Homework Help: Proof Similarity

1. Dec 10, 2007

jesuslovesu

1. The problem statement, all variables and given/known data
Prove B^T ~ A^T (tranpose)
Given: B ~ A
Can anyone check if my proof is correct? Towards the end I'm not quite sure if I can do it like that. Do I have to say what P is exactly? The only matrix I can think that would satisfy that is the identity matrix.

2. Relevant equations

3. The attempt at a solution

B ~ A
B = P^-1 * A * P
B^T = P^T * A^T * (P^-1)^T
so P is a matrix that where P^T = P^-1 therefore (P^T)^-1 = P
B^T = P^-1 * A^T * P
B^T ~ A^T

Last edited: Dec 10, 2007
2. Dec 10, 2007

Dick

You don't need to assume P^(-1)=P^T. That means P is a special kind of matrix. It's enough that (P^T)^(-1)=(P^(-1))^T, which is true for any invertible matrix.