$Ker(T) \subseteq V$ Is A Subspace Let $V,W$ be a vector space over a field $\mathbb F$, and $T$ a linear transformation $T:V \rightarrow W$  
$Ker(T) \subseteq V $ to prove that $Ker(T)$ is a subspace can we say that:
by definition $0\in Ker(T)$ and because V is a vector space, and in any vector space there are two trivial subspace (0,V) therefore $Ker(T)$  is a subspace?

Your reasoning isn't correct. Try the following hints:
To start with, one can show that for any linear map $T$, $T(0) = 0$ and so $0 \in \text{ker }T$ (you should prove this). Thus, since $\text{ker }T$ is non-empty, it suffices to show that it's closed under linear combinations. So suppose $u,v \in \text{ker }T$ and $\lambda, \mu \in F$, and show that $T(\lambda u + \mu v) = 0$ using the fact that $T$ is a linear map.
