In my study of linear algebra, I learned that the definition of linearity is:

A transformation (also: map, mapping, function, etc.) is linear if it preserves additivity and scaling (also: homogeneity).

This means that a linear map $L(\vec{\mathbf{v}})$ is linear if:

$$L(\vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2) = L(\vec{\mathbf{v}}_1) + L(\vec{\mathbf{v}}_2) $$ $$ kL(\vec{\mathbf{v}}) = L(k\vec{\mathbf{v}})$$

While I’m not sure how this relates to the “origin staying fixed” and “grid lines staying parallel and evenly spaced” from 3Blue1Brown’s essence of linear algebra series, I won’t focus on that now.

I also learnt in my study that there are non-linear algebra things that are in a sense, linear maps.

Take the derivative operator $\frac{d}{dx}$. In a calculus class one of the first things we learn about derivatives is

$$ \frac{d}{dx}(f + g) = \frac{d}{dx}(f) + \frac{d}{dx}(g) $$ $$ \frac{d}{dx}(kf) = k\frac{d}{dx}(f) $$

These are the exact conditions for some operator to be linear. The derivative, single or partial, is a linear operator.

Actually, the same applies to the integral operator:

$$ \int (f + g) dx = \int f dx + \int g dx $$ $$ \int(kf) dx = k\int f dx $$

Hence the integral operator $\int dx$ is also a linear operator.

Using these fundamental operators we can also derive the linearity of the Laplace transform and Fourier transform operators.

Question: Except in linear algebra and analysis (calculus, complex analysis, real analysis, Fourier analysis, and everything else), are there other linear operators? E.g. linear operators like the convolution are part of analysis.

It seems really hard to find any!

Help is appreciated. Answers may be inside or outside the scope of elementary mathematics (for me, anything differential equations or below is elementary mathematics).

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    $\begingroup$ How about linearity of expected value in statistics? $\endgroup$
    – Gregory
    Commented Feb 6 at 17:32
  • $\begingroup$ First, a linear operator is a very specific kind of linear transformation. namely ones of a space to itself. $d/dx$ is a linear operator on the space of infinitely differentiable functions. It is not any different from any other linear operator. but it acts on a space of functions, not a traditional space of finite-dimension vectors. $\endgroup$ Commented Feb 6 at 17:39
  • $\begingroup$ You can see $d/dx$ acting on a finite-dimensional space if we act on the space of polynomials of degree less than $n.$ $\endgroup$ Commented Feb 6 at 17:41
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    $\begingroup$ In any event, you need to understand the notion of a "vector space." This includes the intro notion of $\mathbb R^n,$ but it also includes lots of other examples. $\endgroup$ Commented Feb 6 at 17:43
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    $\begingroup$ I can see your point, but I have heard $T$ as an operator and as a transformation $T: \mathbb R^n \to \mathbb R^m$. Come to think of it though, I have seen operator be used without specifying the codomain, so perhaps linear operator is a subset of linear transformation. /musings. tl;dr I agree that expectation is certainly a linear functional. $\endgroup$
    – Gregory
    Commented Feb 6 at 17:54

1 Answer 1


Since by definition a liner map needs to be defined on a vector space, to find exotic examples one needs to consider exotic vector spaces.

For example, for any set $X$ consider $V := {P}(X)$ to be the power set of $X$. Define $$\begin{align*} +: V &\times V \to V, (A,B) \mapsto A \triangle B, \end{align*}$$ and consider the scalar multiplication over $\mathbb{Z}_2$ given by $1 \cdot A = A$ and $0 \cdot A = \emptyset$. Then $V$ is a vector space over $\mathbb{Z}_2$. Now fix any $Y \subseteq X$ and consider the map

$$f: V \to V, A \mapsto A \cap Y.$$

Try to verify that $f$ is indeed a linear map.

  • $\begingroup$ Is there something about the phrase 'linear map' that excludes maps between modules over rings? $\endgroup$
    – Barri
    Commented Feb 6 at 18:11
  • $\begingroup$ No, it's just a matter of definition. Although in my experience, when you only have modules the name "homomorphism" is more common. $\endgroup$
    – Dedekind
    Commented Feb 6 at 19:01

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