kernel and image subspaces proof 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...
 A: 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
$$
au + v = af(x) + f(y) = f(\overbrace{ax + y}^{\text{this is our }"z"})
$$
A: To prove that image of a linear transformation is a subspace of domain vectorial space you need use definitions.
Let $T$ a linear transformation such that $T:U\to V$ and the $Im(T)$ is defined as:
$Im(T)=\{v \in V: \exists u\in U, T(u)=v\}$
So, you should always get a $u \in U$ that holds the definition.
So, to prove that $Im(T)$ is a subspace of $V$ we need to prove that is closed under the  $(+, \cdot)$ operators and have the $0$ vector of $V$.

*

*Have zero vector

Since $T(0_U)=0_V \implies \exists0_U\in U: T(0_U)= 0_V\in V$ so $0 _V \in Im(T)$


*Closed under $(+)$
Let $v_1, v_2 \in Im(K)$, this implies that $T(u_1)=v_1$ for some $u_1 \in U$ and same for $T(u_2)=v_2$ for some $u_2 \in U$.
And $v_2+v_1=T(u_2)+T(u_1)=T(u_1+u_2)$ and here you can see that exist a $(u_1+u_2)$ such that $T(u_1+u_2) = v_1 + v_2$ so is closed under sum.
And to prove $(\cdot)$ is the same process.
