Given two square matrices $A$ and $B$. There are quite some results on the distance between the eigenvalues, e.g.,

$$ | \lambda_A - \lambda_B | \leq \| A - B \|_F, $$ where $A$ and $B$ are Hermitian (see here for more). I am looking for similar results for the argument of the eigenvalues, for example
$$ | arg(\lambda_A) - arg(\lambda_B) | \leq \| A - B \|_? $$ One simple instance I have found: if invertible $A$ is normal, given the polar decomposition $A = UP$, then $| arg(\lambda_A) - arg(\lambda_U) | = 0$. Can this result be extended in terms of perturbation?


1 Answer 1


Uhm, it does not hold even in $\mathbb{R}^{1\times 1}$. $A=\varepsilon, B = -\varepsilon$ gives $LHS=\pi$, $RHS = 2\varepsilon$.

  • $\begingroup$ Right, but please see my normal matrix example. It will certainly need some constraints. $\endgroup$
    – Jiro
    May 23, 2018 at 11:34
  • $\begingroup$ @SebastianSchlecht All 1x1 matrices are normal. What is your definition of argument? Isn't it $arg(-1)=\pi$, $arg(1)=0$? $\endgroup$ May 23, 2018 at 11:38
  • $\begingroup$ Yes, I agree on the definition. What I wanted to say is that in the normal matrix example I have given, the perturbation is $P$. Therefore your example would not occur. $\endgroup$
    – Jiro
    May 23, 2018 at 11:45
  • $\begingroup$ Hmm, that's a different kind of perturbation, though, a multiplicative one. There is some work around on multiplicative perturbation theory (for instance, Froilan Dopico worked on it), maybe it contains something interesting for you. $\endgroup$ May 23, 2018 at 12:00

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