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From Weyl's theorem, i.e.:
Let $A$ and $E$ be $n\times n$ real symmetric matrices. Let $\alpha_1\geq\ldots\geq\alpha_n$ be the eigenvalues of $A$ and $\hat{\alpha}_1\geq\ldots\geq\hat{\alpha}_n$ be the eigenvalues of $\hat{A}=A+E$. Then $|\alpha_i-\hat{\alpha}_i|\leq\|E\|_2$,
we know that the perturbed eigenvalues of a symmetric matrix differ at most $\|E\|_2$ size, which means they can be calculated in a high accuracy.

But here I have the question: what if we assume $E$ be a skew-symmetric matrix, i.e.: $E^T=-E$? Do we also have $|\alpha_i-\hat{\alpha}_i|\leq c\|E\|_2$, for some constant $c$?

If it is, we may conclude that any perturbed matrix to $A$ (not only symmetric perturbation) changes little to $A$'s eigenvalues.

If not, is there any example to show that such $c$ (of $O(1)$ order) does not exists?

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  • $\begingroup$ Is $\|\cdot\|_2$ the Frobenius norm (entry-wise 2-norm) or the operator norm (the vector-induced norm)? $\endgroup$ – Omnomnomnom Dec 22 '14 at 18:19
  • $\begingroup$ @Omnomnomnom Obviously it is vector-induced 2-norm. Frobenius norm is denoted by $\|\cdot\|_F$ $\endgroup$ – gaoxinge Dec 23 '14 at 9:10
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    $\begingroup$ @gaoxinge notation varies. Horn and Johnson, for example, uses differing notation, wherein $\|\cdot\|_2$ denotes the Frobenius norm and $|\|\cdot\||_2$ denotes the induced 2-norm. $\endgroup$ – Omnomnomnom Dec 23 '14 at 14:55
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Since $A$ is diagonalizable, according to the Bauer-Fike theorem, for any eigenvalue $\lambda$ of $A$, there exists an eigenvalue $\mu$ of $A+E$ such that $$ |\lambda-\mu|\leq\kappa_2(V)\|E\|_2, $$ where $\kappa_2(V)$ is the spectral condition number of an eigenvector matrix $V$ of $A$. We can put $\kappa_2(V)=1$ because $A$ is symmetric and thus $V$ can be chosen to be orthogonal. Hence $$|\lambda-\mu|\leq\|E\|_2.$$ Note that $E$ does not need to be symmetric. If $E$ is skew-symmetric and hence normal, $\|E\|_2$ is equal to the spectral radius of $E$.

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  • $\begingroup$ Thanks for answering my question, and this indeed helps me to understand the perturbation theory. But now I doubt that, since we have more property on $E$ (skew-symmetric), can we have a more precise estimate on $|\lambda-\mu|$, say $|\lambda-\mu|\leq c\|E\|_2^2$, for some constant $c$? $\endgroup$ – Gabriel Dec 23 '14 at 14:00
  • $\begingroup$ @Gabriel You cannot replace $\|E\|_2$ with $c\|E\|_2^2$ (with $c$ independent of $E$). Consider $A=0$, then the eigenvalues of $A+E$ are simply those of $E$ which cannot be bounded from above in the absolute value by anything smaller than $\|E\|_2$. $\endgroup$ – Algebraic Pavel Dec 23 '14 at 15:58
  • $\begingroup$ Oh, I'm sorry. I forget to mention that $A$ has different eigenvalues. For space limited, I asked another question just now, can you take a look? math.stackexchange.com/questions/1079423/… $\endgroup$ – Gabriel Dec 24 '14 at 2:25

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