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by whatshisface 2957 days ago
As someone who understands eigenfunctions already, I don't understand the pictures either. Here is the best way to think about it: a matrix is a transformation, a composition of rotation, scaling, etc. Eigensets are lines going through the origin that the matrix moves points along. So a rotation would have no eigenvectors because none of the points move in a straight line, while a scaling along the x axis would have an eigenset that was also along the x axis, consisting of the points that were moved straight up or down.

To imagine finding the eigenset, just ask, could I draw a line through 0,0 such that any point I put on it would stay on it after the matrix acted?

3 comments

> So a rotation would have no eigenvectors because none of the points move in a straight line, while a scaling along the x axis would have an eigenset that was also along the x axis, consisting of the points that were moved straight up or down.

Maybe I'm being pedantic, but rotation matrices have eigenvectors and eigenvalues, but the eigenvalues are imaginary because imaginary numbers are rotations in the complex plane.

It's exactly like saying x^2 = -1 has no solutions: It has two solutions, like any other quadratic, but neither of them are real.

In three dimensions, rotation matrices have three eigenvalues, one of them being 1, and the eigenvector corresponding to that eigenvalue is, naturally, the rotation axis.

https://en.wikipedia.org/wiki/Rotation_matrix

> So a rotation would have no eigenvectors

Rotations have eigenvectors: a 2D rotation has two complex eigenvectors, a 3D rotation has one real and two complex eigenvectors, ...

That's a fair and true catch, but I can cover myself by pointing out that the article was only talking about matrices in R^(m x n). ;)
When your field of interest is the reals, those complex eigenvectors don't matter.
I want to know it. I have to manipulate the mathematical objects computationally using proof techniques to know it. That just takes time. Thanks though.