If $ \ T \ $ is onto , then what is the dimension of the null space of $ \ A \ $ Suppose that $ \ A \ $ is a $ \ 6 \times 13 \ $ matrix and that  $ \ T(x)=Ax \ $.
If $ \ T \ $ is onto , then what is  the dimension of the null space of $ \ A \ $.
Answer:
Since $ \ A \ $ is $ \ 6 \times 13 \ $ matrix , we can write $ \ T : V \to W  \ $ ,
where $ \dim(V)=13 \ $ and $ \ \dim (W)=6 \ $
Now since $ \ T \ $ is onto map , $ \ \operatorname{Im}(T)=W \ $ 
Thus $ \ \operatorname{rank}(T)=\dim (\operatorname{Im}(T))=\dim(W)=6 \ $
Now ,
$ \operatorname{rank}(T)+ \operatorname{Nullity}(T)=13 \ \Rightarrow \operatorname{Nullity}(T)=13-6=7 \ $
Thus , $ \operatorname{Nullity}(T)=7 \ $
Thus ,  Dimension of  Null  space of $ \ A \ $ =7 \ $
But it is given wrong.
I think we have to consider the case $ \ A^T \ $ and $ \ T: V \to W \ $
Am I right?
Help me out
 A: But in this case $ T : \mathbb{R}^{13} \rightarrow \mathbb{R}^7 $.  So, the answer that the poster thinks is wrong is actually correct.
A: Theorem. Let $T:R^{n}→R^{m}$ be a linear transformation. The following are equivalent:
1-T is onto.
2-The equation $T(x)=b$ has solutions for every $b∈R^{m}$.
3-If $A$ is the standard matrix of $T$, then the columns of $A$ span $R^{m}$. That is: every $b∈R^{m}$ is a linear combination of the columns of $A$.
proof of $(3)⇒(2)$ Suppose the columns of $A$ span $R^{m}$ and let $b∈Rm$. We want to show that $T(x)=b$ has at least one solution.
Since the columns of $A$ span $R^{m}$, there exist scalars $α_{1},…,α_{n}$ such that
$$\mathbf{b} = T(\mathbf{a}) = A\mathbf{a} = A\left(\begin{array}{c}a_1\\a_2\\ \vdots\\a_n\end{array}\right) = a_1A_1 + a_2A_2 + \cdots + a_nA_n.$$
in your case the columns of $A$ have to span $W$, which is make sense, since there are 13 vectors for spaning $R^{6}$.
$ rank(T)=6 $ so null space has $7$ dimension. so this is right answer. what is problem with it?
