Find the dimension of $V$ Let $T\in M_{m\times n}(\Bbb{R})$. Let $V$ be the subspace of $M_{n\times p}(\Bbb{R})$ defined by $V=\{X\in M_{n\times p}(\Bbb{R}): TX=0\}$. Find the dimension of $V$.
To find the dimension of  $V$.
Define a map $S:M_{n\times p}(\Bbb{R})\to M_{m\times p}(\Bbb{R})$ by $S(X)=TX$. Clearly, $S$ is a linear transformation. If I'll have $\operatorname{dim}Ker(S)$ then I can easily find $\operatorname{dim}V$.
But how to find $\operatorname{dim}Ker(S)$?
Any help will be appreciated.
 A: We can solve this question in two steps. The first step is reformulating the question to be about linear maps instead of matrices. Then we answer the reformulated question.
A reformulation should look like this:

Let $g:\mathbb R^n\to\mathbb R^m$ be linear. Let $V$ be the subspace of $\operatorname{Hom}(\mathbb R^p,\mathbb R^n)$ defined by $V:=\{f\in\operatorname{Hom}(\mathbb R^p,\mathbb R^n)~:~g\circ f=0\}$. Find the dimension of $V$.

$\operatorname{Hom}(\mathbb R^p,\mathbb R^n)$ is the space of linear maps $\mathbb R^p\to\mathbb R^n$. To get this formulation, we have to note that every matrix in $M_{m\times n}(\mathbb R)$ is just the representation of a linear map $\mathbb R^n\to\mathbb R^m$, and that every such linear map is represented by exactly one such matrix (given fixed bases of the involved vector spaces). We also have to note that if the matrices $A,B$ represent the linear maps $f,g$, then the matrix $BA$ represents the linear map $g\circ f$. Matrix multiplication has deliberately been defined in a way that this is true. With this, we take $g$ to be the linear map represented by $T$, and $f$ the linear map represented by $X$, and then we're essentially done translating.
Now to solving the question. You correctly identified that $V$ is the kernel of the map $S:X\mapsto TX$. In the language of linear maps, we should instead consider $S:f\mapsto g\circ f$. It's kernel is $V$ as we defined it in our reformulated question. Now we have to ask ourselves: What is the kernel of this map? It's the space of all $f\in\operatorname{Hom}(\mathbb R^p,\mathbb R^n)$ whose image is in the kernel of $g$, because then and only then will $g\circ f$ map everything to $0$. If anything outside $\ker g$ were contained in $\operatorname{im}f$, then $g$ would map that to something other than $0$, so $\operatorname{im}f\subseteq\ker g$. And in the other direction, if $\operatorname{im}f\subseteq\ker g$, then obviously $g\circ f=0$.
So now we have to find the dimension of the space of all linear maps $\mathbb R^p\to\mathbb R^n$ whose image is in $\ker g$. That's essentially all the linear maps $\mathbb R^p\to\ker g$, that is, $\operatorname{Hom}(\mathbb R^p,\ker g)$. And the dimension of $\operatorname{Hom}(V,W)$ is just $\dim V\cdot\dim W$ for any vector spaces $V,W$. In this case, $V=\mathbb R^p$ and $W=\ker g$, so $\dim\ker S=p\cdot\dim(\ker g)$.
Translating back, $g$ was the map represented by the matrix $T$. Their kernels are the same, by definition. So in the end we get
$$\dim V=\dim(\ker S)=p\cdot\dim(\ker T).$$
