So I am teaching myself more in-depth about integral operators and every once and awhile I see this little 'factoid', that integral operators are generalizations of matrix multiplications.

In particular, if:

$$Lf(x) = \int_{X} k(x,y) f(y) du(y)$$

So then, naturally, if $\mathbf{A}$ is an $m,n$ matrix with entries $a_{ij}$ and $\mathbf{u} \in \mathcal{C}^{n}$, $\mathbf{A}\mathbf{u} \in \mathcal{C}^{m}$, we have:

$(\mathbf{A}\mathbf{u})_{i} = \sum_{j=1}^{n} a_{ij} u_{j}, i=1 \ldots m$

So I always see something similar to the following statement:

The entries $k(x,y)$ are analogous to the entries $a_{ij}$ of matrix $\mathbf{A}$ and the values $Lf(x)$ are analogous to the entries $(\mathbf{A}\mathbf{u})_{i}$

I have never seen any example or more detail regarding this statement. Can somebody make this a bit clearer?

Here is my thought, the $x$ defines the 'row', while the $y$ defines the column.

So then if we wanted to, we could define a sequence $\{x_{0}, x_{1}, \ldots\}$ to then 'sample' the output of the operator integral $L$. For instance, if $L$ mapped $f$ to a certain region $P$, we would want to define our sequence to exist only in this region $P$ in order to save computation. If in this region, the functions $Lf$ had significant norm/power only in a subspace or section of this region $P_{k}$, we could 'sample' this subspace only -- ie $\{x_{0}, x_{1}, \ldots\} \in P_{k}$ and still get a 'reasonable' approximation to the resulting function/signal $Lf$.

Similarly, we could 'sample' $f$ on its domain with $\{y_{0}, y_{1}, \ldots\}$ and reduce the computation even further, when the sequence captures 'most of the information' about $f$ that is.

  • 1
    $\begingroup$ if $k(x,y)$ is continuous (or piecewise continuous) and $X$ is a bounded subset then the operator $L$ is the limit of a converging (in some sense) sequence of finite rank operators : it is a compact operator $L^p \to L^p$ (see en.wikipedia.org/wiki/Lp_space and en.wikipedia.org/wiki/Mercer%27s_theorem ) $\endgroup$
    – reuns
    Feb 8, 2016 at 17:30
  • $\begingroup$ In the book on quantum mechanics by Dirac, around the place where the author introduces the "delta function", there is something about this "factoid" you mention. You might want to have a look $\endgroup$ Feb 8, 2016 at 23:55

2 Answers 2


Its always a bit hard to guess what another person might find intuitive, but here are my two cents on the topic.

You can interpretate the elements of $\mathbb{R}^n$ as functions from the set $\{1,...,n\}$ to $\mathbb{R}$, where for $f \in \mathbb{R}^{n}$, $f(i)$ would just be the $i$-th component of the vector. We know from linear algebra that any linear operator $L: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ can be written as $L f = A\cdot \vec x$, where $A$ is an $n \times n$-matrix and $\vec x$ is the vector associated with $f$. We could invent a "kernel" function to write this down differently, with $k: \{1,...,n\} \times \{1,...,n\} \to \mathbb{R}$, $k(i,j) := A_{ij}$. We then have the formula

$$Lf(i) = (A\cdot\vec x)_i = \sum_{j=1}^{n} k(i,j) f(j).$$ Now let's replace $\{1,...,n\}$ with some infinite set $X$. Writing down matrices and using the multiplication rules in the same way as in $\mathbb{R}^n$ seems to be a complicated approach here, but it is easy to see what the generalisation of the formula above should be: The values $k(x,y)$ for $x,y \in X$ are the "matrix entries", so we get $$Lf(x) := \sum_{y \in X}k(x,y)f(y)$$ Now for countable $X$ this might still make sense, if we introduce some restrictions on $k$ and $f$ in order to ensure convergence, but for uncountable $X$ (which is the more interesting case) the sum doesn't make sense any more (at least if $k$ is nonzero almost everywhere). The integral is often viewed as a "continuous" analogon to summation (e.g. by physicists, or in measure theory), and as it is itself a limit of sums, it seems only natural to consider operators of the form

$$Lf(x) = \int_{X}k(x,y)f(y) dy$$

  • 1
    $\begingroup$ Excellent answer! $\endgroup$
    – The Dude
    Feb 8, 2016 at 21:58

More thoughts about this. Matrix $A$ can be thought of as a linear operator from $\mathbb{R}^n$ to $\mathbb{R}^n$. In a similar way your integral transform $L$ is an operators from a (Hilbert) space of functions to a different space.

Just like you can define characteristics of $A$ (like eigenvalues and eigenvectors), and talk about basis of its image, so too you can do the same to $L$.

For an in-depth example of continuous and discrete transformations with similar eigenvalues and "related" eigenvectors, look at continuous and discrete Fourier transforms.

  • 2
    $\begingroup$ Indeed, $L$ has eigenvalues and eigenfunctions, while $A$ has eigenvalues and eigenvectors. $\endgroup$
    – The Dude
    Feb 8, 2016 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.