# Condition of an eigenvector problem

Please, somebody help me with this problem.

[Ciarlet 2.3-5] Let ${A}$ and ${B} = {A} + \delta{A}$ be two symmetric matrices with eigenvalues $$\alpha_1\ \leq\ \alpha_2\ \leq\ \ldots\ \leq\ \alpha_n\;\;\; \mbox{ and }\;\;\; \beta_1\ \leq\ \beta_2\ \leq\ \ldots\ \leq\ \beta_n.$$ Let $\alpha_k$ be a simple eigenvalue of the matrix ${A}$ and let ${a}_k$ be an eigenvector (with $\|{a}_k\|_2 = 1$) corresponding to the eigenvalue $\alpha_k$. Show that if $$\|\delta{A}\|_2\ <\ \Delta\ \stackrel{\text{def}}{=}\ \min_{i\neq k}|\alpha_i-\alpha_k|,$$ there exists an eigenvector ${b}_k$ (with $\|{b}_k\|_2 = 1$) corresponding to the eigenvalue $\beta_k$, which satisfies $$\|{a}_k - {b}_k\|_2\ \leq\ \gamma(1 + \gamma^2)^{1/2},\;\;\; \mbox{ with }\;\; \gamma\ =\ \frac{\|\delta{A}\|_2}{\Delta - \|\delta{A}\|_2}.$$

Thanks for the time.

• You'll find the theorems you need here: uta.edu/faculty/rcli/Teaching/math5371/Notes/ch5.pdf Look for Weyl's theorem, and the development that follows. Please make sure to cite! (Not that it's mine.) – Michael Grant Apr 19 '13 at 6:14
• I have not been able to solve this exercise, but these are some things I've done. I know that $Aa_k = \alpha_ka_k$ and I set $\beta_k = \alpha_k+\delta\alpha_k$ and define $b_k = a_k + \delta a_k$. The idea is to prove that $\|\delta a_k\|_2^2 \leq \gamma^2 + \gamma^4$. Next, we know that $(A + \delta A)b_k = \beta_kb_k$, then expanding we have $$(A + \delta A)(a_k + \delta a_k)\ =\ (\alpha_k + \delta\alpha_k)(a_k + \delta a_k),$$ $$A(\delta a_k) + (\delta A)a_k + (\delta A)(\delta a_k) = \alpha_k(\delta a_k) + (\delta\alpha_k)a_k + (\delta\alpha_k)(\delta a_k).$$ – FASCH May 1 '13 at 17:11
• Now, multiplying by $a_k^T$ and recall that $A=A^T$ and $a_k^Ta_k = 1$, we have $$a_k^T(\delta A)a_k + a_k^T(\delta A)(\delta a_k)\ =\ \delta\alpha_k + \delta\alpha_k\left[a_k^T(\delta a_k)\right].\hspace{1cm}(1)$$ Furthermore, we want that $\|b_k\|_2^2 =1$ and we know that $\|a_k\|_2^2=1$, so $$2\left[a_k^T(\delta a_k)\right] + \|\delta a_k\|_2^2\ =\ 0\hspace{1cm}(2).$$ So, I've tried to work with the equations (1) and (2) and Weyl's theorem, in order to reach the conclusion, but I have not yet. Hope this is of help. Thank you. – FASCH May 1 '13 at 17:11