$f$ is a linear map if

$f(v+w)=f(v)+f(w)$ for all $v,w$

$f(\lambda v)=\lambda f(v)$ for all $v$ and all $\lambda \in \mathbf{R}$ .
Recall the following definitions:
$f$ is a linear map if
$f(v+w)=f(v)+f(w)$ for all $v,w$
$f(\lambda v)=\lambda f(v)$ for all $v$ and all $\lambda \in \mathbf{R}$ .
$V$ is a linear subspace if
$v,w\in V$ implies $v+w\in V$
$v\in V$ implies $\lambda v\in V$ for all $\lambda \in \mathbf{R}$ .
These two definitions are very similar. We will exploit this in the next two videos: given a linear map $f:{\mathbf{R}}^{n}\to {\mathbf{R}}^{m}$ , we will associate to it two subspaces $\mathrm{ker}(f)\subset {\mathbf{R}}^{n}$ (the kernel of $f$ ) and $\mathrm{im}(f)$ (the image of $f$ ).
The kernel of $f$ is the set of $v\in {\mathbf{R}}^{n}$ such that $f(v)=0$ . (If $m=n$ then this is just the 0eigenspace of $f$ ).
Let $f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{3}$ be the map $f(v)=Av$ for $A=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ . Note that $f\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right)=\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill 0\hfill \end{array}\right)$ . This is the vertical projection to the $xy$ plane.
The kernel of $f$ is the $z$ axis (blue in the figure; these are the points which project vertically down to the origin). That is $\mathrm{ker}(f)=\{\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill z\hfill \end{array}\right):z\in \mathbf{R}\}$ .
Recall the example $A=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right)$ (going from ${\mathbf{R}}^{3}$ to ${\mathbf{R}}^{2}$ ) from Video 3. This projects vectors into the plane; if we think of ${\mathbf{R}}^{2}$ as the $xy$ plane then we can visualise this map as the projection of vectors in the $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)$ direction until they live in the $xy$ plane.
We described this as projecting light rays in the $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)$ direction. In this case, the kernel of $A$ is precisely the light ray which hits the origin, which is the line $\{\left(\begin{array}{c}\hfill x\hfill \\ \hfill x\hfill \\ \hfill x\hfill \end{array}\right):x\in \mathbf{R}\}$ (light blue in the picture).
The kernel is a subspace.
Given $v,w\in \mathrm{ker}(f)$ , we need to show that $v+w\in \mathrm{ker}(f)$ . Since $v,w\in \mathrm{ker}(f)$ , we know that $f(v)=f(w)=0$ . Therefore $f(v+w)=f(v)+f(w)$ (since $f$ is linear) $=0+0=0$ , so $v+w\in \mathrm{ker}(f)$ . Similarly, $f(\lambda v)=\lambda f(v)=\lambda 0=0$ .
$0\in \mathrm{ker}(f)$ for any linear map $f$ because $f(0)=0$ .
If $f$ is invertible then $\mathrm{ker}(f)=\{0\}$ : if $v\in \mathrm{ker}(f)$ then $v={f}^{1}(0)=0$ .
The "kernel" in a nut is the little bit in the middle that's left when you strip away the husk. If $f(v)=Av$ then we can think of $\mathrm{ker}(f)$ as the space of solutions to the simultaneous equations $Av=0$ , which is the intersection of the hyperplanes $${A}_{11}{v}_{1}+\mathrm{\cdots}+{A}_{1n}{v}_{n}=0,\mathrm{\cdots},{A}_{m1}{v}_{1}+\mathrm{\cdots}+{A}_{mn}{v}_{n}=0.$$ In other words, it's the little bit left over when you've intersected all these hyperplanes.
Consider the simultaneous equations $Av=b$ ($A$ is an $m$ by$n$ matrix and $b\in {\mathbf{R}}^{m}$ ). Let $f(v)=Av$ . The space of solutions to $Av=b$ , if nonempty, is an affine translate of $\mathrm{ker}(f)$ .
If $A=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ (so $f$ is vertical projection to the $xy$ plane) then $Av=b$ has a solution only if $b$ is in the $xy$ plane, and in that case it has a whole vertical line of solutions sitting above $b$ .
This vertical line of solutions is parallel to the kernel of $f$ (the $z$ axis), i.e. it is a translate of the kernel.
We saw this lemma earlier in a different guise in Video 19. Namely, we saw that if ${v}_{0}$ is a solution to $Av=b$ then the set of all solutions is the affine subspace ${v}_{0}+U$ where $U$ is the space of solutions to $Av=0$ . In other words, $U=\mathrm{ker}(f)$ .
In particular, we see that if $Av=b$ has a solution then it has a $k$ dimensional space of solutions, where $k$ is the dimension of $\mathrm{ker}(f)$ .
Remember that the space of solutions has dimension equal to the number of free variables when we put $A$ into reduced echelon form. For example, $A=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ is in reduced echelon form with two leading entries and one free variable, which is why we get 1dimensional solution spaces.
The nullity of $A$ (or of $f$ ) is the dimension of $\mathrm{ker}(f)$ (i.e. the number of free variables of $A$ when put into reduced echelon form).
Our goal for the next video is to prove the ranknullity theorem which gives us a nice formula relating the nullity to another important number called the rank.