given $V$ a vector space, let $\dim{V}=n$.
$T\colon V\longrightarrow V$ is a linear transformtion with $\dim(\operatorname{Im}{T})=r$.
Let $X$ be the set of all linear transformtion $F:V\longrightarrow V$ with $T\circ F=0$.
I need to prove that $X$ is a subspace of the space of all linear transformtion $V\longrightarrow V$ and find its dimension.
It's not a problem to prove that $X$ is a subspace of all the linear transformations $V\longrightarrow V$ as in fact $X=KerT\circ F$.
What I'm confused about is $\dim{X}$. Looks like it's $n-r=V-ImT$ but i'm not sure how to exacly explain this, as I don't know what is $KerF$ and don't know how to prove what is the dim of a LT composition.
1) It doesn't make sense to say $X=\ker{T}\circ F$, because $X$ is the set of such maps $F$. I guess you meant that $X$ is the set of linear transformations $F$ that factor (uniquely) through the kernel of $T$. This is indeed the case by the universal property of the kernel. To show that $X$ is a subspace you must show that the zero map is in $X$ and that $X$ is closed under addition and scalar multiplication.
2) We know that $\dim{\ker{T}}=n-r$. The maps in $X$ are in bijection with the linear maps from $V$ to $\ker{T}$. Hence we want to find the dimension of linear maps from $V\cong \mathbb{K}^{n}$ to $\ker{T}\cong \mathbb{K}^{n-r}$. The dimension of $X$ is therefore $n(n-r)$.