Can someone explain to me in simple terms what a non-linear transformation is in maths?

I know some single-variable calculus, but I read it has to do with multi-variable calculus, which I'm not familiar with.

If someone could explain it in simple words, that would be helpful.

  • 8
    $\begingroup$ A transformation which is not linear. Do you understand the definition of a linear transformation? It has nothing to do with whether you are working with one or many variables; for example, the transformation which sends x to x^2 is not linear. $\endgroup$ Aug 5, 2010 at 9:11
  • $\begingroup$ Is it possible the OP is after an explanation of 'differentiable' in a multivariable setting? $\endgroup$ Aug 5, 2010 at 10:57

3 Answers 3


Let $V_1, V_2$ be two vector spaces over the field $F$. A transformation $T: V_1 \to V_2$ is linear if for every $x, y \in V_1$ and every $\alpha \in F$ it is true that

(*) $T(x + \alpha y) = T(x) + \alpha T(y)$

T is not a linear transformation if there are some $x, y, \alpha$ such that (*) is not true.


In addition to the definition of linear map that Tomer remind you, here are two examples.

For instance, $f(x,y) = x^2y$ is not a linear map $f: \mathbb{R}^2 \longrightarrow \mathbb{R}$ because

$$ f(2x,2y) = 4x^22y \neq 2x^2y = 2f(x,y) \ . $$

More generally, the linear maps $f: \mathbb{R}^m \longrightarrow \mathbb{R}^n$ are necessarily of the form

$$ f(x_1, \dots , x_m) = (a_1^1 x_1 + \dots + a_1^m x_m , \dots , a_n^1 x_1 + \dots + a_n^m x_m) $$

with $a^i_j$ constant coefficients.

So, two more examples:

  1. $f(x,y) = x + 2y$ is a linear map.
  2. $f(x,y,z) = 3x + 1$ is a non-linear map

It's a mapping that is not linearly closed. either: zero vector is not within the image of the transformation, the transformation is not closed under scalar multiplication, the transformation is not closed under addition.

  • 4
    $\begingroup$ This is a technically correct answer, but it's 5 years after the question was posted and it doesn't really add anything to the other correct answers. $\endgroup$ Jun 19, 2015 at 22:13
  • 3
    $\begingroup$ I actually found it very beneficial $\endgroup$
    – DanielSon
    Nov 17, 2016 at 2:59

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