Proof that a Kernel of a Linear Mapping is a Subspace In my linear algebra classes, we often just assume that a kernel is a subspace.
However, how do we prove this?

My current idea is to apply subspace theorem (closed under addition, multiplication, and contains the zero vector). However when I try working it out, I can not seem to get it.

Does someone know how to actually prove this?
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 $V$.
The kernel of F is a subspace of V.
Proof Since F(O) = 0, we see that 0 is in the kernel. Let v, w be in
the kernel. Then F(v + w) = F(v) + F(w) = 0 + 0 = 0, so that v + w is
in the kernel. If c is a number, then F(cv) = cF(v) = 0 so that cv is also
in the kernel. Hence the kernel is a subspace.