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?



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

  • $\begingroup$ 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

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).

  • $\begingroup$ 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
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    $\begingroup$ @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

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