1
$\begingroup$

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

$\endgroup$
4
$\begingroup$

$H$ 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 $M^{-1} H M$.

$\endgroup$
  • $\begingroup$ You mean even if $M^{-1}$ was not obtained using the equation Ax=λx, the eigenvalues would be close to one another? My assumption is, In general, the equality of eigenvalues holds true when we know that columns in $M$ are the eigenvectors. $\endgroup$ – Crimson Aug 13 '16 at 3:34
  • $\begingroup$ you are right. I tried different matrices( M ).So far when matrix M is non-singular, the eigenvalues of $H$ and $N$ are the same. $\endgroup$ – Crimson Aug 13 '16 at 3:54
  • $\begingroup$ I wonder is there any relations between eigenvectors of similar matrices? $\endgroup$ – Crimson Aug 13 '16 at 4:03
  • $\begingroup$ You wonder? Did you read the second sentence of my answer? $\endgroup$ – Robert Israel Aug 14 '16 at 5:45
  • $\begingroup$ I see. $\lambda v = Hv $ => $\lambda v = H(MM^{-1})v$ => $M^{-1} \lambda v = M^{-1} H(MM^{-1})v$ => $M^{-1} \lambda v = (M^{-1} HM)M^{-1}v$. Thanks. $\endgroup$ – Crimson Aug 14 '16 at 14:08
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Not sure if I understand correctly what you meant in the first paragraph of your answer. If the eigenvalues of N and H are the same I think it will not be a counter example. $\endgroup$ – Crimson Aug 13 '16 at 2:06
  • 1
    $\begingroup$ @Zereshki You ask if $M$ would be a good estimation for the eigenvectors of $H$ when the eigenvalues of $N$ and $H$ are (almost) the same. But in general the identity matrix will most likely not be the eigenvectors of $H$, even though then the eigenvalues of $N$ and $H$ are always the same. $\endgroup$ – Kwin van der Veen Aug 13 '16 at 2:12
  • $\begingroup$ I see. Now I understood what you meant. My conclusion likely has a counter example $\endgroup$ – Crimson Aug 13 '16 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.