linear algebra - Is this circular reasoning, when trying to prove that similar matrices have the same eigenvalues and characteristic polynomial?

I am trying to prove that similar nxn matrices A and B share the same eigenvalues and characteristic polynomial.

I am not sure how to start, but I am tempted to start with something like

$$tr(A) = tr(B)$$

and

$$det(A) = det(B)$$

since we know that the matrices are similar.

But, this approach sort of seems to be using what I am trying to prove.

Is this a valid approach? Or, should I head in another direction instead?

Thanks,

Answer

If $A$ is similar to $B$ then there exists invertible $P$ such that $A=P^{-1}BP$.

We can rewrite this as $PA=BP$.

Suppose $\lambda$ is an eigenvalue of $A$ with eignevector $v$. Then $$PAv=P(\lambda v)=\lambda Pv=BPv$$

$\therefore\lambda$ is an eigenvalue of $B$ with eigenvector $Pv$.

Switching the roles of $A$ and $B$ gives you that an eigenvalue of $B$ is an eigenvalue of $A$.