Mathematically, the convolution is an operation that takes two functions, $f$ and $g$, and produces a third function, $h$. Concisely, we can denote the convolution operation as follows
$$f \circledast g = h$$
In the context of computer vision and, in particular, image processing, the convolution is widely used to apply a so-called kernel (aka filter) to an input (typically, an image, but this does not have to be the case). The input (e.g. an image), the kernel, and the output of the convolution, in this context, is usually a matrix or a tensor. In image processing, the convolution is typically used to e.g. blur images or maybe to remove noise.
However, in the beginning, I said that the convolution is an operation that takes two functions (and not matrices) and produces a third one, so these two explanations of the convolution do not seem to be consistent, right?
The answer to this question is that the two explanations are consistent with each other. More precisely, if you have a function $f : X \rightarrow Y$ (assuming that $X$ is discrete/countable), you can represent it in a vector form as follows $\mathbf{f} = [y_1, y_2, \dots, y_n]$, i.e. $\mathbf{f}$ is a vector that contains all outputs of the function $f$ (for all possible inputs).
In image processing, an image and a kernel can also be thought of as a function with a discrete domain (i.e. the pixels), so the matrices that represent the image or the kernel are just the vector forms of the corresponding functions. See this answer for more details about representing an image as a function.
Once you understand that the convolution in image processing is really the convolution operation as defined in mathematics, then you can simply look up the mathematical definition of the convolution operation.
In the discrete case (i.e. you can think of the function as vectors, as explained above), the convolution is defined as
$${\displaystyle h[n] = (f \circledast g)[n]=\sum _{m=-M}^{M}f[n-m]g[m].} \tag{1}\label{1}$$
You can read equation $1$ as follows
- $f \circledast g$ is the convolution of the input function (or matrix) $f$ and the kernel $g$
- $(f \circledast g)[n]$ is the output of the convolution $f \circledast g$ at index (or input position) $n$ (so you need to apply equation \ref{1} for all $n$, if you want to have $h$ and not just $h[n]$)
- So, the result of the convolution at $n$, $h[n]$, is defined as $\sum _{m=-M}^{M}f[n-m]g[m]$, a sum that goes from $m = -M$ to $m = M$. Here $M$ may be half of the length of the kernel matrix. For example, if you use the following Gaussian kernel, then $M = 2$ (and I assume that the center of the kernel is at coordinate $(0, 0)$).
$$
\mathbf{g} =
\frac{1}{273}
\begin{bmatrix}
1 & 4 & 7 & 4 & 1 \\
4 & 16 & 26 & 16 & 4 \\
7 & 26 & 41 & 26 & 7 \\
4 & 16 & 26 & 16 & 4 \\
1 & 4 & 7 & 4 & 1
\end{bmatrix}
\label{2}\tag{2}
$$
Here are some notes:
The kernel \ref{2} is symmetric around the $x$ and $y$ axes: this actually implies that the convolution is equal to the cross-correlation, so you don't even have to worry about their equivalence or not (in case you have ever worried about it, which would have happened only if you already came across the cross-correlation). See this question for more info.
The kernel \ref{2} is the vector form of the function form of the 2d Gaussian kernel (the one in your question): more precisely, an integer-valued approximation of the 2D Gaussian kernel when $\sigma = 1$ (as stated in your slides).
The convolution can be implemented as matrix multiplication. This may not be useful now, but it's something useful to know if you want to implement it. See this question for more info.
Question for you: what is the result of the application of this Gaussian kernel to any input? What does this kernel intuitively do? Once you fully understand the convolution, you can answer this question.
