# Similar Matrices - Proof

1. Jul 31, 2009

### phyzmatix

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

Let A and B be similar matrices. Prove that $$A^T$$ and $$B^T$$ are also similar.

2. The attempt at a solution

We know that, since A and B are similar matrices, there exists an invertible matrix P such that

$$B=P^{-1}AP$$

so, I thought if we transposed both sides, this could lead to a proof, but this gives

$$B^T=\left(P^{-1}AP\right)^T$$

$$B^T=P^TA^T\left(P^{-1}\right)^T$$

$$B^T=P^TA^T\left(P^T\right)^{-1}$$

Which feels close, but how do I get it to the form

$$B^T=\left(P^T\right)^{-1}A^TP^T$$ ?

Provided of course, that this is an actually viable route to take (I have virtually zero experience with mathematical proofs).

Thanks!
phyz

2. Jul 31, 2009

### winter85

hi phyz,
you reached $$B^T=P^TA^T(P^T)^{-1}$$.
if you put $$Q = (P^T)^{-1}$$ , what form would the above statement take?

3. Aug 1, 2009

### phyzmatix

Ha! Easy as that is it?

Thanks winter85!

4. Aug 1, 2009

yep :)