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
Best Answer
But in this case $ T : \mathbb{R}^{13} \rightarrow \mathbb{R}^7 $. So, the answer that the poster thinks is wrong is actually correct.