# Similarity is transitive

1. Nov 22, 2008

### FourierX

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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: Nov 22, 2008
2. Nov 22, 2008

### Dick

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.

3. Nov 22, 2008

### ptr

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.

4. Nov 22, 2008

### FourierX

how do PQ and P-1Q-1 behave ?

5. Nov 22, 2008

### Dick

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.