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

Given a set $\{f_{1},\ldots,f_{k}\}$ of maps $\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ how is it usually defined the notion of linear (in)dependence among them? (Is it the same as for single variable real valued functions?).

Does the notion of linear (in)dependence have an analog, or a generalization to maps between more general types of spaces? (e.g. maps between Banach or Hilbert spaces, or even metric spaces)

share|cite|improve this question
You have to plug your functions into a vector space. The definition of linear dependence is the same in every vector space. – Siminore Sep 24 '12 at 16:27
@Siminore: Considering the single variable case $n=m=1$, if the functions $f_1,\ldots,f_k$ are in $C^{k-1}$ then linear independence in addition requires that $c_{i}f_{i}^{(1)}=c_{i}f_{i}^{(2)}=\cdots=c_{i}f_{i}^{(k-1)}=0$. Doesn´t it? – John Sep 24 '12 at 16:33
The definition in the above link is exactly as Siminore said. The entire section on the Wronskian is a set of conclusions that one can draw about the linear independence of functions if they are all $n-1$-differentiable. – Thomas Andrews Sep 24 '12 at 16:52
@John You should not confuse the definition of linear dependence with the tricks you can do in single, special cases. For functions, the definition of linear dependence is a pointwise one: $\sum_{i=1}^n \alpha_i f_i(x)=0$ at every point $x$. Maybe you can take derivatives to get some additional information, but this depends on the choice of the vector space. – Siminore Sep 24 '12 at 17:13
up vote 2 down vote accepted

It's the same. The set is linearly independent if there is no non-zero solution to $$ a_1f_1+\cdots a_kf_k=0 $$ for $a_1,\ldots ,a_k\in \mathbb{R}$.

share|cite|improve this answer
This is what I though. However, are there no aditional conditions on the partial derivatives? (See my coment to Siminore above) – John Sep 24 '12 at 16:35
@John: linear independence is a linear algebraic property. It does not depend on partial derivatives or anything of the sort. – tomasz Sep 24 '12 at 17:01
@tomasz John is confusing the definition of linear independence in the above Wolfram link with the section on the Wronskian that follows the definition. That section is not the definition of independence, but rather a cute way of showing independence of $n$ $n-1$-differentiable functions – Thomas Andrews Sep 24 '12 at 17:25
@Thomas. +1 to your comment. The point you make precisely is where my confusion came from. The way it is written it seems that both parts are part of the same definition when the functions are $C^{n-1}$. – John Sep 24 '12 at 17:57

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.