Units of eigenvalues

Question

Suppose you have the system $\bf x' = \bf Ax$, where $\bf x$ is a vector and $\bf A$ is a matrix. What are the units of the eigenvalues of $\bf A$? I think they should be $1/t$ but I'm not sure how to verify this. Can you give me a starting point?

Where did $t$ come into the Question? Pure numbers are "dimensionless"; they have no units in the sense of measurement. — hardmath, Feb 05 '14 at 23:01
This belongs to Physics.SE; There are no units of measurement in eigenvalues. — AlexR, Feb 05 '14 at 23:08
TIL mathematicians have a really difficult time grasping the notion of "unit"... — Dustan Levenstein, Feb 06 '14 at 00:48

Ross Millikan · Answer 1 · 2014-02-05T23:21:13.983

6

Your intuition is correct, they should be 1/(the dimension of what you are taking the derivative with respect to). If your derivative is with respect to time, then it should be $1/t$. If your derivative is with respect to something unitless, the eigenvalue will be unitless, too. Write it as $\frac {dx}{dt}=Ax$ The units of $A$ are then $1/t$ and so the eigenvalues will be, too. If you are taking derivatives with respect to time, maybe the question belongs at physics.

edited Feb 05 '14 at 23:21

answered Feb 05 '14 at 23:02

Ross Millikan

374,822

1

This is not sensible. Whatever units you want to attach to the equation, so that a derivative with respect to time has a measurable interpretation, eigenvalues are by definition pure numbers. We cannot say $Ax = \lambda x$ defines an eigenvalue if the value of $\lambda$ is something other than a scalar. – hardmath Feb 05 '14 at 23:05
2

@hardmath: I believe you can. A matrix can have units, for example the inertia matrix of an object has units $m\cdot \ell^2$ The eigenvalue equation shows the eigenvalue has the same units. As A in this problem has units of inverse time, so must $\lambda$ The units will tell me how the numeric entries in the matrix change if I change units. So if we change the units of time from seconds to minutes, the numeric entries in A need to be multiplied by $60$ – Ross Millikan Feb 05 '14 at 23:11

score 2 · Answer 2 · answered Feb 06 '14 at 00:03

The unit of the eigenvalue is the same as the unit of the matrix itself. That becomes evident from the eigenvalue equation $A v = \lambda v$.

So if your matrix $A$ has unit 1/s (I assume in your case the dash denotes time derivative), the eigenvalue $\lambda$ has unit 1/s as well.

The unit of the eigenvectors on the other hand is undetermined - it can be anything you chose.

Units of eigenvalues

2 Answers2