Question: Assuming we have the matrices $M$, and $H$.
I substituted both in the equation $N = M^{-1}HM$ to obtain the matrix $N$. I calculated eigenvalues of $H$ and $N$ separately using the equation $Ax= \lambda A$. 
If I found eigenvalues of $N$ and $H$ are pretty close to one another, can I conclude that $M$ is a good estimation for eigenvectors of $H$? (In other words, if $M$ is a different matrix, this similarity in eigenvalues cannot be guaranteed)
Is there any other conclusions I can draw from this scenario? 
Does anyone have additional insights on this scenario?

Thanks

Answer: 
$H$ and $M^{-1} H M$ always have the characteristic polynomial and the same eigenvalues. In fact, $v$ is an eigenvector for $H$ with eigenvalue $\lambda$ iff $M^{-1} v$ is an eigenvector for $M^{-1} H M$ with eigenvalue $\lambda$. 

Of course, in a numerical example roundoff error may cause small differences between the computed eigenvalues for $H$ and the computed eigenvalues for $M^{-1} H M$.
I can give a counter example. If you choose the identity matrix for $M$, then $N=H$ and so per definition their eigenvalues will also be the same.

I am a bit rusty on this part, but if I remember correctly, then choosing any nonsingular matrix for $M$ will have this property (the actual eigenvalues might differ due to numerical rounding).