Proof: Prove B^T ~ A^T (Transpose)

[jesuslovesu]

Homework Statement

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.

Homework Equations

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

 

[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.