$Ker(T) \subseteq V$ 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?

You have to show, that


*

*$\mathbf{0}_V\in\ker(T)$

*$\mathbf{u},\mathbf{v}\in\ker(T)$ implies $\mathbf{u}+\mathbf{v}\in\ker(T)$

*$a\mathbf{u}\in\ker(T)$ for all $a\in\mathbb{F}$


You've already said that $\mathbf{0}_V\in\ker(T)$. Now, let $\mathbf{u},\mathbf{v}\in\ker(T)$. Using the linearity of $T$, we get
\begin{align*}
T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})=\mathbf{0}_W,
\end{align*}
so $\mathbf{u}+\mathbf{v}\in\ker(T)$. In a similar way, we obtain
\begin{align*}
T(a\mathbf{u})=aT(\mathbf{u})=\mathbf{0}_W.
\end{align*}
