Let $f:U \to V$ be a linear map between two vector spaces $U$ and $V$ over a field $IF$. The kernel is $\ker(f)=\{u \in U \mid f(u)=0\}$ and image $\operatorname{im}(f)=\{v \in V \mid \exists u \in U s.t. f(u)=v\}$
i) Prove that $\ker(f)$ is a subspace of $U$
ii) Prove that $\operatorname{im}(f)$ is a subspace of $V$
For i) I got using the subspace criterion 2: clearly $\ker(f)$ is a subset of U. and clearly $0$ is contained in $\ker(f)$. And let u and v be in $\ker(f)$ and a be in $IF$, so
$f(u)=f(v)=0$. Then $au+v=af(u)+f(v)=a0+0=0$ which is in $\ker(f)$. I am not sure if this is correct because it seems a bit abstract for me. and do i need more justification that $0$ is in $\ker(f)$?
Also for the image, i have no idea…
Best Answer
Except for the typo I pointed out in my comment, your proof that the kernel is a subspace is perfectly fine. Note that it is not necessary to separately show that $0$ is contained in the set, since this is a consequence of closure under scalar multiplication.
As for the image: we say that a vector $u \in V$ is in the image if, for some $x \in U$, $u = f(x)$.
In order to show that this set is a subspace, you need to show that for any constant $a$: if $u$ and $v$ are in the image, then so is $au + v$. That is, for any $u$ and $v$ where $u = f(x)$ and $v = f(y)$ (for some choice of $x$ and $y$ from $U$), we need to show that $$ au + v = f(z) $$ for some vector $z \in U$.
In fact, we note that $$ au + v = af(x) + f(y) = f(\overbrace{ax + y}^{\text{this is our }"z"}) $$