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.
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 $ka\in\ker(T)$