Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The L1 Mat-1.1010 -course here has taught me the linearity conditions $f(a x)=a f(x)$ and $f(a+b)=f(a)+f(b)$. I want to generalize it, some quite irrelevant slow investigation here.

It requires time to verify statements like below, source here.

enter image description here

According to Wikipedia here, $Ly=f$ and

enter image description here

but I cannot see this kind of statements fast, unless going through the conditions one-by-one. So are there fast ways to check whether some differential is linear or not?

share|cite|improve this question
It's handy to remember that sums of linear operators are linear, so you can break stuff down. – Alex Becker Sep 20 '12 at 23:24

Before going any further, let's define different types of linearities first. I use this course book here but I provide translations to English below. Feel free to improve or adapt this answer.

Different types of linearity

  1. Bilinear form here i.e. $V^k \times W^m\mapsto F^n$ where $F^n$ is a field of scalars -- does $k,m,n$ have some constraint?

  2. Sesquilinear, generalizing inner -product to complex numbers here, i.e. $\bar V\times V \mapsto C$.

  3. Differential linearity defined on page 319

enter image description here

Work in progress, trying to improve this answer.


  1. Do not mix up algebraic linear-independence to linearity of differentials, two different things.

The page 288 of the course book outlines:

enter image description here

that means "A finite set of functions $\{f_i, i\in\mathbb Z\}\in V$ is algeberaically linear-independent if $\sum_{i=1}^{n}\lambda_if_i=\bf 0 \rightarrow\lambda_i=0,i\in\mathbb Z$."

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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