I was wondering how to prove the following:
For any square matrix A, prove that A and A^t (transpose) have the same characteristic polynomial (and hence the same eigenvalues)
It is well known, though non trivial at all to prove, that a square matrix is always similar (conjugate) to its transpose, so it's enough to show that similar matrices have the same char. pol.
Assume then that $\displaystyle A\sim B \Longrightarrow A=P^{-1}BP$, for some invertible matrix $\displaystyle P$. Let us denote the determinant by |.|, so:
$\displaystyle |\lambda I-A|=|\lambda I-P^{-1}BP|=|P^{-1}(\lambda I-B)P|=...$...end the argument now.
Here above there're quite a few non-trivial things you must know: distributivity of matrix multiplication, the fact that scalar matrices commute with any other matrix, the multiplicativity of the determinant, etc.
If you don't know these things then the above won't help much.
Tonio
Couldn't you just prove that they are similar like this:
PA(P^-1) = D
((P^-1)^T)(A^T)(P^T) = D^T, transpose of both sides
(Q^-1)(A^T)Q = D, By letting Q = P^T, and transpose of diagonal matrix is still diagonal (also note: (P^-1)^T = (P^T)^-1 = Q^-1)
(P^-1)(Q^-1)(A^T)(Q)(P) = A, so they are similar, share the same eigenvalues and therefore have the same characteristic polynomial.
(Note: (QP)^-1 = (P^-1)(Q^-1), in case you were wondering why the previous equations suggests similar matrices)
Your entire proof seems to be heavily based on the very first line, and this is already a big assumption: that A is diagonalizable and thus $\displaystyle PAP^{-1}=D$ for some diagonal D.
The problems begin when trying to prove this for general, probably non-diagonalizable, matrices...The only way I can think of to prove this is using Jordan Canonical Forms and this already makes the proof a little more involved than the above one.
Tonio