4 ms·

Eigen-things can also be thought of "fixed values" of a "thing" transformation. For example - the eigenfunction of a derivative is e^x since when you run the d

Eigen-things can also be thought of "fixed values" of a "thing" transformation. For example - the eigenfunction of a derivative is e^x since when you run the derivative function on e^x you get.... e^x

marviel 4y ago
After watching the 3blueonebrown video linked below in the comments, I'm inclined to agree with you -- nice way of putting it.
Tainnor 4y ago
They don't have to be fixed, a scalar multiple is allowed too. e^ax is an eigenfunction of the derivative too.
- rmeno12 4y ago
  In this case, we could say that all functions e^ax are eigenfunctions of the derivative, with the eigenvalues being a.
patrick451 4y ago
Another good example is that sinusoids are eigenfunction of a linear dynamical system. If you drive the system with a sinusoid, the output is also a sinusoid of the same frequency, but different amplitude and phase. In other words, the frequency response is the eigenvalue at a given frequency.
monktastic1 4y ago
Functions satisfy the laws of a vector space (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here). I.e., functions are vectors. And the derivative is a linear operator on that space: d/dx(af + bg) = ad/dx(f) + bd/dx(g). Therefore d/dx has eigenvalues, one of which you properly identified as e^x (and more generally, e^ax, as others have pointed out).
- Tainnor 4y ago
  > (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here) For any vector space V over a field F, the set of functions V -> V is a vector space over F. This is essentially because the vector space properties are carried over "point-wise". When you consider only certain classes of functions (such as L^2), the question of whether they form a vector space boils down to the question whether those classes are closed under addition and scalar multiplication. All other vector space properties are satisfied because they are in the larger vector space of all functions.
pclmulqdq 4y ago
This is the best explanation of the eigenthings in my opinion. Too many people, this article included, reach for more advanced concepts to explain something as simple as "eigenvectors are how an operator transforms stuff and eigenvalues are how much."
- kiernanmcgowan 4y ago
  Thank you - its too easy to get into the weeds with mathematics and lose the "why" of something. Connecting these higher level concepts with the algebra that _most_ people have experience with ends up being a more direct way to share knowledge vs diving into the higher level stuff that enthusiasts really enjoy.
- kridsdale2 4y ago
  God knows how I passed 4 years of engineering math without ever being able to describe to anyone what eigen* actually are, but I think that's actually the norm.
  - pclmulqdq 4y ago
    That's completely normal and very unfortunate. I only learned about the geometry of the eigen*s from an EE professor who loved math. Unfortunately, the history of the eigen*s is rooted in the opaque and cryptic (but rigorous) linear algebra definition, and the non-rigorous but meaningful geometry came later.
ukj 4y ago
So… fixed points? https://en.wikipedia.org/wiki/Fixed_point_(mathematics) https://en.wikipedia.org/wiki/Fixed_point_(mathematics)
- bobbylarrybobby 4y ago
  Yes except eigenvalues need not be 1. e^(kx) is an eigenvector of d/dx for all k but is only a fixed point when k=1
  - ukj 4y ago
    Right, in topology a “point” can represent any mathematical object e.g 1 of “e^(kx)”
- sampo 4y ago
  > So… fixed points? Fixed directions. The operator operating on an eigenvector doesn't change the vector's direction, but can change its magnitude (i.e. length). (Except, if you change the length from positive to negative, then you kind of change, or flip, the direction to 180 degrees opposite.)

marviel 4y ago

After watching the 3blueonebrown video linked below in the comments, I'm inclined to agree with you -- nice way of putting it.

Tainnor 4y ago

They don't have to be fixed, a scalar multiple is allowed too. e^ax is an eigenfunction of the derivative too.

rmeno12 4y ago
In this case, we could say that all functions e^ax are eigenfunctions of the derivative, with the eigenvalues being a.

patrick451 4y ago

Another good example is that sinusoids are eigenfunction of a linear dynamical system. If you drive the system with a sinusoid, the output is also a sinusoid of the same frequency, but different amplitude and phase. In other words, the frequency response is the eigenvalue at a given frequency.

monktastic1 4y ago

Functions satisfy the laws of a vector space (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here). I.e., functions are vectors. And the derivative is a linear operator on that space: d/dx(af + bg) = ad/dx(f) + bd/dx(g). Therefore d/dx has eigenvalues, one of which you properly identified as e^x (and more generally, e^ax, as others have pointed out).

Tainnor 4y ago
> (well, certain classes of functions, such as L^2 anyway -- I'm certainly no expert here) For any vector space V over a field F, the set of functions V -> V is a vector space over F. This is essentially because the vector space properties are carried over "point-wise". When you consider only certain classes of functions (such as L^2), the question of whether they form a vector space boils down to the question whether those classes are closed under addition and scalar multiplication. All other vector space properties are satisfied because they are in the larger vector space of all functions.

pclmulqdq 4y ago

This is the best explanation of the eigenthings in my opinion. Too many people, this article included, reach for more advanced concepts to explain something as simple as "eigenvectors are how an operator transforms stuff and eigenvalues are how much."

kiernanmcgowan 4y ago
Thank you - its too easy to get into the weeds with mathematics and lose the "why" of something. Connecting these higher level concepts with the algebra that _most_ people have experience with ends up being a more direct way to share knowledge vs diving into the higher level stuff that enthusiasts really enjoy.
kridsdale2 4y ago
God knows how I passed 4 years of engineering math without ever being able to describe to anyone what eigen* actually are, but I think that's actually the norm.
- pclmulqdq 4y ago
  That's completely normal and very unfortunate. I only learned about the geometry of the eigen*s from an EE professor who loved math. Unfortunately, the history of the eigen*s is rooted in the opaque and cryptic (but rigorous) linear algebra definition, and the non-rigorous but meaningful geometry came later.

ukj 4y ago

So… fixed points? https://en.wikipedia.org/wiki/Fixed_point_(mathematics) https://en.wikipedia.org/wiki/Fixed_point_(mathematics)

bobbylarrybobby 4y ago
Yes except eigenvalues need not be 1. e^(kx) is an eigenvector of d/dx for all k but is only a fixed point when k=1
- ukj 4y ago
  Right, in topology a “point” can represent any mathematical object e.g 1 of “e^(kx)”
sampo 4y ago
> So… fixed points? Fixed directions. The operator operating on an eigenvector doesn't change the vector's direction, but can change its magnitude (i.e. length). (Except, if you change the length from positive to negative, then you kind of change, or flip, the direction to 180 degrees opposite.)