Proving kerT is a subspace of V. and rangeT is a subspace of W. My question is as follows: Suppose $V$ and $W$ are vector spaces, and let $T: V \longrightarrow W$ be a linear transformation.  
Show that $\ker T$ is a subspace of $V$. 
Show that $\operatorname{range} T$ is a subspace of $W$. 
Most of my problem here is I don't think I fully understand the concept of the kernel and the range.  I know that the kernel relates to a null space, and the range to the column space...  I also know that I have to show that each (the kernel and the range) must contain the zero vector, preserve vector addition and scalar multiplication. 
The kernel is then all the solutions to $Ax= \vec 0$, so the zero vector must be there? 
I don't know how to proceed from there and we have hardly worked with the kernel and the range in class, so I really don't know where to go with this problem.  
 A: To show that $\ker T$ is a subspace of $V$, we need to show that it has the following properties:


*

*Has $0$

*Is additively closed

*Is scalar multiplicatively closed


Clearly $T(0)=0$. So we need only show additive and scalar multiplicative closure.
Additive closure: We want to show that if $a,b\in \ker(T)$ then $a+b\in\ker(T)$. E.g. $T(a)=0,T(b)=0\implies T(a+b)=0$. This follows from the property of additivity, $T(a+b)=T(a)+T(b)=0+0=0$, and hence $a+b\in\ker T$.
Scalar multiplicative closure: We want to show that if $a\in\ker(T)$ then $ka\in\ker(T)$ where $k\in\Bbb F$. So $T(a)=0$ and by the property of homogeneity(of degree $1$) we have that $0=k0=kT(a)=T(ka)$ and hence $ka\in\ker(T)$
A: Let $F=\text{ker}(T)$. Then $0_V\in S$ because $T(0_V)=0_W$ and if $x,y\in F$ ($x,y$ are elements of $V$) then $\alpha x+\beta y\in F$ for all scalars $\alpha$ and $\beta$ because
$$
T(\alpha x+\beta y)=\alpha T(x)+\beta T(y)=\alpha 0_W+\beta 0_W=0_W+0_W=0_W
$$
with the first equality above uses the linearity of $T$. Similarly, let $G=\text{range}(T)$. Then $0_W\in G$ because there exists an element in $V$ (i.e. $0_V$) such that this element is mapped by $T$ to $0_W$. And finally, if $u,v\in W$ are s.t. $w,w'\in G$, then $\alpha  w+\beta w'$ is also in $G$. This is because $w,w'\in G$ implies that there exist $v,v'$ in $V$ s.t. $w=T(v)$ and $w'=T(v')$ and so
$$
\alpha w+\beta w'=\alpha T(v)+\beta T(v')=T(\underbrace{\alpha v+\beta v'}_{\in V})\in\text{range}(T)=G.
$$
