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Let $A$ be a square matrix with all eigenvalues negative. What is the relationship between the $\lambda_\max$ of perturbed matrix $A + X$ and the norm of the perturbation $\|X\|$?

PS: I know that the spectral radius of $A+X$ is bounded by the spectral radii of $A$ and $X$, but it only considers $\lambda$ with maximum absolute value, and in my case I'm worried about the sign of $\lambda_\max$.

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  • $\begingroup$ Is the perturbation symmetric? $\endgroup$ – Yiorgos S. Smyrlis Feb 2 '14 at 22:07
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Given a (real symmetric) $n\times n$ matrix $A$, for each $k$, $1\le k\le n$, define the numbers \begin{equation} \lambda_k \,=\, \lambda_k(A) \,=\, \sup_{V_{k-1}} \bigg( \inf_{\substack{ \boldsymbol{u}\perp V_{k-1} \\ \|\boldsymbol{u}\|=1}} \langle\boldsymbol{u},A\boldsymbol{u}\rangle\bigg),\tag{1} \end{equation} where the supremum is taken over all $(k-1)$-planes (i.e., $(k-1)$-dimensional subspaces) $V_{k-1}$. By convention $V_0=\{0\}$ so $\lambda_1=\inf_{\|\boldsymbol{u}\|=1}\langle\boldsymbol{u},A\boldsymbol{u}\rangle$.

Theorem. (Courant, Rayleigh & Ritz). The quantities $\lambda_k$, as defined by $(1)$ are increasing $$ \lambda_1 \,\le\, \lambda_2 \,\le\, \cdots \,\le\, \lambda_n, $$ and are precisely the eigenvalues of $A$, with appropriate multiplicities.

Hence $$ |\lambda_{j}(A+X)-\lambda_{j}(A)|\le \max_{\|\boldsymbol{u}\|=1}\langle\boldsymbol{u},X\boldsymbol{u}\rangle=\|X\|. $$

All these, provided that the perturbation is symmetric!

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  • $\begingroup$ Thanks, that's a totally cool theorem, but in my case A and X are jacobians and hessians, so they are not symmetric in general. I tested numerically and found that for some random matrices this relation is broken (although extremely rarely), probably because of the lack of symmetry. Is there any hope for general matrices? $\endgroup$ – dmytro Feb 3 '14 at 12:22
  • $\begingroup$ A negative definite matrix is also symmetric. $\endgroup$ – Yiorgos S. Smyrlis Feb 3 '14 at 12:27
  • $\begingroup$ Sorry, apparently I made a mistake in the statement of the question. I edited the question. The point is that matrix has all negative eigenvalues (a global stable fixed point) $\endgroup$ – dmytro Feb 3 '14 at 12:40

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