Condition of an eigenvector problem #2 This one of the problem, where the only thing I can do is ask for help.



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*Let $A$ be a diagonalisable matrix, $\lambda_i\in\mathbb{R}$ a simple eigenvalue of $A$, and $B$ any matrix. Show that, for $\varepsilon > 0$ suficiently small, there exists a unique eigenvalue $\lambda_i(\varepsilon)$ of the 'perturbed' matrix $A+\varepsilon B$ satisfying
$$\lambda_i(\varepsilon)\ =\ \lambda_i + \varepsilon\frac{q_i^*Bp_i}{q^*_ip_i} + \mathcal{O}(\varepsilon^2),$$
where $p_i$ and $q_i$ are vectors which satisfy
$$\|p_i\|_2\ =\ 1,\;\; Ap_i\ =\ \lambda_ip_i,$$
$$\|q_i\|_2\ =\ 1,\;\; A^*q_i\ =\ \lambda_iq_i.$$
What do we obtain for Hermitian matrices?

*The inequality
$$\left|\frac{q^*_iBp_i}{q^*_ip_i}\right|\ \leq\ \frac{\|B\|_2}{\sigma_i},\;\; \mbox{ where}\; \sigma_i\ =\ |q_i^*p_i|,$$
shows that the sensitivityof the $i$th eigenvalue to perturbations is 'measured' by the number $\sigma^{-1}_i$ (in fact, it is possible to limit consideration to perturbations which are of the form $A+\varepsilon B$ with $\|B|_2=1$). Prove that the number $\sigma_i^{-1}$ is invariant under unitary transformations (of the matrix $A$) and that it satisfies the inequality
$$1\ \leq\ \sigma_i^{-1}\ \leq\ \Gamma_2(A)\ \stackrel{\text{def}}{=}\ \inf\{\mbox{cond}_2(P) : P^{-1}AP = \mbox{diag}(\lambda_k)\}.$$

*Prove that the result of question (1.) is stilltrue even if the matrix $A$ is not diagonalisable, while the assumption continues to be made that the eigenvalue $\lambda_i$ is real and simple. To this end, good use can be made of the Jordan canonical form of the matrix $A$.



Thanx so much in advance.
 A: Hints:
(1) Let $\lambda_i(\varepsilon)\in\mbox{sp}(A + \varepsilon B)$, and set $\lambda_i(\varepsilon) = \lambda_i + \varepsilon\delta\lambda_i$. Also consider $p_i(\varepsilon) = p_i + \varepsilon\delta p_i$ an eigenvector of $A + \varepsilon B$ corresponding to $\lambda_i(\varepsilon)$. Then,
$$(A + \varepsilon B)(p_i + \varepsilon\delta p_i)\ =\ (\lambda_i + \varepsilon\delta\lambda_i)(p_i + \varepsilon \delta p_i),$$
expand it and concludes that
$$\varepsilon\delta\lambda_i\ =\ \varepsilon\frac{q_i^*Bp_i}{q_i^*p_i} + \mathcal{O}(\varepsilon^2).$$
Finally, if $A$ is Hermitian, show that $$|\lambda_i(\varepsilon) - \lambda_i|\ \leq\ \left|\frac{p_i^*(\varepsilon B)p_i}{\|p_i\|_2}\right|\ \leq\ \|\varepsilon B\|_2\ =\ \|\delta A\|_2.$$
(2) the invariance proof is easy, so consider the inequality and


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*the first inequality is consequence of Cauchy-Schwarz's inequality.

*for the second inequality, let $Q = (q_1, \ldots, q_n)$ and $P = (p_1,\ldots,p_n)$, then show that $Q^*P = \mbox{diag}(q_k^*p_k)$; that is, $q_j^*p_i = 0$ if $i\neq j$.

*Deduce form (2) that $\|P^{-1}\|_2\|(Q^*)^{-1}\|_2 \geq \sigma_i^{-1}$

*Show that $Q^*=P^{-1}$

*Concludes that $\sigma_i^{-1}$ is lower bound of the set $\{\mbox{cond}_2(P) : P^{-1}AP = \mbox{diag}(\lambda_k)\}$


(3) Use Jordan canonical form of the matrix $A$ and the fact of $\lambda_i\in\mathbb{R}$ is simple. Follow the proof of (1).
