Suppose $T$ is a linear transformation $T:V \rightarrow W$
To show $Ker(T)$ is a subspace, you need to show three things:
1) Show it is closed under addition.
2) Show it is closed under scalar multiplication.
3) Show that the vector $0_v$ is in the kernel.
To show 1, suppose $x,y\in Ker(T)$. Then $T(x+y)=T(x)+T(y)=0_w+0_w=0_w$ Hence $x+y$ also exists in the kernel and so the kernel is closed on addition.
To show 2, Assume $\lambda \in F,x\in Ker(T)$ so it follows that $T(\lambda x)=\lambda T(x)=\lambda 0_w=0_w$ So again $\lambda v\in Ker(T)$ so it is closed on scalar multiplication.
Finally it is simple to see 3, $\forall v \in V,T(0_v)=T(v+(-v))=T(v)+T(-v)=T(v)-T(v)=0_w$ so indeed $0_v$ is in the kernel.
So you have shown that $Ker(T)$ is a subspace of $V$.