Q:

Finding eigenvalues of $A=PDP^{-1}$ for $3\times3$ matrices

If I have the equation $A=PDP^{-1}$ for $3 \times 3$ matrices, where $P$ and $D$ are known and $D$ is a diagonal matrix, how can I use this information to find the eigenvalues of $A$?

A:

For any two square $n \times n$ matrices $M$ and $N$ such that
$M = PNP^{-1}, \tag{1}$
the eigenvalues are the same.  For if
$N \vec v =\mu \vec v, \tag{2}$
with $\vec v \ne 0$, then
$M(P\vec v) = PNP^{-1}(P\vec v) = PN\vec v = P(\mu \vec v) = \mu P \vec v, \tag{3}$
showing that $\mu$ is an eigenvalue of $M$ with eigenvector $Pv \ne 0$.   Thus every eigenvalue of $N$ is also one of $M$;  reversing the roles of $M$ and $N$ so that 
$N = QMQ^{-1} \tag{4}$
where
$Q = P^{-1} \tag{5}$
shows the eigenvalues of $M$ are shared by $N$.  Thus the eigenvalues of $M$ and $N$ are the same.  Now take $N = D$ and $M = A$.  Since $D$ is diagonal, the eigenvalues of $D$ are simply its diagonal entries.

