How would you prove that the kernel of a linear mapping $\theta\colon V \rightarrow W$ is a subspace of $V$?

Question

In my lecture notes, it says the following:

The kernel of a linear mapping $\theta\colon V \rightarrow W$ is a subspace of $V$.

Proof: Straightforward

I can't say that I see this as being particularly obvious to prove, so I was hoping that someone could explain how it would be done.

Answer 1

First, $\ker\theta$ is non-empty because $\theta(0)=0$ by linearity.

If $v,w\in\ker \theta \subset V$, then $$\theta(av+bw)=a\theta(v)+b\theta(w) = a(0)+b(0)=0$$ so $av+bw\in \ker \theta$. Hence $\ker \theta$ is a subset of $V$ which is closed under scalar multiplication and addition as required for $\ker\theta$ to be a subspace of $V$.

Answer 2

First of all, $0$ is in $\text{Ker } \theta$ since $\theta$ is linear.

Next, let $v$ and $w$ be in $\text{Ker } \theta$. We wish to show that $v + w$ is in $\text{Ker } \theta$. But $\theta(v + w) = \theta(v) + \theta(w)$, because $\theta$ is linear. Since $v$ and $w$ are in the kernel of $\theta$, the right hand side of the above equation is $0 + 0 = 0$, so indeed $v+w$ is in the kernel of $\theta$.

Finally, let $v$ be in $\text{Ker } \theta$ and let $\lambda$ be an arbitrary scalar. We wish to show that $\lambda v$ is in $\text{Ker} \theta$. But $\theta(\lambda v) = \lambda \theta(v)$, because $\theta$ is linear. Since $v$ is in the kernel of $\theta$, the right hand side of the above equation is $\lambda \cdot 0 = 0$, so indeed $\lambda v$ is in the kernel of $\theta$.