# Condition number of two perburbation matrix regarding limit and quadtrature integration rules

I have a question regarding the condition number of two different perturbation matrices. To start with let $$A$$ be a spd matrix with elements defined by $$a_{i,j} = \int\limits_{\Omega\subset \mathbb{R}^d} \nabla \varphi_i\cdot \nabla\varphi_j \,d\Omega$$ arising in Finite Element Methods with $$\varphi_i$$ basis functions of the approximation space. Furthermore let $$cond_2(A)$$ be the condition number of $$A$$.

1. My first question is regarding the numerical integration of the matrix elements and how do they affect the condition number. My Idea was to write $$\tilde A = A + E$$ where the elements of $$E$$ illustrate the error between the numerical integration and the exact integration. Is there any way to give a inequality relation between $$cond_2(\tilde{A})$$ and $$cond_2(A)$$?

2. My second question is about the limit inside the condition number. Assuming a matrix $$E$$ with elements $$e_{i,j} = \int\limits_{\tilde{\Omega}\subset \mathbb{R}^d} \nabla \tilde{\varphi}_i\cdot \nabla\tilde{\varphi}_j \,d\Omega$$ with $$\tilde{\Omega}\cap\Omega = \emptyset$$. How does the condition number reacts to limit of matrices i.e. does it hold that $$cond_2(A + \lim_{\alpha\to 0} \alpha E) = cond_2(A)$$ or is there any inequality relation between these two? I first tried to use $$cond_2(A + \lim_{\alpha\to 0} \alpha E) \leq cond_2(A) + cond_2(\lim_{\alpha\to 0} \alpha E)$$ but I realized that the condition number of zero Matrix is defined as $$\infty$$. And therefore the inequality is useless.

1. For a matrix $$A$$ with distinct eigenvalues, adding a perturbation $$\delta A$$ results [1] in a change to eigenvalues of magnitude (to first order) $$\delta\lambda_i = (X^{-1}\delta A X)_{ii},$$ so $$|\delta \lambda_i| \leq \kappa(X)\|\delta A\|$$. The change in condition number $$\kappa(A)$$ is then given by $$\delta(\kappa(A)) = \kappa(A)\big(\delta(\log \lambda_n) - \delta(\log \lambda_1)\big) \leq \kappa(A)\|\delta A\|\big(|\lambda_n|^{-1} + |\lambda_1|^{-1}\big),$$ where $$\lambda_1,\lambda_n$$ are the smallest and largest eigenvalues by absolute value. Since this is proportional to $$\|\delta A\|$$, the only issue you can get is if your matrix is almost singular and the perturbation turns it singular, then $$\delta(\kappa(A))$$ would be large.
2. It isn't true that $$\kappa(A+B)\leq \kappa(A) + \kappa(B)$$. Consider $$A=\mathrm{diag}(1,1+\epsilon)$$ and $$B=\mathrm{diag}(1,-1)$$, then both are well-conditioned, $$\kappa(A)=1+\epsilon$$, $$\kappa(B)=1$$, but $$\kappa(A+B)=\epsilon^{-1}$$.
• For 1: Is there also an upper bound for $\delta cond_2(A)$? For 2: I understand your counter example. But it should be true that: $cond_2(A +\lim_{\alpha \ to 0} E) = cond_2(A)$ – Kerem Dec 15 '18 at 16:02
• @Kerem (1) You might be able to get a rigorous bound instead of the first-order expansion by looking more carefully at the perturbation theory of eigenvalues. (2) Of course it is true that $\lim_{\epsilon\to0}\kappa(A+\epsilon E) = \kappa(A)$ (the way you placed the limits in that expression doesn't make sense btw), because $\kappa(A)$ is a continuous function of the entries of $A$. – Kirill Dec 15 '18 at 20:23