The rank-nullity theorem says that if linear transformation $T: V\rightarrow W$ then $$ \\ \dim(Im(T))+\dim(Ker(T))=\dim(V)$$ and equivalently, $$\\ rank(T) + nullity(T) = \dim(V) $$ when seeing $T$ as a $m \times n $ matrix.
This makes perfect sense to me whenever $V$ and $W$ are vector spaces of column vectors. However, in terms of vector space of matrices, I got confused easily.
The question is: does rank of $T$ always equal to the dimension of its image? Also how about the relationship between nullity and $\dim (ker(T))$?
Say if $V$ is the vector space of all $2 \times 3$ matrices, and $W$ is the vector space of all $4 \times 1$ matrices, and if T is onto, then what would T look like? (i.e. what are m and n?)
Following the first expression of RNT, it suggests that T should have a corresponding range of 4, and the kernel should have a dimension of 2, but does it mean that $rank(T) = 4$ and $nullity(T) = 2$? If yes, then shouldn't $T$ have definitely 2 columns since $V$ is $2 \times 3$?