3

Two matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ for which $B = PAP^{-1}$.

This definition seems presented without any motivation behind it. However, I did notice that it seems to be very much like the automorphism conjugation, $f(x) = axa^{-1}$ for some $a$. Is this just a coincidence, or is there some connection?

Michael Hardy
  • 1
Joseph DiNatale
  • 2,845
  • 22
  • 36
  • 2
    Related: What is the motivation defining Matrix Similarity?. – Git Gud Aug 03 '14 at 00:33
  • 1
    From Wikipedia : "A transformation $A\mapsto P^{-1} A P$ is called a similarity transformation or conjugation of the matrix $A$. In the general linear group, similarity is therefore the same as conjugacy, and similar matrices are also called conjugate; however in a given subgroup $H$ of the general linear group, the notion of conjugacy may be more restrictive than similarity, since it requires that P can be chosen to lie in $H$." – Semiclassical Aug 03 '14 at 00:45
  • "Conjugation automorphism," (since adjective goes before noun), not "automorphism conjugation." What exactly do you feel is the coincidence here? As I see it, we're not looking at two things coinciding improbably, we're looking at a single thing occurring once in a broad definition and another place in a more specific context. – anon Aug 03 '14 at 04:32

1 Answers1

2

The motivation is that two matrices are similar if and only if they both represent the same linear transformation with respect to two different bases. So you convert from one basis to another by multiplying by $P^{-1}$, then apply the matrix $A$ which represents the transformation in that other basis, then convert back to the original basis by multiplying by $P$, and get the same result as if you just applied the linear transformation by multiplying by the original matrix $B$, which represents the linear transformation in the original basis.

And that's what conjugation is: transforming something, then doing something with the transformed object, then transforming back again.

So it's not "coincidental" at all.

Michael Hardy
  • 1