Let $F:V\longrightarrow W$ be a linear map. The image of $F$ is the set
$$\mathrm{Im}F=\{w\in W: F(v)=w\ \mbox{for some}\ v\in V\}.$$

Proposition. The image of $F$ is a subspace of $W$.

Proof. The proof is straightforward. It is left for an exercise.

The preimage of the identity element $O$ under the linear map $F$ i.e. the set of elements $v\in V$ such that $F(v)=O$ is called the kernel of $F$ and is denoted by $\ker F$.

Proposition. The kernel of $F$ is a subspace of $V$.

Proof. It is straightforward and left for an exercise.

Example. Let $L: \mathbb{R}^3\longrightarrow\mathbb{R}$ be the map defined by
$$L(x,y,z)=3x-2y+z.$$
If we write $A=(3,-2,1)$, then $L(X)$ may be written as
$$L(X)=X\cdot A.$$
So, the kernel of $L$ is the set of all $X$ that are perpendicular to $A$.

Example. Let $A$ be an $m\times n$ matrix and let $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ be the linear map defined by
$$L_A(X)=AX.$$
The kernel of $L_A$ is the subspace of solutions $X$ of the system of linear equations
$$AX=O.$$

Example. Let $\mathcal{F}$ be the vector space of smooth functions. Let $a_1,\cdots,a_m$ be numbers and let
$$L=a_m\frac{d^m}{dx^m}+a_{m-1}\frac{d^{m-1}}{dx^{m-1}}+\cdots+a_1.$$
Then $L:\mathcal{F}\longrightarrow\mathcal{F}$ is a linear map. $\ker L$ is the space of solutions of the homogeneous linear differential equation
$$a_m\frac{d^mf}{dx^m}+a_{m-1}\frac{d^{m-1}f}{dx^{m-1}}+\cdots+a_1f=0.$$
If there exists one solution $h_0$ for the non-homogeneous linear differential equation $L(h)=g$, then any solution $h$ may be written as $h=f+h_0$ where $f$ is a solution of the homogeneous equation $L(f)=0$. The proof is left as an exercise.

Theorem. Let $F: V\longrightarrow W$ be a linear map such that $\ker F=\{O\}$. If $v_1,\cdots,v_n$ are linearly independent elements of $V$, then $F(v_1),\cdots,F(v_n)$ are linearly independent elements of $W$.

Proof. The proof is straightforward and is left for an exercise.

Theorem. Let $F: V\longrightarrow W$ be a linear map. $F$ is one-to-one if and only if $\ker F=\{O\}$.

Proof. Suppose that $F$ is one-to-one. Let $v\in\ker F$. Then $F(v)=O=F(O)$. Since $F$ is one-to-one, $v=O$. So, $\ker F=\{O\}$. Suppose that $\ker F=\{O\}$. Let $F(v_1)=F(v_2)$. Then $F(v_1-v_2)=O$ and so $v_1-v_2\in\ker F=\{O\}$ i.e. $v_1=v_2$. Hence, $F$ is one-to-one.

Given a linear map $L: V\longrightarrow W$, there is a relationship between the dimensions of $V$, $\ker L$, and $\mathrm{Im}L$, namely
$$\dim V=\dim\ker L+\dim\mathrm{Im}L.$$
We will not prove it here but those interested may find proof in [1].

Example. Consider the linear map $L:\mathbb{R}^3\longrightarrow\mathbb{R}$ given by
$$L(x,y,z)=3x-2y+z.$$
The image is not $\{O\}$, so it is $\mathbb{R}$. Therefore the dimension of $\ker L$, the space of all solutions of $3x-2y+z=0$ is 2. $3x-2y+z=0$ is indeed equation of a plane through the origin and we know that the dimension of a plane as a vector space is 2.

References:

[1] Serge Lang, Introduction to Linear Algebra, Second Edition, Undergraduate Texts in Mathematics, Springer, 1986