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

An eigenvector is a vector in the preimage of the transformation whose direction is not changed when the transformation is applied. It seems like the concept of eigenvectors and eigenvalues would extend beyond matrices (linear transformations) to vector fields (nonlinear transformations). For example, for the vector field $\langle y^2,x^2\rangle$, any vector whose components are identical is an eigenvector of this vector field (if the term applies here). I only figured this by observation, but is there a general way to find these "eigenvectors" for a vector field? Also, it seems the "eigenvalues" would be a function rather than a finite set (e.g. the eigenvalue for $\langle1,1\rangle$ is 1, but the eigenvalue for $\langle3,3\rangle$ is 3, and so on).

What is the interpretation of this? What uses does/might it have in mathematics or engineering?

share|cite|improve this question
Yeah, all the vectors $(a,a)$ are going to wind up being a scalar multiple of themselves, but the scalar is going to be a function of $a$, as you noticed. For another example, $(0,a)$ and $(a,0)$ are also like an eigenvectors for the example you gave. It's interesting that this eigenvectorish idea allows the sum of eigenvectors to still be eigenvectors. Someone will probably write in soon that this is a real thing :) – rschwieb May 10 '13 at 14:17
There are nonlinear eigenproblems, but that's a different kind of nonlinearity. Here it's not clear how $\langle y^2,x^2\rangle $ actually acts on vectors. (But it is clear that $<\cdot,\cdot>$ look much worse than proper angle brackets $\langle \cdot,\cdot \rangle $.) – 75064 May 15 '13 at 23:58
@75064 The input vector is $<x,y>$, so for example an input of $<2,3>$ gives an output of $<9,4>$. – Ataraxia May 16 '13 at 0:08
OK. But the choice of the coordinate origin is arbitrary. And the concept of eigenvector you consider depends on that arbitrary choice. For example, after a coordinate shift the same vector field is $\langle (y-2)^2, (x-7)^2\rangle $, and $\langle 1,1\rangle $ is no longer an eigenvector. ¶ I would not expect such a thing to be of use, in particular in engineering. – 75064 May 16 '13 at 0:12
@75064 I understand what you're saying, I think. Can you expound on this some more and post it as an answer? Specifically, can you explain more on why this makes it less useful in engineering? – Ataraxia May 16 '13 at 0:57

Your Answer


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

Browse other questions tagged or ask your own question.