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? 
 A: 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 $V$.
A: This is a simple applicaiton of properties of linear transformations.  As a sketch of the proof:
(1) Contains zero:
$$f(0)=f(0_F 0_V)=0_Ff(0_V)=0$$
(2)Scalar multiplication:
$$f(cv)=cf(v)=c0_V=0$$
(3) Addition
$$f(v_1+v_2)=f(v_1)+f(v_2)=0+0=0$$
A: Let $V$ and $W$ linear spaces and $f:V \to W$ linear. Let $K:=ker(f)$
From $f(0)=0$ we get $ 0 \in K$.
Now suppose that $x,y \in K$ and $\alpha, \beta$ are scalars. Use linearity to derive
$f(\alpha x+ \beta y)= \alpha f(x) + \beta f(y)$. Do you now see that $\alpha x+ \beta y \in K$ ?
A: Let $f : E\rightarrow F$ be a linear mapping. we have :


*

*$0_E \in Kerf$

*$\forall x,y\in Kerf,\ \forall \alpha \in \mathbb{K}, f(\alpha x +y)=\alpha f(x)+f(y)=0_F \Rightarrow \alpha x +y\in Kerf$
because $f(x)=f(y)=O_F$.
Then $Kerf$ is a subspace of $E$.

A: 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.
