Let $T\in M_{m\times n}(R)$. Let $V$ be a subspace of $M_{n\times p}(R)$ defined by $\{X\;|\;TX=0\}$ What is the dimension of $V$? Let $T\in M_{m\times n}(R)$. Let $V$ be a subspace of $M_{n\times p}(R)$ defined by $\{X\;|\;TX=0\}$. What is the dimension of $V$?
Efforts: 
Define $\gamma: M_{n\times p}(R)\rightarrow M_{m\times p}(R)$ by $$X \mapsto TX $$
So kernel of this map is equal to $V$ and $\mathrm{rank}(T)+\mathrm{null}(T)=n\cdot p$.
But how do I find the rank?
I guess somewhere we would see an expression involving rank of $T$.
Thanks for help. 
 A: It comes down to thinking of "multiplication by $T$" as a matrix. That is, each matrix $X \in M_{n \times p}$ can be written in the basis $E_{ij}$ of matrices with a $1$ in position $i,j$ and zero elsewhere. Similarly, for $m \times p$ matrices we'd have a similar basis.
Now, we need to understand the "matrix of $T$" in this situation. For that, we need to note that it is an $mp \times np$ matrix. The entry in row $(k,l)$ column $(i,j)$, is the answer to the question : when I multiply $T$ by $E_{ij}$, what is the $k,l$ th entry of that?
In other words, what is $(TE_{ij})_{kl}$? It is $\sum_{t=1}^p T_{kt}(E_{ij})_{tl}$. So it is $0$ if $j \neq l$, and $T_{ki}$ if $j = l$. Using this, we can actually draw up the matrix for multiplication by $T$.Indeed, if we let that matrix be $M_T$ then $(M_T)_{(k,l),(i,j)} = 0$ if $l \neq j$, and is $T_{ki}$ if $l=j$. Let us, for example, consider $m=n=p=2$ and the $2 \times 2$ matrix $T = \begin{pmatrix}4\quad 1 \\ 2 \quad 3\end{pmatrix}$ , and put up its multiplication matrix $M_T$. The rows will be indexed in the order $(1,1),(2,1),(1,2),(2,2)$ and the columns also the same, and you will see why.
For example what is $(M_T)_{(1,1),(1,1)}$?  As per our rules, since the second components match the answer is $T_{11}=4$. Similarly, $(M_T)_{(1,1),(2,1)}$ will be $T_{12} = 1$. We would have $(M_T)_{(1,1),(1,2)} = 0$, and the last entry would be zero.
Similarly, $(M_T)_{(2,1),(1,1)} = T_{21} = 2$, and then $(M_T)_{(2,1),(2,1)} = T_{22} = 3$ etc. Now doing this for everything, I think you should verify that:
$$
M_T = \begin{bmatrix}
4 & 1& 0 & 0 \\
2& 3& 0 & 0 \\
0 & 0 & 4 & 1\\
0 & 0 & 2& 3\\
\end{bmatrix}
$$
Stare at $T$. Stare at $M_T$. Get a conclusion.
Now, I leave you to see how things look in general : how does $M_T$ depend on $T$?
Finally, the rank of $M_T$ is just the rank of $T$ times one of $m,n,p$(because the first "block" of rows is linearly independent of the second "block" of rows and so on.So any linear dependence must come from these blocks themselves, but then that is governed by the rank of $T$). Figure out which one it is, and then use the rank nullity formula accordingly, which you have written correct.
