Proving Transitivity of Similarity in Linear Algebra

  • Thread starter Thread starter FourierX
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around proving the transitivity of similarity in linear algebra, specifically focusing on the relationships between matrices and their similarity transformations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of similarity transformations, with attempts to express the relationships between matrices A, B, and C using invertible matrices. Questions arise regarding the behavior of products of matrices and their inverses.

Discussion Status

The discussion is active, with participants providing various formulations and exploring the properties of matrix products and inverses. There is no explicit consensus, but several lines of reasoning are being examined.

Contextual Notes

Participants are working within the constraints of linear algebra principles and the definitions of similarity, with some questioning the implications of matrix behavior in the context of similarity transformations.

FourierX
Messages
73
Reaction score
0

Homework Statement



In linear algebra, how to prove that the Similarity is transitive?

Homework Equations



[T]C = P-1 [T]B P

The Attempt at a Solution


My logic: if a is similar to b and b is similar to c, then a is similar to c. But how do you implement that in case of similarity ?
 
Last edited:
Physics news on Phys.org
If A is similar to B, A=PBP^(-1). If B is similar to C, B=QCQ^(-1). Where P and Q are invertible. Put the second equation into the first equation.
 
If A = P B Pinv and B = K C Kinv, where Pinv is the inverse of P, and mutatis mutandis for K, use the property that (PK)inv is (Kinv)(Pinv). and hence A = (PK) C (PK)inv.
 
how do PQ and P-1Q-1 behave ?
 
You should have asked how do PQ and Q^(-1)P^(-1) behave. Whatever 'behave' means. They don't do anything, they are just invertible matrices. And one is the inverse of the other, see ptr's post.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K