row vectors at least have some hope of forming a basis: there are $n$ of them, and each is a vector in $\mathbb R^n$. But the proof that they actually DO form a basis is not completely trivial. (@BaronVT gives a nice explanation of it, and you can see that it has little to do with the "Because" clause of your author.)
